Yearly Archives: 2022

Exercise MCQ

Question 1

Which of the following is not a criterion for congruence of triangles?

(a) SSA

(b) SAS

(c) ASA

(d)SSSSolution 1

Correct option: (a)

SSA is not a criterion for congruence of triangles.Question 2

If AB = QR, BC = RP and CA = PQ, then which of the following holds?

(a) ∆ABC ≅ ∆PQR

(b) ∆CBA ≅ ∆PQR

(c) ∆CAB ≅ ∆PQR

(d) ∆BCA ≅ ∆PQRSolution 2

Correct option: (c)

Question 3

If ∆ABC ≅ ∆PQR then which of the following is not true?

(a) BC = PQ

(b) AC = PR

(c) BC = QR

(d) AB = PQSolution 3

Question 4

In Δ ABC, AB = AC and ∠B = 50°. Then, ∠A = ?

(a) 40° 

(b) 50° 

(c) 80° 

(d) 130° Solution 4

Correct option: (c)

In ΔABC,

AB = AC

⇒ ∠C = ∠B (angles opposite to equal sides are equal)

⇒ ∠C = 50° 

Now, ∠A + ∠B + ∠C = 180° 

⇒ ∠A + 50° + 50° = 180° 

⇒ ∠A = 80° Question 5

In Δ ABC, BC = AB and ∠B = 80°. Then, ∠A = ?

(a) 50° 

(b) 40° 

(c) 100° 

(d) 80° Solution 5

Correct option: (a)

In ΔABC,

BC = AB

⇒ ∠A = ∠C (angles opposite to equal sides are equal)

Now, ∠A + ∠B + ∠C = 180° 

⇒ ∠A + 80° + ∠A = 180° 

⇒ 2∠A = 100° 

⇒ ∠A = 50°  Question 6

In ΔABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?

(a) 4 cm

(b) 5 cm

(c) 8 cm

(d) 2.5 cmSolution 6

Correct option: (a)

In ΔABC,

∠C = ∠A

⇒ AB = BC (sides opposite to equal angles are equal)

⇒ AB = 4 cm Question 7

Two sides of a triangle are of length 4 cm and 2.5 cm. The length of the third side of the triangle cannot be

(a) 6 cm

(b) 6.5 cm

(c) 5.5 cm

(d) 6.3 cmSolution 7

Correct option: (b)

The sum of any two sides of a triangle is greater than the third side.

Since, 4 cm + 2.5 cm = 6.5 cm

The length of third side of a triangle cannot be 6.5 cm. Question 8

In ΔABC, if ∠C > ∠B, then

(a) BC > AC

(b) AB > AC

(c) AB < AC

(d) BC < ACSolution 8

Correct option: (b)

We know that in a triangle, the greater angle has the longer side opposite to it.

In ΔABC,

∠C > ∠B

⇒ AB >AC  Question 9

It is given that ∆ABC ≅ ∆FDE in which AB = 5 cm, ∠B = 40^o, ∠A = 80^o and FD = 5 cm. Then which of the following is true?

(a) ∠D = 60^o

(b) ∠E = 60^o

(c) ∠F = 60^o

(d) ∠D = 80^oSolution 9

Question 10

In ∆ABC, ∠A = 40^o and ∠B = 60^o. Then the longest side of ∆ABC is

(a) BC

(b) AC

(c) AB

(d) Cannot be determinedSolution 10

Question 11

In the given figure AB > AC. Then, which of the following is true?

(a) AB < AD

(b) AB = AD

(c) AB > AD

(d) Cannot be determined

Solution 11

Correct option: (c)

Question 12

In the given figure AB > AC. If BO and CO are the bisectors of ∠B and ∠C respectively, then

(a) OB = OC

(b) OB > OC

(c) OB < OC

Solution 12

Question 13

In the given figure, AB = AC and OB = OC. Then, ∠ABO : ∠ACO = ?

(a) 1 :1

(b) 2 : 1

(c) 1 :2

(d) None of these 

Solution 13

Question 14

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is

(a) Equilateral

(b) Isosceles

(c) Scalene

(d) Right-angledSolution 14

Question 15

In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have

(a) ∠A = ∠D

(b) ∠B = ∠E

(c) ∠C = ∠F

(d) None of these

Solution 15

Question 16

In ∆ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABC ≅ ∆DEF, we must have

(a) AB = DF

(b) AC = DE

(c) BC = EF

(d) ∠A = ∠D

Solution 16

Question 17

In ∆ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are

(a) Isosceles but not congruent

(b) Isosceles but congruent

(c) Congruent but not isosceles

(d) Neither congruent nor isosceles

Solution 17

Question 18

Which is true ?

(a) A triangle can have two right angles.

(b) A triangle can have two obtuse angles.

(c) A triangle can have two acute angles.

(d) An exterior angle of a triangle is less than either of the interior opposite angles.

Solution 18

Question 19

Fill in the blanks with

(a) (Sum of any two sides of a triangle)……(the third side)

(b) (Difference of any two sides of a triangle)…..(the third side)

(c) (Sum of three altitudes of a triangle) ……. (sum of its three sides

(d) (Sum of any two sides of a triangle)….. (twice the median to the 3^rd side)

(e) (Perimeter of a triangle)……(sum of its medians)Solution 19

Question 20

Fill in the blanks

(a) Each angle of an equilateral triangles measures …….

(b) Medians of an equilateral triangle are ……….

(c) In a right triangle the hypotenuse is the ….. side

(d) Drawing a ∆ABC with AB = 3cm, BC= 4 cm and CA = 7 cm is ……..Solution 20

Exercise Ex. 9B

Question 1(i)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

5 cm, 4 cm, 9 cmSolution 1(i)

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 5 cm and 4 cm, is not greater than the third side, 9 cm. Question 1(ii)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

8 cm, 7 cm, 4 cmSolution 1(ii)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(iii)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

10 cm, 5 cm, 6 cmSolution 1(iii)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(iv)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

2.5 cm, 5 cm, 7 cmSolution 1(iv)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(v)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

3 cm, 4 cm, 8 cmSolution 1(v)

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 3 cm and 4 cm, is not greater than the third side, 8 cm. Question 2

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the triangle.Solution 2

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ 50° + 60° + ∠C = 180° 

⇒ ∠C = 70° 

Thus, we have

∠A < ∠B < ∠C

⇒ BC < AC < AB

Hence, the longest side is AB and the shortest side is BC. Question 3(iii)

In ΔABC, ∠A = 100° and ∠C = 50°. Which is its shortest side?Solution 3(iii)

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ 100° + ∠B + 50° = 180° 

⇒ ∠B = 30° 

Thus, we have

∠B < ∠C < ∠A

⇒ AC < AB < BC

Hence, the shortest side is AC. Question 3(i)

In ABC, if A = 90^o, which is the longest side?Solution 3(i)

Question 3(ii)

In ABC, if A = B = 45^o, name the longest side.Solution 3(ii)

Question 4

In ABC, side AB is produced to D such that BD = BC. If B = 60^o and A = 70^o, prove that (i) AD > CD and (ii) AD > AC.

Solution 4

Question 5

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution 5

In ΔAOB,

∠B < ∠A

⇒ AO < BO ….(i)

In ΔCOD,

∠C < ∠D

⇒ DO < CO ….(ii)

Adding (i) and (ii),

AO + DO < BO + CO

⇒ AD < BC Question 6

AB and CD are respectively the smallest and largest sides of quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Solution 6

Construction: Join AC and BD.

In ΔABC,

BC > AB

⇒ ∠BAC > ∠ACB ….(i)

In ΔACD,

CD > AD

⇒ ∠CAD > ∠ACD ….(ii)

Adding (i) and (ii), we get

∠BAC + ∠CAD > ∠ACB + ∠ACD

⇒ ∠A > ∠C 

In ΔADB,

AD > AB

⇒ ∠ABD > ∠ADB ….(iii)

In ΔBDC,

CD > BC

⇒ ∠CBD > ∠BDC ….(iv)

Adding (iii) and (iv), we get

∠ABD + ∠CBD > ∠ADB + ∠BDC

⇒ ∠B > ∠D Question 7

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) > (AC + BD).Solution 7

In ΔABC,

AB + BC > AC  ….(i)

In ΔACD,

DA + CD > AC  ….(ii)

In ΔADB,

DA + AB > BD  ….(iii)

In ΔBDC,

BC + CD > BD  ….(iv)

Adding (i), (ii), (iii) and (iv), we get

AB + BC + DA + CD + DA + AB + BC + CD > AC + AC + BD + BD

⇒ 2(AB + BC + CD + DA) > 2(AC + BD)

⇒ AB + BC + CD + DA > AC + BD Question 8

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) < 2(BD + AC).Solution 8

In ΔAOB,

AO + BO > AB  ….(i)

In ΔBOC,

BO + CO > BC  ….(ii)

In ΔCOD,

CO + DO > CD  ….(iii)

In ΔAOD,

DO + AO > DA  ….(iv)

Adding (i), (ii), (iii) and (iv), we get

AO + BO + BO + CO + CO + DO + DO + AO > AB + BC + CD + DA

⇒ 2(AO + CO) + 2(BO + DO) > AB + BC + CD + DA

⇒ 2AC + 2BD > AB + BC + CD + DA

⇒ 2(AC + BD) > AB + BC + CD + DA

⇒ AB + BC + CD + DA < 2(AC + BD) Question 9

In ABC, B = 35^o, C = 65^o and the bisector of BAC meets BC in X. Arrange AX, BX and CX in descending order.

Solution 9

Question 10

In the given figure, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R respectively. Show that SQ > SR.

Solution 10

In ΔPQR,

PQ > PR

⇒ ∠PRQ > ∠PQR

⇒ ∠SRQ > ∠SQR

⇒ SQ > SRQuestion 11

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.Solution 11

In ΔABC,

AB = AC

⇒ ∠ABC = ∠ACB ….(i)

Now, ∠ABC = ∠ABD + ∠DBC

⇒ ∠ABC > ∠DBC

⇒ ∠ACB > ∠DBC [From (i)]

⇒ ∠DCB > ∠DBC

⇒ BD > CD

i.e. CD < BD Question 12

Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than  of a right angle.Solution 12

Let PQR be the required triangle.

Let PR be the longest side.

Then, PR > PQ

⇒ ∠Q > ∠R ….(i)

Also, PR > QR

⇒ ∠Q > ∠P ….(ii)

Adding (i) and (ii), we get

2∠Q > ∠R + ∠P

⇒ 2∠Q + ∠Q > ∠P + ∠Q + ∠R (adding ∠Q to both sides)

⇒ 3∠Q > 180° 

⇒ ∠Q > 60° 

Question 13(i)

In the given figure, prove that CD + DA + AB > BC

Solution 13(i)

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AC + AB > BC ….(ii)

Adding (i) and (ii), we get

CD + DA + AC + AB > AC + BC

Subtracting AC from both sides, we get

CD + DA + AB > BC Question 13(ii)

In the given figure, prove that

CD + DA + AB + BC > 2AC.

Solution 13(ii)

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AB + BC > AC ….(ii)

Adding (i) and (ii), we get

CD + DA + AB + BC > AC + AC

⇒ CD + DA + AB + BC > 2AC Question 14(i)

If O is a point within ABC, show that:

AB + AC > OB + OCSolution 14(i)

Given : ABC is a triangle and O is appoint insideit.

To Prove : (i) AB+AC > OB +OCQuestion 14(ii)

If O is a point within ABC, show that:

AB + BC + CA > OA + OB + OCSolution 14(ii)

AB+BC+CA > OA+OB+OCQuestion 14(iii)

If O is a point within ABC, show that:

OA + OB + OC > (AB + BC + CA)Solution 14(iii)

OA+OB+OC> (AB+BC+CA)

Proof:

(i)InABC,

AB+AC>BC.(i)

And in , OBC,

OB+OC>BC.(ii)

Subtracting (i) from (i) we get

(AB+AC)-(OB+OC)> (BC-BC)

i.e.AB+AC>OB+OC

(ii)AB+AC> OB+OC[proved in (i)]

Similarly,AB+BC > OA+OC

AndAC+BC> OA +OB

Addingboth sides of these three inequalities, we get

(AB+AC) +(AC+BC) +(AB+BC)>OB+OC+OA+OB+OA+OC

i.e.2(AB+BC+AC)> 2(OA+OB+OC)

Therefore, we have

AB+BC+AC > OA+OB+OC

(iii)InOAB

OA+OB > AB(i)

InOBC,

OB+OC > BC(ii)

And, in OCA,

OC+OA>CA

Adding (i), (ii) and (iii)we get

(OA+OB) + (OB+OC)+(OC+OA)> AB+BC+CA

i.e2(OA+OB+OC) > AB+BC+CA

OA+OB+OC> (AB+BC+CA)Question 15

In the given figure, AD ⊥ BC and CD > BD. Show that AC > AB.

Solution 15

Construction: Mark a point S on BC such that BD = SD. Join AS.

In ΔADB and ΔADS,

BD = SD (by construction)

∠ADB = ∠ADS   (Each equal to 90°)

AD = AD (common)

∴ ΔADB ≅ ΔADS (by SAS congruence criterion)

⇒ AB = AS (c.p.c.t.)

Now, in ΔABS,

AB = AS

⇒ ∠ASB = ∠ABS ….(i)(angles opposite to equal sides are equal)

In ΔACS,

∠ASB > ∠ACS ….(ii)

From (i) and (ii), we have

∠ABS > ∠ACS

⇒ ∠ABC > ∠ACB

⇒ AC > ABQuestion 16

In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE. Prove that AB + AC > BE.

Solution 16

In ΔABC,

AB + AC > BC

⇒ AB + AC >BD + DC

⇒ AB + AC >BD + DE ….(i) [since CD = DE]

In ΔBED,

BD + DE > BE ….(ii)

From (i) and (ii), we have

AB + AC > BE

Exercise Ex. 9A

Question 1

In the given figure, AB ∥ CD and O is the midpoint of AD.

Show that (i) Δ AOB ≅ Δ DOC (ii) O is the midpoint of BC.

Solution 1

(i) In ΔAOB and ΔDOC,

∠BAO = ∠CDO (AB ∥ CD, alternate angles)

AO = DO (O is the mid-point of AD)

∠AOB = ∠DOC (vertically opposite angles)

∴ ΔAOB ≅ ΔDOC (by ASA congruence criterion)

(ii) Since ΔAOB ≅ ΔDOC,

BO = CO (c.p.c.t.)

⇒ O is the mid-point of BC.Question 2

In the given figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisect AB.

Solution 2

In ΔAOD and ΔBOC,

∠AOD = ∠BOC (vertically opposite angles)

∠DAO = ∠CBO (Each 90°)

AD = BC (given)

∴ ΔAOD ≅ BOC (by AAS congruence criterion)

⇒ AO = BO (c.p.c.t.)

⇒ CD bisects AB. Question 3

In the given figure, two parallels lines l and m are intersected by two parallels lines p and q. Show that Δ ABC ≅ Δ CDA.

Solution 3

In ΔABC and ΔCDA

∠BAC = ∠DCA  (alternate interior angles for p ∥ q)

AC = CA  (common)

∠BCA = ∠DAC (alternate interior angles for l ∥ m)

∴ ΔABC ≅ ΔCDA (by ASA congruence rule)Question 4

AD is an altitude of an isosceles ΔABC in which AB = AC.

Show that (i) AD bisects BC, (ii) AD bisects ∠A.

Solution 4

(i) In ΔBAD and ΔCAD

∠ADB = ∠ADC  (Each 90° as AD is an altitude)

AB = AC (given)

AD = AD (common)

∴ ΔBAD ≅ ΔCAD (by RHS Congruence criterion)

⇒ BD = CD (c.p.c.t.)

Hence AD bisects BC.

(ii) Also, ∠BAD = ∠CAD (c.p.c.t.)

 Hence, AD bisects ∠A.Question 5

In the given figure, BE and CF are two equal altitudes of ΔABC.

Show that (i) ΔABE ≅ ΔACF, (ii) AB = AC.

Solution 5

(i) In ΔABE and ΔACF,

∠AEB = ∠AFC (Each 90°)

BE = CF (given)

∠BAE = ∠CAF (common ∠A)

∴ ΔABE ≅ ACF (by ASA congruence criterion)

(ii) Since ΔABE ≅ ΔACF,

AB = AC (c.p.c.t.)Question 6

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABE ≅ ΔACE

(iii) AE bisects ∠A as well as ∠D

(iv) AE is the perpendicular bisector of BC.

Solution 6

(i) In ΔABD and ΔACD,

AB = AC  (equal sides of isosceles ΔABC)

DB = DC  (equal sides of isosceles ΔDBC)

AD = AD (common)

∴ ΔABD ≅ ACD (by SSS congruence criterion)

(ii) Since ΔABD ≅ ΔACD,

∠BAD = ∠CAD (c.p.c.t.)

⇒ ∠BAE = ∠CAE ….(1)

Now, in ΔABE and ΔACE

AB = AC (equal sides of isosceles ΔABC)

∠BAE = ∠CAE [From (1)]

AE = AE (common) 

∴ ΔABE ≅ ACE (by SAS congruence criterion)

(iii) Since ΔABD ≅ ΔACD,

∠BAD = ∠CAD (c.p.c.t.)

⇒ ∠BAE = ∠CAE 

Thus, AE bisects ∠A.

In ΔBDE and ΔCDE,

BD = CD (equal sides of isosceles ΔABC)

BE = CE (c.p.c.t. since ΔABE ≅ ACE) 

DE = DE (common)

∴ ΔBDE ≅ CDE (by SSS congruence criterion)

⇒ ∠BDE = ∠CDE (c.p.c.t.)

Thus, DE bisects ∠D, i.e., AE bisects ∠D.

Hence, AE bisects ∠A as well as ∠D.

(iv) Since ΔBDE ≅ ΔCDE,

BE = CE and ∠BED = ∠CED (c.p.c.t.)

⇒ BE = CE and ∠BED = ∠CED = 90° (since ∠BED and ∠CED form a linear pair) 

⇒ DE is the perpendicular bisector of BC.

⇒ AE is the perpendicular bisector of BC.Question 7

In the given figure, if x = y and AB = CB, then prove that AE = CD.

Solution 7

Question 8

In the given figure, line l is the bisector of an angle ∠A and B is any point on l. If BP and BQ are perpendiculars from B to the arms of ∠A, show that

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Solution 8

(i) In ΔAPB and ΔAQB,

∠APB = ∠AQC (Each 90°)

∠BAP = ∠BAQ (line l is the bisector of ∠A)

AB = AB (common)

∴ ΔAPB ≅ AQB (by AAS congruence criterion)

(ii) Since ΔAPB ≅ ΔAQB,

BP = BQ (c.p.c.t.) Question 9

ABCD is a quadrilateral such that diagonal AC bisect the angles ∠A and ∠C. Prove that AB = AD and CB = CD.Solution 9

In ΔABC and ΔADC,

∠BAC = ∠DAC (AC bisects ∠A)

AC = AC (common)

∠BCA = ∠DCA (AC bisects ∠C)

∴ ΔABC ≅ ADC (by ASA congruence criterion)

⇒ AB = AD and CB = CD (c.p.c.t.) Question 10

ΔABC is a right triangle right angled at A such that AB = AC and bisector of ∠C intersect the side AB at D. Prove that AC + AD = BC.Solution 10

Construction: Draw DE ⊥ BC.

In ΔDAC and ΔDEC,

∠DAC = ∠DEC (Each 90°)

∠DCA = ∠DCE (CD bisects ∠C)

CD = CD (common)

∴ ΔDAC ≅ ΔDEC (by AAS congruence criterion)

⇒ DA = DE (c.p.c.t.) ….(i)

and AC = EC (c.p.c.t.)  ….(ii)

Given, AB = AC

⇒ ∠B = ∠C (angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

∠A + ∠B + ∠C = 180° 

⇒ 90° + ∠B + ∠B = 180° 

⇒ 2∠B = 90° 

⇒ ∠B = 45° 

In ΔBED,

∠BDE + ∠B = 90° (since ∠BED = 90°)

⇒ ∠BDE + 45° = 90° 

⇒ ∠BDE = 45° 

⇒ ∠BDE = ∠DBE = 45° 

⇒ DE = BE ….(iii)

From (i) and (iii),

DA = DE = BE ….(iv) 

Now, BC = BE + EC

⇒ BC = DA + AC [From (ii) and (iv)

⇒ AC + AD = BCQuestion 11

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX (ii) AX = BX.

Solution 11

Question 12

In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

Solution 12

Question 13

In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

Solution 13

Question 14

In the given figure, ABCD is square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

Solution 14

Question 15

In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

Solution 15

ΔOAB is an equilateral triangle.

⇒ ∠OAB = ∠OBA = AOB = 60° 

ABCD is a square.

⇒ ∠A = ∠B = ∠C = ∠D = 90° 

Now, ∠A = ∠DAO + ∠OAB

⇒ 90° = ∠DAO + 60° 

⇒ ∠DAO = 90° – 60° = 30° 

Similarly, ∠CBO = 30° 

In ΔOAD and ΔOBC,

AD = BC  (sides of a square ABCD)

∠DAO = ∠CBO = 30° 

OA = OB (sides of an equilateral ΔOAB)

∴ ΔOAD ≅ ΔOBC (by SAS congruence criterion)

⇒ OD = OC (c.p.c.t.)

Hence, ΔOCD is an isosceles triangle.Question 16

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ABC such that AX = AY. Prove that CX = BY.

Solution 16

Question 17

In ABC, D is the midpoint of BC. If DL AB and DM AC such that DL = DM, prove that AB = AC.

Solution 17

Question 18

In ABC, AB = AC and the bisectors of _B and C meet at a point O. Prove that BO = CO and the ray AO is the bisector A.

Solution 18

Question 19

The line segments joining the midpoints M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.Solution 19

Construction: Join AN and BN.

In ΔANM and ΔBNM

AM = BM (M is the mid-point of AB)

∠AMN = ∠BMN (Each 90°)

MN = MN (common)

∴ ΔANM ≅ ΔBNM (by SAS congruence criterion)

⇒ AN = BN (c.p.c.t.) ….(i)

And, ∠ANM = ∠BNM (c.p.c.t.)

⇒ 90° – ∠ANM = 90° – ∠BNM

⇒ ∠AND = ∠BNC ….(ii) 

In ΔAND and DBNC,

AN = BN [From (i)]

∠AND = ∠BNC [From (ii)]

DN = CN (N is the mid-point of DC)

∴ ΔAND ≅ ΔBNC (by SAS congruence criterion)

⇒ AD = BC (c.p.c.t.)Question 20

The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that ∠MOC = ∠ABC.Solution 20

In ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

Now, by exterior angle property,

∠MOC = ∠OBC + ∠OCB

⇒ ∠MOC = 2∠OBC [From (i)]

⇒ ∠MOC = ∠ABC (OB is the bisector of ∠ABC)  Question 21

The bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to ∠ABC is equal to ∠BOC.Solution 21

In ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

In ΔBOC, by angle sum property,

∠BOC + ∠OBC + ∠OCB = 180° 

⇒ ∠BOC + 2∠OBC = 180° [From (i)]

⇒ ∠BOC + ∠ABC = 180° 

⇒ ∠BOC + (180° – ∠ABP) = 180° (∠ABC and ∠ABP form a linear pair)

⇒ ∠BOC + 180° – ∠ABP = 180° 

⇒ ∠BOC – ∠ABP = 0

⇒ ∠BOC = ∠ABP Question 22

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meets BC at Q, prove that ΔBPQ is an isosceles triangle.Solution 22

AB ∥ PQ and BP is a transversal.

⇒ ∠ABP = ∠BPQ (alternate angles) ….(i)

BP is the bisector of ∠ABC.

⇒ ∠ABP = ∠PBC

⇒ ∠ABP = ∠PBQ ….(ii)

From (i) and (ii), we have

∠BPQ = ∠PBQ

⇒ PQ = BQ (sides opposite to equal angles are equal)

⇒ ΔBPQ is an isosceles triangle.Question 23

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

Solution 23

To prove that the image is as far behind the mirror as the object is in front of the mirror, we need to prove that AT = BT. 

We know that angle of incidence = angle of reflection.

⇒ ∠ACN = ∠DCN ….(i)

AB ∥ CN and AC is the transversal.

⇒ ∠TAC = ∠ACN (alternate angles) ….(ii)

Also, AB ∥ CN and BD is the transversal.

⇒ ∠TBC = ∠DCN (corresponding angles) ….(iii)

From (i), (ii) and (iii),

∠TAC = ∠TBC ….(iv)

In ΔACT and ΔBCT,

∠TAC = ∠TBC [From (iv)]

∠ATC = ∠BTC (Each 90°)

CT = CT (common)

∴ ΔACT ≅ ΔBCT (by AAS congruence criterion)

⇒ AT = BT (c.p.c.t.)Question 24

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

Solution 24

_{Let AB be the breadth of a river. Now take a point M on that bank of the river where point B is situated. Through M draw a perpendicular and take point N on it such that point, A, O and N lie on a straight line where point O is the mid point of BM.}

Question 25

In a ΔABC, D is the midpoint of side AC such that BD = . Show that ∠ABC is a right angle.Solution 25

D is the mid-point of AC.

⇒ AD = CD =  

Given, BD =  

⇒ AD = CD = BD 

Consider AD = BD

⇒ ∠BAD = ∠ABD (i)(angles opposite to equal sides are equal)

Consider CD = BD

⇒ ∠BCD = ∠CBD (ii)(angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

∠ABC + ∠BAC + ∠BCA = 180° 

⇒ ∠ABC + ∠BAD + ∠BCD = 180° 

⇒ ∠ABC + ∠ABD + ∠CBD = 180° [From (i) and (ii)]

⇒ ∠ABC + ∠ABC = 180° 

⇒ 2∠ABC = 180° 

⇒ ∠ABC = 90° 

Hence, ∠ABC is a right angle.Question 26

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent.” Is the statement true? Why?Solution 26

The given statement is not true.

Two triangles are congruent if two sides and the included angle of one triangle are equal to corresponding two sides and the included angle of another triangle. Question 27

“If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent.” Is the statement true? Why?Solution 27

The given statement is not true.

Two triangles are congruent if two angles and the included side of one triangle are equal to corresponding two angles and the included angle of another triangle. 

Exercise MCQ

Question 1

If (x + 1) is a factor of the polynomial (2x² + kx) then the value of k is

(a) -2

(b) -3

(c) 2

(d) 3Solution 1

Correct option: (c)

Let p(x) = 2x² + kx

Since (x + 1) is a factor of p(x),

P(-1) = 0

⇒ 2(-1)² + k(-1) = 0

⇒ 2 – k = 0

⇒ k = 2Question 2

The value of (249)² – (248)² is

(a) 1²

(b) 477

(c) 487

(d) 497Solution 2

Correct option: (d)

(249)² – (248)²

= (249 + 248)(249 – 248)

= 497 × 1

= 497Question 3

Solution 3

Question 4

If a + b + c = 0, then a³ + b³ + c³= ?

(a) 0

(b) abc 

(c) 2abc

(d)3abcSolution 4

Question 5

(a) 0

(b) 

(c) 

(d) Solution 5

Correct option: (c)

Question 6

The coefficient of x in the expansion of (x + 3)³ is

(a) 1

(b) 9

(c) 18

(d)27Solution 6

Question 7

Which of the following is a factor of (x + y)³ – (x³ + y³)?

(a) x² + y² + 2xy

(b) x² + y² – xy

(c) xy²

(d) 3xySolution 7

Correct option: (d)

(x + y)³ – (x³ + y³)

= (x + y)³ – [(x + y)(x² – xy + y²)]

= (x + y)[(x + y)² – (x² – xy + y²)]

= (x + y)[x² + y² + 2xy – x² + xy – y²]

= (x + y)(3xy) Question 8

One of the factors of (25x² – 1) + (1 + 5x)² is

(a) 5 +x

(b) 5 – x

(c) 5x – 1

(d) 10xSolution 8

Correct option: (d)

(25x² – 1) + (1 + 5x)²

= [(5x)² – (1)²] + (1 + 5x)²

= (5x – 1)(5x + 1) + (1 + 5x)²

= (1 + 5x)[(5x – 1) + (1 + 5x)]

= (1 + 5x)(5x – 1 + 1 + 5x)

= (1 + 5x)(10x)Question 9

If (x + 5) is a factor of p(x) = x³ – 20x + 5k then k = ?

(a) -5

(b) 5

(c) 3

(d) -3Solution 9

Correct option: (b)

Since (x + 5) is a factor of p(x) = x³ – 20x + 5k,

p(-5) = 0

⇒ (-5)³ – 20(-5) + 5k = 0

⇒ -125 + 100 + 5k = 0

⇒ -25 + 5k = 0

⇒ 5k = 25

⇒ k = 5 Question 10

If (x + 2) and (x – 1) are factors of (x³ + 10x² + mx + n), then

(a) m = 5, n = -3

(b) m = 7, n = -18

(c) m = 17, n = -8

(d)m = 23, n = -19Solution 10

Question 11

104 ⨯ 96 =?

(a) 9894

(b) 9984

(c) 9684

(d)9884Solution 11

Question 12

305 ⨯ 308 = ?

(a) 94940

(b) 93840

(c) 93940

(d)94840Solution 12

Question 13

207 ⨯ 193 = ?

(a) 39851

(b) 39951

(c) 39961

(d)38951Solution 13

Question 14

4a² + b² + 4ab + 8a + 4b + 4 =?

(a) (2a + b + 2)²

(b) (2a – b + 2)²

(c) (a + 2b + 2)²

(d)None of theseSolution 14

Question 15

(x² – 4x – 21) =?

(a) (x – 7)(x – 3)

(b) (x + 7)(x – 3)

(c) (x – 7)(x + 3)

(d)none of theseSolution 15

Question 16

(4x² + 4x -3) = ?

(a) (2x – 1)(2x – 3)

(b) (2x + 1)(2x – 3)

(c) (2x + 3)(2x – 1)

(d)none of theseSolution 16

Question 17

6x² + 17x + 5 =?

(a) (2x +3)(3x + 5)

(b) (2x + 5)(3x + 1)

(c) (6x + 5)(x + 1)

(d)none of theseSolution 17

Question 18

(x + 1) is a factor of the polynomial

(a) x³ – 2x² + x + 2

(b) x³ + 2x² + x – 2

(c) x³ + 2x² – x – 2

(d)x³ + 2x² – x + 2Solution 18

Question 19

3x³ + 2x² + 3x + 2 =?

(a) (3x – 2)(x² – 1)

(b) (3x – 2)(x² + 1)

(c) (3x + 2)(x² – 1)

(d)(3x + 2)(x² + 1)Solution 19

Question 20

(a) 1

(b) 0

(c) -1

(d)3Solution 20

Question 21

If x + y + z =9 and xy + yz + zx = 23, then the value of (x³ + y³ + z³ – 3xyz) = ?

(a) 108

(b) 207

(c) 669

(d)729Solution 21

Question 22

then (a³ – b³) = ?

(a) -3

(b) -2

(c) -1

(d) 0Solution 22

Correct option: (d)

Exercise Ex. 3C

Question 1

Factorise:

x² + 11x + 30Solution 1

x² + 11x + 30

= x² + 6x + 5x + 30

= x (x + 6) + 5 (x + 6)

= (x + 6) (x + 5).Question 2

Factorise:

x² + 18x + 32Solution 2

x² + 18x + 32

= x² + 16x + 2x + 32

= x (x + 16) + 2 (x + 16)

= (x + 16) (x + 2).Question 3

Factorise:

x² + 20x – 69Solution 3

x² + 20x – 69

= x² + 23x – 3x – 69

= x(x + 23) – 3(x + 23)

= (x + 23)(x – 3) Question 4

Factorise:

x² + 19x – 150Solution 4

x² + 19x – 150

= x² + 25x – 6x – 150

= x(x + 25) – 6(x + 25)

= (x + 25)(x – 6) Question 5

Factorise:

x² + 7x – 98Solution 5

x² + 7x – 98

= x² + 14x – 7x – 98

= x(x + 14) – 7(x + 14)

= (x + 14)(x – 7) Question 6

Factorise:

Solution 6

Question 7

Factorise:

x² – 21x + 90Solution 7

x² – 21x + 90

= x² – 6x – 15x + 90

= x(x – 6) – 15(x – 6)

= (x – 6)(x – 15) Question 8

Factorise:

x² – 22x + 120Solution 8

x² – 22x + 120

= x² – 10x – 12x + 120

= x(x – 10) – 12(x – 10)

= (x – 10)(x – 12) Question 9

Factorise:

x² – 4x + 3Solution 9

x² – 4x + 3

= x² – 3x – x + 3

= x(x – 3) – 1(x – 3)

= (x – 3)(x – 1) Question 10

Factorise:

Solution 10

Question 11

Factorise:

Solution 11

Question 12

Factorise:

Solution 12

Question 13

Factorise:

Solution 13

Question 14

Factorise:

x² – 24x – 180Solution 14

x² – 24x – 180

= x² – 30x + 6x – 180

= x(x – 30) + 6(x – 30)

= (x – 30)(x + 6) Question 15

Factorise:

z² – 32z – 105Solution 15

z² – 32z – 105

= z² – 35z + 3z – 105

= z (z – 35) + 3 (z – 35)

= (z – 35) (z + 3)Question 16

Factorise:

x² – 11x – 80Solution 16

x² – 11x – 80

= x² – 16x + 5x – 80

= x (x – 16) + 5 (x – 16)

= (x – 16) (x + 5).Question 17

Factorise:

6 – x – x²Solution 17

6 – x – x²

= 6 + 2x – 3x – x²

= 2(3 + x) – x (3 + x)

= (3 + x) (2 – x).Question 18

Factorise:

Solution 18

Question 19

Factorise:

40 + 3x – x²Solution 19

40 + 3x – x²

= 40 + 8x – 5x – x²

= 8 (5 + x) -x (5 + x)

= (5 + x) (8 – x).Question 20

Factorise:

x² – 26x + 133Solution 20

x² – 26x + 133

= x² – 19x – 7x + 133

= x(x – 19) – 7(x – 19)

= (x – 19)(x – 7) Question 21

Factorise:

Solution 21

Question 22

Factorise:

Solution 22

Question 23

Factorise:

Solution 23

Question 24

Factorise:

Solution 24

Question 25

Factorise:

x² – x – 156Solution 25

x² – x – 156

= x² – 13x + 12x – 156

= x (x – 13) + 12 (x – 13)

= (x – 13) (x + 12).Question 27

Factorise:

9x² + 18x + 8Solution 27

9x² + 18x + 8

= 9x² + 12x + 6x + 8

= 3x (3x+ 4) +2 (3x + 4)

= (3x + 4) (3x + 2).Question 28

Factorise:

6x² + 17x + 12Solution 28

6x² + 17x + 12

= 6x² + 9x + 8x + 12

= 3x (2x + 3) + 4(2x + 3)

= (2x + 3) (3x + 4).Question 29

Factorise:

18x² + 3x – 10Solution 29

18x² + 3x – 10

= 18x² – 12x + 15x – 10

= 6x (3x – 2) + 5 (3x – 2)

= (6x + 5) (3x – 2).Question 30

Factorise:

2x² + 11x – 21Solution 30

2x² + 11x – 21

= 2x² + 14x – 3x – 21

= 2x (x + 7) – 3 (x + 7)

= (x + 7) (2x – 3).Question 31

Factorise:

15x² + 2x – 8Solution 31

15x² + 2x – 8

= 15x² – 10x + 12x – 8

= 5x (3x – 2) + 4 (3x – 2)

= (3x – 2) (5x + 4).Question 32

Factorise:

21x² + 5x – 6Solution 32

21x² + 5x – 6

= 21x² + 14x – 9x – 6

= 7x(3x + 2) – 3(3x + 2)

= (3x + 2)(7x – 3) Question 33

Factorise:

24x² – 41x + 12Solution 33

24x² – 41x + 12

= 24x² – 32x – 9x + 12

= 8x (3x – 4) – 3 (3x – 4)

= (3x – 4) (8x – 3).Question 34

Factorise:

3x² – 14x + 8Solution 34

3x² – 14x + 8

= 3x² – 12x – 2x +8

= 3x (x – 4) – 2(x – 4)

= (x – 4) (3x – 2).Question 35

Factorise:

2x² + 3x – 90Solution 35

2x² + 3x – 90

= 2x² – 12x + 15x – 90

= 2x (x – 6) + 15 (x – 6)

= (x – 6) (2x + 15).Question 37

Factorise:

Solution 37

Question 38

Factorise:

Solution 38

Question 39

Factorise:

Solution 39

Question 40

Factorise:

Solution 40

Question 41

Factorise:

Solution 41

Question 42

Factorise:

Solution 42

Question 43

Factorise:

Solution 43

Question 44

Factorise:

15x² – x – 28Solution 44

15x² – x – 28

= 15x² + 20x – 21x – 28

= 5x (3x + 4) – 7 (3x + 4)

= (3x + 4) (5x – 7).Question 45

Factorise:

6x² – 5x – 21Solution 45

6x² – 5x – 21

= 6x² + 9x – 14x – 21

= 3x (2x + 3) – 7 (2x + 3)

= (3x – 7) (2x + 3).Question 46

Factorise:

2x² – 7x – 15Solution 46

2x² – 7x – 15

= 2x² – 10x + 3x – 15

= 2x (x – 5) + 3 (x – 5)

= (x – 5) (2x + 3).Question 47

Factorise:

5x² – 16x – 21Solution 47

5x² – 16x – 21

= 5x² + 5x – 21x – 21

= 5x (x + 1) -21 (x + 1)

= (x + 1) (5x – 21).Question 48

Factorise:

6x² – 11x – 35Solution 48

6x² – 11x – 35

= 6x² – 21x + 10x – 35

= 3x(2x – 7) + 5(2x – 7)

= (2x – 7)(3x + 5) Question 49

Factorise:

9x² – 3x – 20Solution 49

9x² – 3x – 20

= 9x² – 15x + 12x – 20

= 3x(3x – 5) + 4(3x – 5)

= (3x – 5)(3x + 4) Question 50

Factorise:

10x² – 9x – 7Solution 50

10x² – 9x – 7

= 10x² + 5x – 14x – 7

= 5x (2x + 1) – 7 (2x+ 1)

= (2x + 1) (5x – 7).Question 51

Factorise:

x² – 2x + Solution 51

x² – 2x + 

Question 52

Factorise:

Solution 52

Question 53

Factorise:

Solution 53

Question 54

Factorise:

Solution 54

Question 55

Factorise:

Solution 55

Question 56

Factorise:

Solution 56

Question 57

Factorise:

Solution 57

Question 58

Factorise:

Solution 58

Question 59

Factorise:

Solution 59

Letx + y = z

Then, 2 (x+ y)² – 9 (x + y) – 5

Now, replacing z by (x + y), we get

Question 60

Factorise:

9 (2a – b)² – 4 (2a – b) – 13Solution 60

Let2a – b = c

Then,9 (2a – b)² – 4 (2a – b) -13

Now, replacing c by (2a – b) , we get

9 (2a – b)² – 4 (2a – b) – 13

Question 61

Factorise:

7 (x – 2y)² – 25 (x – 2y) + 12Solution 61

Let x – 2y = z

Then, 7 (x – 2y)² – 25 (x – 2y) + 12

Now replace z by (x – 2y), we get

7 (x – 2y)² – 25 (x – 2y) + 12

Question 62

Factorise:

Solution 62

Question 63

Factorise:

Solution 63

Question 64

Factorise:

(a + 2b)² + 101(a + 2b) + 100Solution 64

Given equation: (a + 2b)² + 101(a + 2b) + 100

Let (a + 2b) = x

Then, we have

x² + 101x + 100

= x² + 100x + x + 100

= x(x + 100) + 1(x + 100)

= (x + 100)(x + 1)

= (a + 2b + 100)(a + 2b + 1) Question 65

Factorise:

Solution 65

Let x² = y

Then, 4x⁴ + 7x² – 2

Now replacing y by x², we get

Question 66

Evaluate {(999)² – 1}.Solution 66

{(999)² – 1}

= {(999)² – (1)²}

= {(999 – 1)(999 + 1)}

= 998 × 1000

= 998000 

Exercise Ex. 3D

Question 1(i)

Expand:

(a + 2b + 5c)²Solution 1(i)

We know:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(i) (a + 2b + 5c)²

= (a)² + (2b)² + (5c)² + 2(a) (2b) + 2 (2b) (5c) + 2(5c) (a)

= a² + 4b² + 25c² + 4ab + 20bc + 10acQuestion 1(ii)

Expand:

(2a – b + c)²Solution 1(ii)

We know:

(2a – b + c)²

= (2a)² + (-b)² + (c)² + 2 (2a) (-b) + 2(-b) (c) + 2 (c) (2a)

= 4a² + b² + c² – 4ab – 2bc + 4ac.Question 1(iii)

Expand:

(a – 2b – 3c)²Solution 1(iii)

We know:

(a – 2b – 3c)²

= (a)² + (-2b)² + (-3c)²+ 2(a) (-2b) + 2(-2b) (-3c) + 2 (-3c) (a)

= a² + 4b² + 9c² – 4ab + 12bc – 6ac.Question 2(i)

Expand:

(2a – 5b – 7c)²Solution 2(i)

We know:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(2a – 5b – 7c)²

= (2a)² + (-5b)² + (-7c)² + 2 (2a) (-5b) + 2 (-5b) (-7c) + 2 (-7c) (2a)

= 4a² + 25b² + 49c² – 20ab + 70bc – 28ac.Question 2(ii)

Expand:

(ii) (-3a + 4b – 5c)²Solution 2(ii)

(-3a + 4b – 5c)²

= (-3a)² + (4b)² + (-5c)² + 2 (-3a) (4b) + 2 (4b) (-5c) + 2 (-5c) (-3a)

= 9a² + 16b² + 25c² – 24ab – 40bc + 30ac.Question 2(iii)

Expand:

Solution 2(iii)

= Question 3

Factorise: 4x² + 9y² + 16z² + 12xy – 24yz – 16xz.Solution 3

4x² + 9y² + 16z² + 12xy – 24yz – 16xz

= (2x)² + (3y)² + (-4z)² + 2 (2x) (3y) + 2(3y) (-4z) + 2 (-4z) (2x)

= (2x + 3y – 4z)²Question 4

Factorise: 9x² + 16y² + 4z² – 24xy + 16yz – 12xzSolution 4

9x² + 16y² + 4z² – 24xy + 16yz – 12xz

= (-3x)² + (4y)² + (2z)² + 2 (-3x) (4y) + 2 (4y) (2z) + 2 (2z) (-3x)

= (-3x + 4y + 2z)².Question 5

Factorise: 25x² + 4y² + 9z² – 20xy – 12yz + 30xz.Solution 5

25x² + 4y² + 9z² – 20xy – 12yz + 30xz

= (5x)² + (-2y)² + (3z)² + 2(5x) (-2y) + 2(-2y) (3z) + 2(3z) (5x)

= (5x – 2y + 3z)²Question 6

16x² + 4y² + 9z² – 16xy – 12yz + 24xzSolution 6

16x² + 4y² + 9z² – 16xy – 12yz + 24xz

= (4x)² + (-2y)² + (3z)² + 2(4x)(-2y) + 2(-2y)(3z) + 2(4x)(3z)

= [4x + (-2y) + 3z]²

= (4x – 2y + 3z)²

= (4x – 2y + 3z)(4x – 2y + 3z) Question 7(i)

Evaluate

(99)²Solution 7(i)

(99)²

= (100 – 1)²

= (100)² – 2(100) (1) + (1)²

= 10000 – 200 + 1

= 9801.Question 7(ii)

Evaluate

(995)²Solution 7(ii)

(995)²

= (1000 – 5)²

= (1000)² + (5)² – 2(1000)(5)

= 1000000 + 25 – 10000

= 990025 Question 7(iii)

Evaluate

(107)²Solution 7(iii)

(107)²

= (100 + 7)²

= (100)² + (7)² + 2(100)(7)

= 10000 + 49 + 1400

= 11449 

Exercise Ex. 3E

Question 1(i)

Expand:

(3x + 2)³Solution 1(i)

(3x + 2)³

= (3x)³ + (2)³ + 3 3x 2 (3x + 2)

= 27x³ + 8 + 18x (3x + 2)

= 27x³ + 8 + 54x² + 36x.Question 1(ii)

Expand:

Solution 1(ii)

Question 1(iii)

Expand

Solution 1(iii)

Question 2(i)

Expand

(5a – 3b)³Solution 2(i)

(5a – 3b)³

= (5a)³ – (3b)³ – 3(5a)(3b)(5a – 3b)

= 125a³ – 27b³ – 45ab(5a – 3b)

= 125a³ – 27b³ – 225a²b + 135ab² Question 2(ii)

Expand

Solution 2(ii)

Question 2(iii)

Expand

Solution 2(iii)

Question 3

Factorise 

8a³ + 27b³ + 36a²b + 54ab²Solution 3

8a³ + 27b³ + 36a²b + 54ab²

= (2a)³ + (3b)³ + 3 × (2a)² × (3b) + 3 × (2a) × (3b)²

= (2a + 3b)³

= (2a + 3b)(2a + 3b)(2a + 3b) Question 4

Factorise 

64a³ – 27b³ – 144a²b + 108ab²Solution 4

64a³ – 27b³ – 144a²b + 108ab²

= (4a)³ – (3b)³ – 3 × (4a)² × (3b) + 3 × (4a) × (3b)²

= (4a – 3b)³

= (4a – 3b)(4a – 3b)(4a – 3b) Question 5

Factorise 

Solution 5

Question 6

Factorise 

125x³ – 27y³ – 225x²y + 135xy²Solution 6

125x³ – 27y³ – 225x²y + 135xy²

= (5x)³ – (3y)³ – 3 × (5x)² × (3y) + 3 × (5x) × (3y)²

= (5x – 3y)³

= (5x – 3y)(5x – 3y)(5x – 3y)  Question 7

Factorise 

a³x³ – 3a²bx² + 3ab²x – b³Solution 7

a³x³ – 3a²bx² + 3ab²x – b³

= (ax)³ – 3 × (ax)² × b + 3 × (ax) × (b)² – (b)³

= (ax – b)³

= (ax – b)(ax – b)(ax – b)  Question 8

Factorise 

Solution 8

Question 9

Factorise 

a³ – 12a(a – 4) – 64Solution 9

a³ – 12a(a – 4) – 64

= a³ – 3 × a × 4(a – 4) – (4)³

= (a – 4)³

= (a – 4)(a – 4)(a – 4) Question 10(i)

Evaluate

(103)³Solution 10(i)

(103)³

= (100 + 3)³

= (100)³ + (3)³ + 3 × 100 × 3 × (100 + 3)

= 1000000 + 27 + 900(103)

= 1000000 + 27 + 92700

= 1092727 Question 10(ii)

Evaluate

(99)³Solution 10(ii)

(99)³

= (100 – 1)³

= (100)³ – (1)³ – 3 × 100 × 1 × (100 – 1)

= 1000000 – 1 – 300(99)

= 1000000 – 1 – 29700

= 970299  

Exercise Ex. 3F

Question 1

Factorise:

x³ + 27Solution 1

x³ + 27

= x³ + 3³

= (x + 3) (x² – 3x + 9)

Question 2

Factorise 

27a³ + 64b³Solution 2

27a³ + 64b³

= (3a)³ + (4b)³

= (3a + 4b)[(3a)² – (3a)(4b) + (4b)²]

= (3a + 4b)(9a² – 12ab + 16b²) Question 3

Factorise:

125a³ + Solution 3

125a³ + 

We know that 

Let us rewrite

Question 4

Factorise:

216x³ + Solution 4

216x³ + 

We know that 

Let us rewrite

Question 5

Factorise:

16x⁴ + 54xSolution 5

16x ⁴ + 54x

= 2x (8x ³ + 27)

= 2x [(2x)³ + (3)³]

= 2x (2x + 3) [(2x)² – 2x3 + 3²]    

=2x(2x+3)(4x² -6x +9)
Question 6

Factorise:

7a³ + 56b³Solution 6

7a³ + 56b³

= 7(a³ + 8b³)

= 7 [(a)³ + (2b)³]

= 7 (a + 2b) [a² – a 2b + (2b)²]    

= 7 (a + 2b) (a² – 2ab + 4b²).Question 7

Factorise:

x⁵ + x²Solution 7

x⁵ + x²

= x²(x³ + 1)

= x² (x + 1) [(x)² – x 1 + (1)²]     

= x² (x + 1) (x² – x + 1).Question 8

Factorise:

a³ + 0.008Solution 8

a³ + 0.008

= (a)³ + (0.2)³

= (a + 0.2) [(a)² – a 0.2 + (0.2)²]     

= (a + 0.2) (a² – 0.2a + 0.04).Question 9

Factorise:

1 – 27x³Solution 9

1 – 27x³

= (1)³ – (3x)³

= (1 – 3x) [(1)² + 1 3x + (3x)²]       

= (1 – 3x) (1 + 3x + 9x²).Question 10

Factorise 

64a³ – 343Solution 10

64a³ – 343

= (4a)³ – (7)³

= (4a – 7)[(4a)² + (4a)(7) + (7)²]

= (4a – 7)(16a² + 28a + 49) Question 11

Factorise:

x³ – 512Solution 11

x³ – 512

= (x)³ – (8)³

= (x – 8) [(x)² + x 8 + (8)²]       

= (x – 8) (x² + 8x + 64).Question 12

Factorise:

a³ – 0.064Solution 12

a³ – 0.064

= (a)³ – (0.4)³

= (a- 0.4) [(a)² + a 0.4 + (0.4)²]     

= (a – 0.4) (a² + 0.4 a + 0.16).Question 13

Factorise:

Solution 13

We know that 

Let us rewrite

Question 14

Factorise 

Solution 14

Question 15

Factorise:

x – 8xy³Solution 15

x – 8xy³

= x (1 – 8y³)

= x [(1)³ – (2y)³]

= x (1 – 2y) [(1)² + 1 2y + (2y)²]    

= x (1 – 2y) (1 + 2y + 4y²).Question 16

Factorise:

32x⁴ – 500xSolution 16

32x⁴ – 500x

= 4x (8x³ – 125)

= 4x [(2x)³ – (5)³]

= 4x [(2x – 5) [(2x)² + 2x 5 + (5)²]      

= 4x (2x – 5) (4x² + 10x + 25).Question 17

Factorise:

3a⁷b – 81a⁴b⁴Solution 17

3a⁷b – 81a⁴b⁴

= 3a⁴b (a³ – 27b³)

= 3a⁴b [(a)³ – (3b)³]

= 3a⁴b (a – 3b) [(a)² + a 3b + (3b)²]      

= 3a⁴b (a – 3b) (a² + 3ab + 9b²).Question 18

Factorise 

x⁴y⁴ – xySolution 18

x⁴y⁴ – xy

= xy(x³y³ – 1)

= xy[(xy)³ – (1)³]

= xy{(xy – 1)[(xy)² + (xy)(1) + (1)²]}

= xy(xy – 1)(x²y² + xy + 1) Question 19

Factorise 

8x²y³ – x⁵Solution 19

8x²y³ – x⁵

= x² (8y³ – x³)

= x² [(2y)³ – x³]

= x² [(2y – x)[(2y)² + (2y)(x) + x²]

= x² (2y – x)(4y² + 2xy + x²) Question 20

Factorise 

1029 – 3x³Solution 20

1029 – 3x³

= 3(343 – x³)

= 3[(7)³ – x³]

= 3[(7 – x)(7² + 7x + x²)]

= 3(7 – x)(49 + 7x + x²) Question 21

Factorise:

x⁶ – 729Solution 21

x⁶ – 729

= (x²)³ – (9)³

= (x² – 9) [(x²)² + x² 9 + (9)²]      

= (x² – 9) (x⁴ + 9x² + 81)

= (x + 3) (x – 3) [(x² + 9)² – (3x)²]

= (x + 3) (x – 3) (x² + 3x + 9) (x² – 3x + 9).Question 22

Factorise 

x⁹ – y⁹Solution 22

x⁹ – y⁹

= (x³)³ – (y³)³ 

= [(x³ – y³)][(x³)² + x³y³ + (y³)²]

= [(x – y)(x² + xy + y²)(x⁶ + x³y³ + y⁶)  Question 23

Factorise:

(a + b)³ – (a – b)³Solution 23

We know that, 

Therefore,

(a + b)³ – (a – b)³

= [a + b – (a – b)] [ (a + b)² + (a + b) (a – b) + (a – b)²] 

= (a + b – a + b) [ a² + b² + 2ab + a² – b² + a² + b² – 2ab]

= 2b (3a² + b²).Question 24

Factorise:

8a³ – b³ – 4ax + 2bxSolution 24

8a³ – b³ – 4ax + 2bx

= 8a³ – b³ – 2x (2a – b)

= (2a)³ – (b)³ – 2x (2a – b)

= (2a – b) [(2a)² + 2a b + (b)²] – 2x (2a – b)    

= (2a – b) (4a² + 2ab + b²) – 2x (2a – b)

= (2a – b) (4a² + 2ab + b² – 2x).Question 25

Factorise:

a³ + 3a²b + 3ab² + b³ – 8Solution 25

Question 26

Factorise:

Solution 26

We know that 

Question 27

Factorise:

2a³ + 16b³ – 5a – 10bSolution 27

2a³ + 16b³ – 5a – 10b

= 2 (a³ + 8b³) – 5 (a + 2b)

= 2 [(a)³ + (2b)³] – 5 (a + 2b)

= 2 (a + 2b) [(a)² – a 2b + (2b)² ] – 5 (a + 2b)     

= (a+ 2b) [2(a² – 2ab + 4b²) – 5]Question 28

Factorise 

a⁶ + b⁶Solution 28

a⁶ + b⁶

= (a²)³ + (b²)³ 

= (a² + b²)[(a²)² – (a²b²) + (b²)²]

= (a² + b²)(a⁴ – a²b² + b⁴) Question 29

Factorise 

a¹² – b¹²Solution 29

a¹² – b¹²

= (a⁶)² – (b⁶)² 

= (a⁶ – b⁶)(a⁶ + b⁶)

= [(a³)² – (b³)²][(a²)³ + (b²)³]

= (a³ – b³)(a³ + b³)[(a² + b²)(a⁴ – a²b² + b⁴)] 

= (a – b)(a² + ab + b²)(a + b)(a² – ab + b²)(a² + b²)(a⁴ – a²b² + b⁴)

= (a – b)(a + b)(a² + b²)(a² + ab + b²)(a² – ab + b²)(a⁴ – a²b² + b⁴)Question 30

Factorise 

x⁶ – 7x³ – 8Solution 30

Given equation is x⁶ – 7x³ – 8.

Putting x³ = y, we get

y² – 7y – 8

= y² – 8y + y – 8

= y(y – 8) + 1(y – 8)

= (y – 8)(y + 1)

= (x³ – 8)(x³ + 1)

= (x³ – 2³)(x³ + 1³)

= (x – 2)(x² + 2x + 4)(x + 1)(x² – x + 1)

= (x – 2)(x + 1)(x² + 2x + 4)(x² – x + 1) Question 31

Factorise 

x³ – 3x² + 3x + 7Solution 31

x³ – 3x² + 3x + 7

= x³ – 3x² + 3x – 1 + 8

= (x³ – 3x² + 3x – 1) + 8

= (x – 1)³ + 2³

= (x – 1 + 2)[(x – 1)² – (x – 1)(2) + 2²]

= (x + 1)(x² – 2x + 1 – 2x + 2 + 4)

= (x + 1)(x² – 4x + 7) Question 32

Factorise 

(x + 1)³ + (x – 1)³Solution 32

(x + 1)³ + (x – 1)³

= (x + 1 + x – 1)[(x + 1)² – (x + 1)(x – 1) + (x – 1)²]

= 2x(x² + 2x + 1 – x² + 1 + x² – 2x + 1)

= 2x(x² + 3)  Question 33

Factorise 

(2a + 1)³ + (a – 1)³Solution 33

(2a + 1)³ + (a – 1)³

= (2a + 1 + a – 1)[(2a + 1)² – (2a + 1)(a – 1) + (a – 1)²]

= (3a)[4a² + 4a + 1 – 2a² + 2a – a + 1 + a² – 2a + 1]

= 3a(3a² + 3a + 3)

= 9a(a² + a + 1)  Question 34

Factorise 

8(x + y)³ – 27(x – y)³Solution 34

8(x + y)³ – 27(x – y)³

= [2³ (x + y)³] – [3³ (x – y)³]

= [2(x + y) – 3(x – y)]{[2(x + y)]² + 2(x + y)3(x – y) + [3(x – y)]²}

= (2x + 2y – 3x + 3y){[4(x² + y² + 2xy)] + 6(x² – y²) + [9(x² + y² – 2xy]}

= (-x + 5y){4x² + 4y² + 8xy + 6x² – 6y² + 9x² + 9y² – 18xy}

= (-x + 5y)(19x² + 7y² – 10xy) Question 35

Factorise 

(x + 2)³ + (x – 2)³Solution 35

(x + 2)³ + (x – 2)³

= [(x + 2) + (x – 2)][(x + 2)² – (x + 2)(x – 2) + (x – 2)²]

=(2x)(x² + 4x + 4 – x² + 4 + x² – 4x + 4)

= 2x(x² + 12) Question 36

Factorise 

(x + 2)³ – (x – 2)³Solution 36

(x + 2)³ – (x – 2)³

= [(x + 2) – (x – 2)][(x + 2)² + (x + 2)(x – 2) + (x – 2)²]

= 4(x² + 4x + 4 + x² – 4 + x² – 4x + 4)

= 4(3x² + 4) Question 37

Prove that

Solution 37

Question 38

Prove that

Solution 38

Exercise Ex. 3G

Question 1

Find the product:

(x + y – z) (x² + y² + z² – xy + yz + zx)Solution 1

(x + y – z) (x² + y² + z² – xy + yz + zx)

= [x + y + (-z)] [(x)² + (y)² + (-z)² – (x) (y) – (y) (-z) – (-z) (x)]

= x³ + y³ – z³ + 3xyz.Question 2

Find the product.

(x – y – z)(x² + y² + z² + xy – yz + xz)Solution 2

(x – y – z)(x² + y² + z² + xy – yz + xz)

= x³ + xy² + xz² + x²y  – xyz + x²z – x²y – y³ – yz² – xy² + y²z – xyz – x²z – y²z – z³ – xyz + yz² – xz²

= x³ – y³ – z³ – xyz – xyz – xyz

= x³ – y³ – z³ – 3xyz Question 3

Find the product:

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)Solution 3

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)

= [x + (-2y) + 3] [(x)² + (-2y)² + (3) – (x) (-2y) – (-2y) (3) – (3) (x)]

= (a + b + c) (a² + b² + c² – ab – bc – ca)

= a³ + b³ + c³ – 3abc

Where, x = a, (-2y) = b and 3 = c

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)

= (x)³ + (-2y)³ + (3)² – 3 (x) (-2y) (3)

= x³ – 8y³ + 27 + 18xy.Question 4

Find the product.

(3x – 5y + 4)(9x² + 25y² + 15xy + 20y – 12x + 16)

*Modified the questionSolution 4

(3x – 5y + 4)(9x² + 25y² + 15xy + 20y – 12x + 16)

= (3x + (-5y) + 4)[(3x)² + (-5y)² + (4)² – (3x)(-5y) – (-5y)(4) – (3x)(4)]

= (3x)³ + (-5y)³ + (4)³ – 3(3x)(-5y)(4)

= 27x³ – 125y³ + 64 + 180xy Question 5

Factorise:

125a³ + b³ + 64c³ – 60abcSolution 5

125a³ + b³ + 64c³ – 60abc

= (5a)³ + (b)³ + (4c)³ – 3 (5a) (b) (4c)

= (5a + b + 4c) [(5a)² + b² + (4c)² – (5a) (b) – (b) (4c) – (5a) (4c)]

[ a³ + b³ + c³ – 3abc = (a+ b + c) (a² + b² + c² – ab – bc – ca)]

= (5a + b + 4c) (25a² + b² + 16c² – 5ab – 4bc – 20ac).Question 6

Factorise:

a³ + 8b³ + 64c³ – 24abcSolution 6

a³ + 8b³ + 64c³ – 24abc

= (a)³ + (2b)³ + (4c)³ – 3 a 2b 4c

= (a + 2b + 4c) [a² + 4b² + 16c² – 2ab – 8bc – 4ca).Question 7

Factorise:

1 + b³ + 8c³ – 6bcSolution 7

1 + b³ + 8c³ – 6bc

= 1 + (b)³ + (2c)³ – 3 (b) (2c)

= (1 + b + 2c) [1 + b² + (2c)² – b – b 2c – 2c]

= (1 + b + 2c) (1 + b² + 4c² – b – 2bc – 2c).Question 8

Factorise:

216 + 27b³ + 8c³ – 108bcSolution 8

216 + 27b³ + 8c³ – 108bc

= (6)³ + (3b)³ + (2c)² – 3 6 3b 2c

= (6 + 3b + 2c) [(6)² + (3b)² + (2c)² – 6 3b – 3b 2c – 2c 6]

= (6 + 3b + 2c) (36 + 9b² + 4c² – 18b – 6bc – 12c).Question 9

Factorise:

27a³ – b³ + 8c³ + 18abcSolution 9

27a³ – b³ + 8c³ + 18abc

= (3a)³ + (-b)³ + (2c)³ + 3(3a) (-b) (2c)

= [3a + (-b) + 2c] [(3a)² + (-b)² + (2c)² – 3a (-b) – (-b) (2c) – (2c) (3a)]

= (3a – b + 2c) (9a² + b² + 4c² + 3ab + 2bc – 6ca).Question 10

Factorise:

8a³ + 125b³ – 64c³ + 120abcSolution 10

8a³ + 125b³ – 64c³ + 120abc

= (2a)³ + (5b)³ + (-4c)³ – 3 (2a) (5b) (-4c)

= (2a + 5b – 4c) [(2a)² + (5b)² + (-4c)² – (2a) (5b) – (5b) (-4c) – (-4c) (2a)]

= (2a + 5b – 4c) (4a² + 25b² + 16c² – 10ab + 20bc + 8ca).Question 11

Factorise:

8 – 27b³ – 343c³ – 126bcSolution 11

8 – 27b³ – 343c³ – 126bc

= (2)³ + (-3b)³ + (-7c)³ – 3(2) (-3b) (-7c)

= (2 – 3b – 7c) [(2)² + (-3b)² + (-7c)² – (2) (-3b) – (-3b) (-7c) – (-7c) (2)]

= (2 – 3b – 7c) (4 + 9b² + 49c² + 6b – 21bc + 14c).Question 12

Factorise:

125 – 8x³ – 27y³ – 90xySolution 12

125 – 8x³ – 27y³ – 90xy

= (5)³ + (-2x)³ + (-3y)³ – 3 (5) (-2x) (-3y)

= (5 – 2x – 3y) [(5)² + (-2x)² + (-3y)² – (5) (-2x) – (-2x) (-3y) – (-3y) (5)]

= (5 – 2x – 3y) (25 + 4x² + 9y² + 10x – 6xy + 15y).Question 13

Factorise:

Solution 13

Question 14

Factorise:

27x³ – y³ – z³ – 9xyzSolution 14

27x³ – y³ – z³ – 9xyz

= (3x)³ – y³ – z³ – 3(3x)(y)(z)

= (3x)³ + (-y)³ + (-z)³ – 3(3x)(-y)(-z)

= [3x + (-y) + (-z)][(3x)² + (-y)² + (-z)² – (3x)(-y) – (-y)(-z) – (3x)(-z)]

= (3x – y – z)(9x² + y² + z² + 3xy – yz + 3zx) Question 15

Factorise:

Solution 15

Question 16

Factorise:

Solution 16

Question 17

Factorise:

(a – b)³ + (b – c)³ + (c – a)³Solution 17

Putting (a – b) = x, (b – c) = y and (c – a) = z, we get,

(a – b)³ + (b – c)³ + (c – a)³

= x³ + y³ + z³, where (x + y + z) = (a – b) + (b – c) + (c – a) = 0

= 3xyz[ (x + y + z)= 0 (x³ + y³ + z³) = 3xyz]

= 3(a – b) (b – c) (c – a).Question 18

Factorise:

(a – 3b)³ + (3b – c)³ + (c – a)³Solution 18

Given equation is (a – 3b)³ + (3b – c)³ + (c – a)³

Now,

(a – 3b) + (3b – c) + (c – a)

= a – a – 3b + 3b – c + c

= 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(a – 3b)³ + (3b – c)³ + (c – a)³

= 3(a – 3b)(3b – c)(c – a)Question 19

Factorise:

(3a – 2b)³ + (2b – 5c)³ + (5c – 3a)³Solution 19

We have:

(3a – 2b) + (2b – 5c) + (5c – 3a) = 0

So, (3a – 2b)³ + (2b – 5c)³ + (5c – 3a)³

= 3(3a – 2b) (2b – 5c) (5c – 3a).Question 20

Factorise:

(5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³Solution 20

Given equation is (5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³

Now,

(5a – 7b) + (7b – 9c) + (9c – 5a)

= 5a – 7b + 7b – 9c + 9c – 5a

= 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³

= 3(5a – 7b)(7b – 9c)(9c – 5a) Question 21

Factorise:

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³Solution 21

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

= [a (b – c)]³ + [b (c – a)]³ + [c (a – b)]³

Now, since, a (b – c) + b (c -a) + c (a – b)

= ab – ac + bc – ba + ca – bc = 0

So, a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

= 3a (b – c) b (c – a) c (a – b)

= 3abc (a – b) (b – c) (c – a).Question 22(i)

Evaluate

(-12)³ + 7³ + 5³Solution 22(i)

Given equation is (-12)³ + 7³ + 5³

Now,

-12 + 7 + 5 = -12 + 12 = 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(-12)³ + 7³ + 5³

= 3(-12)(7)(5)

= -1260 Question 22(ii)

Evaluate

(28)³ + (-15)³ + (-13)³Solution 22(ii)

Given equation is (28)³ + (-15)³ + (-13)³

Now,

28 + (-15) + (-13) = 28 – 28 = 0

We know that if x + y + z = 0, then x³ +  y³ + z³ = 3xyz

Hence,

(28)³ + (-15)³ + (-13)³

= 3(28)(-15)(-13)

= 16380 Question 23

Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a)Solution 23

L.H.S. = (a + b + c)³ – a³ – b³ – c³

= [(a + b) + c]³ – a³ – b³ – c³

= (a + b)³ + c³ + 3(a + b)c × [(a + b) + c] – a³ – b³ – c³

= a³ + b³ + 3ab(a + b) + c³ + 3(a + b)c × [(a + b) + c] – a³ – b³ – c³

= 3ab(a + b) + 3(a + b)c × [(a + b) + c]

= 3(a + b)[ab + c(a + b) + c²]

= 3(a + b)[ab + ac + bc + c²]

= 3(a + b)[a(b + c) + c(b + c)]

= 3(a + b)[(b + c)(a + c)]

= 3(a + b)(b + c)(c + a)

= R.H.S.Question 24

If a, b, c are all nonzero and a + b + c = 0, prove that

.Solution 24

Question 25

If a + b + c = 9 and a² + b² + c² = 35, find the value of (a³ + b³ + c³ – 3abc).Solution 25

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

= (a + b + c)[(a² + b² + c²) – (ab + bc + ca)] ….(i)

Now,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (9)² = 35 + 2(ab + bc + ca)

⇒ 81 = 35 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 46

⇒ ab + bc + ca = 23

Substituting in (i), we get

a³ + b³ + c³ – 3abc = (9)[35 – 23] = 9 × 12 = 108 

Exercise Ex. 3B

Question 1

Factorise:

9x² – 16y²Solution 1

9x² – 16y²

= (3x)² – (4y)²

= (3x + 4y)(3x – 4y) Question 2

Factorise:

Solution 2

Question 3

Factorise:

81 – 16x²Solution 3

81 – 16x²

= (9)² – (4x)²

= (9 – 4x)(9 + 4x) Question 4

Factorise:

5 – 20x²Solution 4

5 – 20x²

= 5(1 – 4x²)

= 5[(1)² – (2x)²]

= 5[(1 – 2x)(1 + 2x)]

= 5(1 – 2x)(1 + 2x) Question 5

Factorise:

2x⁴ – 32Solution 5

2x⁴ – 32

= 2(x⁴ – 16)

= 2[(x²)² – (4)²]

= 2[(x² – 4)(x² + 4)]

= 2[(x² – 2²)(x² + 4)]

= 2[(x – 2)(x + 2)(x² + 4)]

= 2(x – 2)(x + 2)(x² + 4) Question 6

Factorise:

3a³b – 243ab³Solution 6

3a³b – 243ab³

= 3ab (a² – 81 b²)

= 3ab [(a)² – (9b)²]

= 3ab (a + 9b) (a – 9b)Question 7

Factorise:

3x³ – 48xSolution 7

3x³ – 48x

= 3x (x² – 16)

= 3x [(x)² – (4)²]

= 3x (x + 4) (x – 4)Question 8

Factorise:

27a² – 48b²Solution 8

27a² – 48b²

= 3 (9a² – 16b²)

= 3 [(3a)² – (4b)²]

= 3(3a + 4b) (3a – 4b)Question 9

Factorise:

x – 64x³Solution 9

x – 64x³

= x (1 – 64x²)

= x[(1)² – (8x)²]

= x (1 + 8x) (1 – 8x)Question 10

Factorise:

8ab² – 18a³Solution 10

8ab² – 18a³

= 2a (4b² – 9a²)

= 2a [(2b)² – (3a)²]

= 2a (2b + 3a) (2b – 3a)Question 11

Factorise:

150 – 6x²Solution 11

150 – 6x²

= 6 (25 – x²)

= 6 (5² – x²)

= 6 (5 + x) (5 – x)Question 12

Factorise:

2 – 50x²Solution 12

2 – 50x²

= 2 (1 – 25x²)

= 2 [(1)² – (5x)²]

= 2 (1 + 5x) (1 – 5x)Question 13

Factorise:

20x² – 45Solution 13

20x² – 45

= 5(4x² – 9)

= 5 [(2x)² – (3)²]

= 5 (2x + 3) (2x – 3)Question 14

Factorise:

(3a + 5b)² – 4c²Solution 14

(3a + 5b)² – 4c²

= (3a + 5b)² – (2c)²

= (3a + 5b – 2c)(3a + 5b + 2c) Question 15

Factorise:

a² – b² – a – bSolution 15

a² – b² – a – b

= a² – b² – (a + b)

= (a – b)(a + b) – (a + b)

= (a + b)(a – b – 1) Question 16

Factorise:

4a² – 9b² – 2a – 3bSolution 16

4a² – 9b² – 2a – 3b

= (2a)² – (3b)² – (2a + 3b)

= (2a – 3b)(2a + 3b) – (2a + 3b)

= (2a + 3b)(2a – 3b – 1) Question 17

Factorise:

a² – b² + 2bc – c²Solution 17

a² – b² + 2bc – c²

= a² – (b² – 2bc + c²)

= a² – (b – c)²

= [a – (b – c)][a + (b – c)]

= (a – b + c)(a + b – c) Question 18

Factorise:

4a² – 4b² + 4a + 1Solution 18

4a² – 4b² + 4a + 1

= (4a² + 4a + 1) – 4b²

= [(2a)² + 2 × 2a × 1 + (1)²] – (2b)²

= (2a + 1)² – (2b)²

= (2a + 1 – 2b)(2a + 1 + 2b)

= (2a – 2b + 1)(2a + 2b + 1) Question 19

Factorise:

a² + 2ab + b² – 9c²Solution 19

a² + 2ab + b² – 9c²

= (a + b)² – (3c)²

= (a + b + 3c) (a + b – 3c)Question 20

Factorise:

108a² – 3(b – c)²Solution 20

108a² – 3(b – c)²

= 3 [(36a² – (b -c)²]

= 3 [(6a)² – (b – c)²]

= 3 (6a + b – c) (6a – b + c)Question 21

Factorise:

(a + b)³ – a – bSolution 21

(a + b)³ – a – b

= (a + b)³ – (a + b)

= (a + b) [(a + b)² – 1²]

= (a + b) (a + b + 1) (a + b – 1)Question 22

Factorise:

x² + y² – z² – 2xySolution 22

x² + y² – z² – 2xy

= (x² + y² – 2xy) – z²

= (x – y)² – z²

= (x – y – z)(x – y + z) Question 23

Factorise:

x² + 2xy + y² – a² + 2ab – b²Solution 23

x² + 2xy + y² – a² + 2ab – b²

= (x² + 2xy + y²) – (a² – 2ab + b²)

= (x + y)² – (a – b)²

= [(x + y) – (a – b)][(x + y) + (a – b)]

= (x + y – a + b)(x + y + a – b) Question 24

Factorise:

25x² – 10x + 1 – 36y²Solution 24

25x² – 10x + 1 – 36y²

= (25x² – 10x + 1) – 36y²

= [(5x)² – 2(5x)(1) + (1)²] – (6y)²

= (5x – 1)² – (6y)²

= (5x – 1 – 6y)(5x – 1 + 6y) Question 25

Factorise:

a – b – a² + b²Solution 25

a – b – a² + b²

= (a – b) – (a² – b²)

= (a – b) – (a – b) (a + b)

= (a – b) (1 – a – b)Question 26

Factorise:

a² – 4ac + 4c² – b²Solution 26

a² – 4ac + 4c² – b²

= a² – 4ac + 4c² – b²

= a² – 2 a 2c + (2c)² – b²

= (a – 2c)² – b²

= (a – 2c + b) (a – 2c – b)Question 27

Factorise:

9 – a² + 2ab – b²Solution 27

9 – a² + 2ab – b²

= 9 – (a² – 2ab + b²)

= 3² – (a – b)²

= (3 + a – b) (3 – a + b)Question 28

Factorise:

x³ – 5x² – x + 5Solution 28

x³ – 5x² – x + 5

= x² (x – 5) – 1 (x – 5)

= (x – 5) (x² – 1)

= (x – 5) (x + 1) (x – 1)Question 29

Factorise:

1 + 2ab – (a² + b²)Solution 29

1 + 2ab – (a² + b²)

= 1 – (a² + b² – 2ab)

= (1)² – (a – b)²

= [1 – (a – b)][1 + (a – b)]

= (1 – a + b)(1 + a – b) Question 30

Factorise:

9a² + 6a + 1 – 36b²Solution 30

9a² + 6a + 1 – 36b²

= (9a² + 6a + 1) – 36b²

= [(3a)² + 2(3a)(1) + (1)²] – (6b)²

= (3a + 1)² – (6b)²

= (3a + 1 – 6b)(3a + 1 + 6b) Question 31

Factorise:

x² – y² + 6y – 9Solution 31

x² – y² + 6y – 9

= x² – (y² – 6y + 9)

= x² – (y² – 2 y 3 + 3²)

= x² – (y – 3)²

= [x + (y – 3)] [x – (y – 3)] 

= (x + y – 3) (x – y + 3)Question 32

Factorise:

4x² – 9y² – 2x – 3ySolution 32

4x² – 9y² – 2x – 3y

= (2x)² – (3y)² – (2x + 3y)

= (2x + 3y) (2x – 3y) – (2x + 3y) 

= (2x + 3y) (2x – 3y – 1)Question 33

Factorise:

9a² + 3a – 8b – 64b²Solution 33

9a² + 3a – 8b – 64b²

= 9a² – 64b² + 3a – 8b

= (3a)² – (8b)² + (3a – 8b)

= (3a + 8b) (3a – 8b) + (3a – 8b)

= (3a – 8b) (3a + 8b + 1)Question 34

Factorise:

Solution 34

Question 35

Factorise:

Solution 35

Question 36

Factorise:

Solution 36

Question 37

Factorise:

x⁸ – 1Solution 37

x⁸ – 1

= (x⁴)² – (1)²

= (x⁴ – 1)(x⁴ + 1)

= [(x²)² – (1)²)(x⁴ + 1)

= (x² – 1)(x² + 1)(x⁴ + 1)

= (x – 1)(x + 1)(x² + 1)[(x²)² + (1)² + 2x² – 2x²

= (x – 1)(x + 1)(x² + 1)[(x²)² + (1)² + 2x²) – 2x²

Question 38

Factorise:

16x⁴ – 1Solution 38

16x⁴ – 1

= (4x²)² – (1)²

= (4x² – 1)(4x² + 1)

= [(2x)² – (1)²](4x² + 1)

= (2x – 1)(2x + 1)(4x² + 1) Question 39

Factorise:

81x⁴ – y⁴Solution 39

81x⁴ – y⁴

= (9x²)² – (y²)²

= (9x² – y²)(9x² + y²)

= [(3x)² – y²](9x² + y²)

= (3x – y)(3x + y)(9x² + y²) Question 40

Factorise:

x⁴ – 625Solution 40

x⁴ – 625

= (x²)² – (25)²

= (x² + 25) (x² – 25)

= (x² + 25) (x² – 5²)

= (x² + 25) (x + 5) (x – 5)

Exercise Ex. 3A

Question 1

Factorise:

9x² + 12xySolution 1

9x² + 12xy = 3x (3x + 4y)Question 2

Factorise:

18x²y – 24xyzSolution 2

18x²y – 24xyz = 6xy (3x – 4z)Question 3

Factorise:

27a³b³ – 45a⁴b²Solution 3

27a³b³ – 45a⁴b² = 9a³b² (3b – 5a)Question 4

Factorise:

2a (x+ y) – 3b (x + y)Solution 4

2a (x + y) – 3b (x + y) = (x + y) (2a – 3b)Question 5

Factorise:

2x (p² + q²) + 4y (p² + q²)Solution 5

2x (p² + q²) + 4y (p² + q²)

= (2x + 4y) (p² + q²)

= 2(x+ 2y) (p² + q²)Question 6

Factorise:

x (a – 5) + y (5 – a)Solution 6

x (a – 5) + y (5 – a)

= x (a – 5) + y (-1) (a – 5)

= (x – y) (a – 5)Question 7

Factorise:

4 (a + b) – 6 (a + b)²Solution 7

4 (a + b) – 6 (a + b)²

= (a + b) [4 – 6 (a + b)]

= 2 (a + b) (2 – 3a – 3b)

= 2 (a + b) (2 – 3a – 3b)Question 8

Factorise:

8 (3a – 2b)² – 10 (3a – 2b)Solution 8

8 (3a – 2b)² – 10 (3a – 2b)

= (3a – 2b) [8(3a – 2b) – 10]

= (3a – 2b) 2[4 (3a – 2b) – 5]

= 2 (3a – 2b) (12 a – 8b – 5)Question 9

Factorise:

x (x + y)³ – 3x²y (x + y)Solution 9

x (x + y)³ – 3x²y (x + y)

= x (x + y) [(x + y)² – 3xy]

= x (x + y) (x² + y² + 2xy – 3xy)

= x (x + y) (x² + y² – xy)Question 10

Factorise:

x³+ 2x² + 5x + 10Solution 10

x³+ 2x² + 5x + 10

= x² (x + 2) + 5 (x + 2)

= (x² + 5) (x + 2)Question 11

Factorise:

x² + xy – 2xz – 2yzSolution 11

x² + xy – 2xz – 2yz

= x (x + y) – 2z (x + y)

= (x+ y) (x – 2z)Question 12

Factorise:

a³b – a²b + 5ab – 5bSolution 12

a³b – a²b + 5ab – 5b

= a²b (a – 1) + 5b (a – 1)

= (a – 1) (a²b + 5b)

= (a – 1) b (a² + 5)

= b (a – 1) (a² + 5)Question 13

Factorise:

8 – 4a – 2a³ + a⁴Solution 13

8 – 4a – 2a³ + a⁴

= 4(2 – a) – a³ (2 – a)

= (2 – a) (4 – a³)Question 14

Factorise:

x³ – 2x²y + 3xy² – 6y³Solution 14

x³ – 2x²y + 3xy² – 6y³

= x² (x – 2y) + 3y² (x – 2y)

= (x – 2y) (x² + 3y²)Question 15

Factorise:

px – 5q + pq – 5xSolution 15

px + pq – 5q – 5x

= p(x + q) – 5 (q + x)

= (x + q) (p – 5)Question 16

Factorise:

x² + y – xy – xSolution 16

x² – xy + y – x

= x (x – y) – 1 (x – y)

= (x – y) (x – 1)Question 17

Factorise:

(3a – 1)² – 6a + 2Solution 17

(3a – 1)² – 6a + 2

= (3a – 1)² – 2 (3a – 1)

= (3a – 1) [(3a – 1) – 2]

= (3a – 1) (3a – 3)

= 3(3a – 1) (a – 1)Question 18

Factorise:

(2x – 3)² – 8x + 12Solution 18

(2x – 3)² – 8x + 12

= (2x – 3)² – 4 (2x – 3)

= (2x – 3) (2x – 3 – 4)

= (2x – 3) (2x – 7)Question 19

Factorise:

a³ + a – 3a² – 3Solution 19

a³ + a – 3a² – 3

= a(a² + 1) – 3 (a² + 1)

= (a – 3) (a² + 1)Question 20

Factorise:

3ax – 6ay – 8by + 4bxSolution 20

3ax – 6ay – 8by + 4bx

= 3a (x – 2y) + 4b (x – 2y)

= (x – 2y) (3a + 4b)Question 21

Factorise:

abx² + a²x + b²x +abSolution 21

abx² + a²x + b²x +ab

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)Question 22

Factorise:

x³ – x² + ax + x – a – 1Solution 22

x³ – x² + ax + x – a – 1

= x³ – x² + ax – a + x – 1

= x² (x – 1) + a (x – 1) + 1 (x – 1)

= (x – 1) (x² + a + 1)Question 23

Factorise:

2x + 4y – 8xy – 1Solution 23

2x + 4y – 8xy – 1

= 2x – 1 – 8xy + 4y

= (2x – 1) – 4y (2x – 1)

= (2x – 1) (1 – 4y)Question 24

Factorise:

ab (x² + y²) – xy (a² + b²)Solution 24

ab (x² + y²) – xy (a² + b²)

= abx² + aby² – a²xy – b²xy

= abx² – a²xy + aby² – b²xy

= ax (bx – ay) + by(ay – bx)

= (bx – ay) (ax – by)Question 25

Factorise:

a² + ab (b + 1) + b³Solution 25

a² + ab (b + 1) + b³

= a² + ab² + ab + b³

= a² + ab + ab² + b³

= a (a + b) + b² (a + b)

= (a + b) (a + b²)Question 26

Factorise:

a³ + ab (1 – 2a) – 2b²Solution 26

a³ + ab (1 – 2a) – 2b²

= a³ + ab – 2a²b – 2b²

= a (a² + b) – 2b (a² + b)

= (a² + b) (a – 2b)Question 27

Factorise:

2a² + bc – 2ab – acSolution 27

2a² + bc – 2ab – ac

= 2a² – 2ab – ac + bc

= 2a (a – b) – c (a – b)

= (a – b) (2a – c)Question 28

Factorise:

(ax + by)² + (bx – ay)²Solution 28

(ax + by)² + (bx – ay)²

= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy

= a²x² + b²y² + b²x² + a²y²

= a²x² + b²x² + b²y² + a²y²

= x² (a² + b²) + y²(a² + b²)

= (a² + b²) (x² + y²)Question 29

Factorise:

a (a + b – c) – bcSolution 29

a (a + b – c) – bc

= a² + ab – ac – bc

= a(a + b) – c (a + b)

= (a – c) (a + b)Question 30

Factorise:

a(a – 2b – c) + 2bcSolution 30

a(a – 2b – c) + 2bc

= a² – 2ab – ac + 2bc

= a (a – 2b) – c (a – 2b)

= (a – 2b) (a – c)Question 31

Factorise:

a²x² + (ax² + 1)x + aSolution 31

a²x² + (ax² + 1)x + a

= a²x² + ax³ + x + a

= ax² (a + x) + 1 (x + a)

= (ax² + 1) (a + x)Question 32

Factorise:

ab (x² + 1) + x (a² + b²)Solution 32

ab (x² + 1) + x (a² + b²)

= abx² + ab + a²x + b²x

= abx² + a²x + ab + b²x

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)Question 33

Factorise:

x² – (a + b) x+ abSolution 33

x² – (a + b) x+ ab

= x² – ax – bx + ab

= x (x – a) – b(x – a)

= (x – a) (x – b)Question 34

Factorise:

Solution 34

Exercise MCQ

Question 1

Which of the following expressions is a polynomial in one variable?

Solution 1

Question 2

Which of the following expressions is a polynomial?

Solution 2

Question 3

Which of the following is a polynomial?

Solution 3

Question 4

Which of the following is a polynomial?

Solution 4

Question 5

Which of the following is a polynomial?

(a) x^-2 + x^-1 + 3

(b) x + x^-1 + 2

(c) x^-1

(d)0Solution 5

Question 6

Which of the following is a quadratic polynomial?

(a) x + 4

(b) x³ + x

(c) x³ + 2x + 6

(d)x² + 5x + 4Solution 6

Question 7

Which of the following is a linear polynomial?

(a) x + x²

(b) x + 1

(c) 5x² – x + 3

(d)Solution 7

Question 8

Which of the following is a binomial?

(a) x² + x + 3

(b) x² + 4

(c) 2x²

(d)Solution 8

Question 9

(a) 

(b) 2

(c) 1

(d)0Solution 9

Question 10

Degree of the zero polynomial is

(a) 1

(b) 0

(c) not defined

(d)none of theseSolution 10

Question 11

Zero of the zero polynomial is

(a) 0

(b) 1

(c) every real number

(d)not definedSolution 11

Question 12

If p(x) = x + 4, then p(x) + p(-x) = ?

(a) 0

(b) 4

(c) 2x

(d)8Solution 12

Question 13

Solution 13

Question 14

If p(x) = 5x – 4x² + 3 then p(-1) = ?

(a) 2

(b) -2

(c) 6

(d) -6Solution 14

Correct option: (d)

P(x) = 5x – 4x² + 3

⇒ p(-1) = 5(-1) – 4(-1)² + 3 = -5 – 4 + 3 = -6Question 15

If (x⁵¹ + 51) is divided by (x + 1) then the remainder is

(a) 0

(b) 1

(c) 49

(d) 50Solution 15

Correct option: (d)

Let f(x) = x⁵¹ + 51

By the remainder theorem, when f(x) is divided by (x + 1), the remainder is f(-1).

Now, f(-1) = [(-1)ⁿ + 51] = -1 + 51 = 50Question 16

If (x + 1) is a factor of the polynomial (2x² + kx) then k = ?

(a)   4

(b)   -3

(c)  2

(d) -2Solution 16

Correct option: (c)

Let p(x) = 2x² + kx

Since (x + 1) is a factor of p(x),

P(–1) = 0

⇒ 2(–1)² + k(–1) = 0

⇒ 2 – k = 0

⇒ k = 2Question 17

When p(x) = x⁴ + 2x³ – 3x² + x – 1 is divided by (x – 2), the remainder is

(a) 0

(b) -1

(c) -15

(d)21Solution 17

Question 18

When p(x) = x³ – 3x² + 4x + 32 is divided by (x + 2), the remainder is

(a) 0

(b) 32

(c) 36

(d)4Solution 18

Question 19

When p(x) = 4x³ – 12x² + 11x – 5 is divided by (2x – 1), the remainder is

(a) 0

(b) -5

(c) -2

(d)2Solution 19

Question 20

When p(x) =x³-ax²+x is divided by (x-a), the remainder is

(a) 0

(b) a

(c) 2a

(d)3aSolution 20

Question 21

When p(x) = x³ + ax² + 2x + a is divided by (x + a), the remainder is

(a) 0

(b) a

(c) -a

(d)2aSolution 21

Question 22

(x + 1) is a factor of the polynomial

(a) x³ – 2x² + x + 2

(b) x³ + 2x² + x – 2

(c) x³ + 2x² – x – 2

(d)x³ + 2x² – x + 2Solution 22

Question 23

Zero of the polynomial p(x) = 2x + 5 is

(a) 

(b) 

(c) 

(d) Solution 23

Correct option: (b)

p(x) = 2x + 5

Now, p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

Question 24

The zeroes of the polynomial p(x) = x² + x – 6 are

(a) 2, 3

(b) -2, 3

(c) 2, -3

(d)-2, -3Solution 24

Question 25

The zeroes of the polynomial p(x) = 2x² + 5x – 3 are

Solution 25

Question 26

The zeros of the polynomial p(x) = 2x² + 7x – 4 are

(a) 

(b) 

(c) 

(d) Solution 26

Correct option: (c)

p(x) = 2x² + 7x – 4

Now, p(x) = 0

⇒ 2x² + 7x – 4 = 0

⇒ 2x² + 8x – x – 4 = 0

⇒ 2x(x + 4) – 1(x + 4) = 0

⇒ (x + 4)(2x – 1) = 0

⇒ x + 4 = 0 and 2x – 1 = 0

⇒ x = -4 and x =  Question 27

If (x + 5) is a factor of p(x) = x³ – 20x + 5k, then k =?

(a) -5

(b) 5

(c) 3

(d)-3Solution 27

Question 28

If (x + 2) and (x – 1) are factors of (x³ + 10x² + mx + n), then

(a) m = 5, n = -3

(b) m = 7, n = -18

(c) m = 17, n = -8

(d)m = 23, n = -19Solution 28

Question 29

If (x¹⁰⁰ + 2x⁹⁹ + k) is divisible by (x + 1), then the value of k is

(a) 1

(b) 2

(c) -2

(d)-3Solution 29

Question 30

For what value of k is the polynomial p(x) = 2x³ – kx² + 3x + 10 exactly divisible by (x + 2)?

Solution 30

Question 31

The zeroes of the polynomial p(x) = x² – 3x are

(a) 0, 0

(b) 0, 3

(c) 0, -3

(d)3, -3Solution 31

Question 32

The zeros of the polynomial p(x) = 3x² – 1 are

(a) 

(b) 

(c) 

(d) Solution 32

Correct option: (d)

p(x) = 3x² – 1

Now, p(x) = 0

⇒ 3x² – 1 = 0

⇒ 3x² = 1

Exercise Ex. 2A

Question 1(v)

Which of the expressions are polynomials?

Solution 1(v)

It is a polynomial, Degree = 2.Question 1(vi)

Which of the expressions are polynomials?

Solution 1(vi)

It is not a polynomial.Question 1(vii)

Which of the expressions are polynomials?

1Solution 1(vii)

It is a polynomial, Degree = 0.Question 1(viii)

Which of the expressions are polynomials?

Solution 1(viii)

It is a polynomial, Degree = 0.Question 1(i)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(i)

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 5. So, it is a polynomial of degree 5. Question 1(ii)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(ii)

The given expression is an expression having only non-negative integral powers of y. So, it is a polynomial.

The highest power of y is 3. So, it is a polynomial of degree 3. Question 1(iii)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(iii)

The given expression is an expression having only non-negative integral powers of t. So, it is a polynomial.

The highest power of t is 2. So, it is a polynomial of degree 2. Question 1(iv)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

X¹⁰⁰ – 1Solution 1(iv)

X¹⁰⁰ – 1 

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 100. So, it is a polynomial of degree 100. Question 1(ix)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(ix)

The given expression can be written as  

It contains a term having negative integral power of x. So, it is not a polynomial.Question 1(x)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(x)

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 2. So, it is a polynomial of degree 2. Question 1(xi)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(xi)

The given expression can be written as 2x^-2.

It contains a term having negative integral power of x. So, it is not a polynomial. Question 1(xii)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(xii)

The given expression contains a term containing x^1/2, where ½ is not a non-negative integer.

So, it is not a polynomial. Question 1(xiii)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(xiii)

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 2. So, it is a polynomial of degree 2. Question 1(xiv)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

x⁴ – x^3/2 + x – 3Solution 1(xiv)

x⁴ – x^3/2 + x – 3 

The given expression contains a term containing x^3/2, where 3/2 is not a non-negative integer.

So, it is not a polynomial. Question 1(xv)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(xv)

The given expression can be written as 2x³ + 3x² + x^1/2 – 1. 

The given expression contains a term containing x^1/2, where ½ is not a non-negative integer.

So, it is not a polynomial. Question 2(i)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-7 + xSolution 2(i)

-7 + x

The degree of a given polynomial is 1.

Hence, it is a linear polynomial.Question 2(ii)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

6ySolution 2(ii)

6y

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. Question 2(iii)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-z³Solution 2(iii)

-z³

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. Question 2(iv)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

1 – y – y³Solution 2(iv)

1 – y – y³

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. Question 2(v)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

x – x³ + x⁴Solution 2(v)

x – x³ + x⁴

The degree of a given polynomial is 4.

Hence, it is a quartic polynomial. Question 2(vi)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

1 + x + x²Solution 2(vi)

1 + x + x²

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. Question 2(vii)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-6x²Solution 2(vii)

-6x²

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. Question 2(viii)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-13Solution 2(viii)

-13

The given polynomial contains only one term namely constant.

Hence, it is a constant polynomial. Question 2(ix)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-pSolution 2(ix)

-p

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. Question 3(i)

Write the coefficient of x³ in x + 3x² – 5x³ + x⁴Solution 3(i)

The coefficient of x³ in x + 3x² – 5x³ + x⁴ is -5. Question 3(ii)

Write the coefficient of x in .Solution 3(ii)

The coefficient of x in . Question 3(iii)

Write the coefficient of x² in 2x – 3 + x³.Solution 3(iii)

The given polynomial can be written as x³ + 0x² + 2x – 3.

Hence, the coefficient of x² in 2x – 3 + x³ is 0. Question 3(iv)

Write the coefficient of x in .Solution 3(iv)

The coefficient of x in . Question 3(v)

Write the constant term in .Solution 3(v)

The constant term in . Question 4(i)

Determine the degree of each of the following polynomials.

Solution 4(i)

Hence, the degree of a given polynomial is 2. Question 4(ii)

Determine the degree of each of the following polynomials.

y²(y – y³)Solution 4(ii)

y²(y – y³)

= y³ – y⁵

Hence, the degree of a given polynomial is 5. Question 4(iii)

Determine the degree of each of the following polynomials.

(3x – 2)(2x³ + 3x²)Solution 4(iii)

(3x – 2)(2x³ + 3x²)

= 6x⁴ + 9x³ – 4x³ – 6x²

= 6x⁴ + 5x³ – 6x²

Hence, the degree of a given polynomial is 4. Question 4(iv)

Determine the degree of each of the following polynomials.

Solution 4(iv)

The degree of a given polynomial is 1. Question 4(v)

Determine the degree of each of the following polynomials.

-8Solution 4(v)

-8

This is a constant polynomial.

The degree of a non-zero constant polynomial is zero. Question 4(vi)

Determine the degree of each of the following polynomials.

x^-2(x⁴ + x²)Solution 4(vi)

x^-2(x⁴ + x²)

= x^-2.x²(x² + 1)

= x⁰ (x² + 1)

= x² + 1

Hence, the degree of a given polynomial is 2. Question 5(i)

Give an example of a monomial of degree 5.Solution 5(i)

Example of a monomial of degree 5:

3x⁵ Question 5(ii)

Give an example of a binomial of degree 8.Solution 5(ii)

Example of a binomial of degree 8:

x – 6x⁸ Question 5(iii)

Give an example of a trinomial of degree 4.Solution 5(iii)

Example of a trinomial of degree 4:

7 + 2y + y⁴ Question 5(iv)

Give an example of a monomial of degree 0.Solution 5(iv)

Example of a monomial of degree 0:

7 Question 6(i)

Rewrite each of the following polynomials in standard form.

x – 2x² + 8 + 5x³Solution 6(i)

x – 2x² + 8 + 5x³ in standard form:

5x³ – 2x² + x + 8 Question 6(ii)

Rewrite each of the following polynomials in standard form.

Solution 6(ii)

Question 6(iii)

Rewrite each of the following polynomials in standard form.

6x³ + 2x – x⁵ – 3x²Solution 6(iii)

6x³ + 2x – x⁵ – 3x² in standard form:

-x⁵ + 6x³ – 3x² + 2x Question 6(iv)

Rewrite each of the following polynomials in standard form.

2 + t – 3t³ + t⁴ – t²Solution 6(iv)

2 + t – 3t³ + t⁴ – t² in standard form:

t⁴ – 3t³ – t² + t + 2 

Exercise Ex. 2B

Question 1

If p(x) = 5 – 4x + 2x², find

(i) p(0)

(ii) p(3)

(iii) p(-2)Solution 1

p(x) = 5 – 4x + 2x²

(i) p(0) = 5 – 4  0 + 2  0² = 5

(ii) p(3) = 5 – 4  3 + 2  3²

= 5 – 12 + 18

= 23 – 12 = 11

(iii) p(-2) = 5 – 4(-2) + 2 (-2)²

= 5 + 8 + 8 = 21Question 2

If p(y) = 4 + 3y – y² + 5y³, find

(i) p(0)

(ii) p(2)

(iii) p(-1)Solution 2

p(y) = 4 + 3y – y² + 5y³

(i) p(0) = 4 + 3  0 – 0² + 5  0³

= 4 + 0 – 0 + 0 = 4

(ii) p(2) = 4 + 3  2 – 2² + 5  2³

= 4 + 6 – 4 + 40

= 10 – 4 + 40 = 46

(iii) p(-1) = 4 + 3(-1) – (-1)² + 5(-1)³

= 4 – 3 – 1 – 5 = -5Question 3

If f(t) = 4t² – 3t + 6, find

(i) f(0)

(ii) f(4)

(iii) f(-5)Solution 3

f(t) = 4t² – 3t + 6

(i) f(0) = 4  0² – 3  0 + 6

= 0 – 0 + 6 = 6

(ii) f(4) = 4(4)² – 3  4 + 6

= 64 – 12 + 6 = 58

(iii) f(-5) = 4(-5)² – 3(-5) + 6

= 100 + 15 + 6 = 121Question 4

If p(x) = x³ – 3x² + 2x, find p(0), p(1), p(2). What do you conclude?Solution 4

p(x) = x³ – 3x² + 2x

Thus, we have

p(0) = 0³ – 3(0)² + 2(0) = 0

p(1) = 1³ – 3(1)² + 2(1) = 1 – 3 + 2 = 0

p(2) = 2³ – 3(2)² + 2(2) = 8 – 12 + 4 = 0

Hence, 0, 1 and 2 are the zeros of the polynomial p(x) = x³ – 3x² + 2x.  Question 5

If p(x) = x³ + x² – 9x – 9, find p(0), p(3), p(-3) and p(-1). What do you conclude about the zeros of p(x)? Is 0 a zero of p(x)?Solution 5

p(x) = x³ + x² – 9x – 9

Thus, we have 

p(0) = 0³ + 0² – 9(0) – 9 = -9 

p(3) = 3³ + 3² – 9(3) – 9 = 27 + 9 – 27 – 9 = 0

p(-3) = (-3)³ + (-3)² – 9(-3) – 9 = -27 + 9 + 27 – 9 = 0

p(-1) = (-1)³ + (-1)² – 9(-1) – 9 = -1 + 1 + 9 – 9 = 0

Hence, 0, 3 and -3 are the zeros of p(x).

Now, 0 is not a zero of p(x) since p(0) ≠ 0. Question 6(i)

Verify that:

4 is a zero of the polynomial p(x) = x – 4.Solution 6(i)

p(x) = x – 4

Then, p(4) = 4 – 4 = 0

 4 is a zero of the polynomial p(x).Question 6(ii)

Verify that:

-3 is a zero of the polynomial p(x) = x – 3.Solution 6(ii)

p(x) = x – 3

Then,p(-3) = -3 – 3 = -6

 -3 is not a zero of the polynomial p(x).Question 6(iii)

Verify that:

is a zero of the polynomial p(y) = 2y + 1.Solution 6(iii)

p(y) = 2y + 1

Then, 

is a zero of the polynomial p(y).Question 6(iv)

Verify that:

is a zero of thepolynomial p(x) = 2 – 5x.Solution 6(iv)

p(x) = 2 – 5x

Then, 

 is a zero of the polynomial p(x).Question 7(i)

Verify that:

1 and 2 are the zeros of the polynomial p(x) = (x – 1) (x – 2).Solution 7(i)

p(x) = (x – 1) (x – 2)

Then,p(1)= (1 – 1) (1 – 2) = 0 -1 = 0

 1 is a zero of the polynomial p(x).

Also, p(2) = (2 – 1)(2 – 2) = 1 0 = 0

 2 is a zero of the polynomial p(x).

 Hence,1 and 2 are the zeroes of the polynomial p(x).Question 7(ii)

Verify that:

2 and -3 are the zeros of the polynomial p(x) = x² + x – 6.Solution 7(ii)

p(x) = x² + x – 6

Then, p(2) = 2² + 2 – 6

= 4 + 2 – 6

= 6 – 6 = 0

 2 is a zero of the polynomial p(x).

 Also, p(-3) = (-3)² – 3 – 6

= 9 – 3 – 6 = 0

 -3 is a zero of the polynomial p(x).

 Hence, 2 and -3 are the zeroes of the polynomial p(x).Question 7(iii)

Verify that:

0 and 3 are the zeros of the polynomial p(x) = x² – 3x.Solution 7(iii)

p(x) = x² – 3x.

Then,p(0) = 0² – 3 0 = 0

p(3) = (3)²– 3 3 = 9 – 9 = 0

0 and 3 are the zeroes of the polynomial p(x).Question 8(i)

Find the zero of the polynomial:

p(x) = x – 5Solution 8(i)

p(x) = 0

x – 5 = 0

x = 5

_{5 is the zero of the polynomial p(x).}Question 8(ii)

Find the zero of the polynomial:

q(x) = x + 4Solution 8(ii)

q(x) = 0

 x + 4 = 0

x= -4

 _{-4 is the zero of the polynomial q(x).} Question 8(iv)

Find the zero of the polynomial:

f(x) = 3x + 1Solution 8(iv)

f(x) = 0

 3x + 1= 0

3x=-1

 x =

  x =is the zero of the polynomial f(x).Question 8(v)

Find the zero of the polynomial:

g(x) = 5 – 4xSolution 8(v)

g(x) = 0

 5 – 4x = 0

 -4x = -5

 x =

 x =  is the zero of the polynomial g(x).Question 8(vii)

Find the zero of the polynomial:

p(x) = ax, a  0Solution 8(vii)

p(x) = 0

 ax = 0

x = 0

0 is the zero of the polynomial p(x).Question 8(viii)

Find the zero of the polynomial:

q(x) = 4xSolution 8(viii)

q(x) = 0

4x = 0

 x = 0

0 is the zero of the polynomial q(x).Question 8(iii)

Find the zero of the polynomial:

r(x) = 2x + 5Solution 8(iii)

r(x) = 2x + 5

Now, r(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

Question 8(vi)

Find the zero of the polynomial:

h(x) = 6x – 2Solution 8(vi)

h(x) = 6x – 2

Now, h(x) = 0

⇒ 6x – 2 = 0

⇒ 6x = 2

Question 9

If 2 and 0 are the zeros of the polynomial f(x) = 2x³ – 5x² + ax + b then find the values of a and b.

HINT f(2) = 0 and f(0) = 0.Solution 9

f(x) = 2x³ – 5x² + ax + b

Now, 2 is a zero of f(x).

⇒ f(2) = 0

⇒ 2(2)³ – 5(2)² + a(2) + b = 0

⇒ 16 – 20 + 2a + b = 0

⇒ 2a + b – 4 = 0 ….(i)

Also, 0 is a zero of f(x).

⇒ f(0) = 0

⇒ 2(0)³ – 5(0)² + a(0) + b = 0

⇒ 0 – 0 + 0 + b = 0

⇒ b = 0

Substituting b = 0 in (i), we get

2a + 0 – 4 = 0

⇒ 2a = 4

⇒ a = 2

Thus, a = 2 and b = 0. 

Exercise Ex. 2D

Question 1

Using factor theorem, show that:

(x – 2) is a factor of (x³ – 8)Solution 1

f(x) = (x³ – 8)

By the Factor Theorem, (x – 2) will be a factor of f(x) if f(2) = 0.

Here, f(2) = (2)³ – 8

= 8 – 8 = 0

 (x – 2) is a factor of (x³ – 8).
Question 2

Using factor theorem, show that:

(x – 3) is a factor of (2x³ + 7x² – 24x – 45)Solution 2

f(x) = (2x³ + 7x² – 24x – 45)

By the Factor Theorem, (x – 3) will be a factor of f(x) if f(3) = 0.

Here, f(3) = 2  3³ + 7  3² – 24  3 – 45

= 54 + 63 – 72 – 45

= 117 – 117 = 0

 (x – 3) is a factor of (2x³ + 7x² – 24x – 45).Question 3

Using factor theorem, show that:

(x – 1) is a factor of (2x⁴ + 9x³ + 6x² – 11x – 6)Solution 3

f(x) = (2x⁴ + 9x³ + 6x² – 11x – 6)

By the Factor Theorem, (x – 1) will be a factor of f(x) if f(1) = 0.

Here, f(1) = 2  1⁴ + 9  1³ + 6  1² – 11  1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17 = 0

 (x – 1) is factor of (2x⁴ + 9x³ + 6x² – 11x – 6).Question 4

Using factor theorem, show that:

(x + 2) is a factor of (x⁴ – x² – 12)Solution 4

f(x) = (x⁴ – x² – 12)

By the Factor Theorem, (x + 2) will be a factor of f(x) if f(-2) = 0.

Here, f(-2) = (-2)⁴ – (-2)² – 12

= 16 – 4 – 12

= 16 – 16 = 0

 (x + 2) is a factor of (x⁴ – x² – 12).Question 5

p(x) = 69 + 11x – x² + x³, g(x) = x + 3Solution 5

By the factor theorem, g(x) = x + 3 will be a factor of p(x) if p(-3) = 0.

Now, p(x) = 69 + 11x – x² + x³

⇒ p(-3) = 69 + 11(-3) – (-3)² + (-3)³

= 69 – 33 – 9 – 27

= 0

Hence, g(x) = x + 3 is a factor of the given polynomial p(x). Question 6

Using factor theorem, show that:

(x + 5) is a factor of (2x³ + 9x² – 11x – 30)Solution 6

f(x) = 2x³ + 9x² – 11x – 30

By the Factor Theorem, (x + 5) will be a factor of f(x) if f(-5) = 0.

Here, f(-5) = 2(-5)³ + 9(-5)² – 11(-5) – 30

= -250 + 225 + 55 – 30

= -280 + 280 = 0

 (x + 5) is a factor of (2x³ + 9x² – 11x – 30).Question 7

Using factor theorem, show that:

(2x – 3) is a factor of (2x⁴ + x³ – 8x² – x + 6)Solution 7

f(x) = (2x⁴ + x³ – 8x² – x + 6)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here, 2x – 3 = 0  x = 

 is a factor of .Question 8

p(x) = 3x³ + x² – 20x + 12, g(x) = 3x – 2Solution 8

By the factor theorem, g(x) = 3x – 2 will be a factor of p(x) if  = 0.

Now, p(x) = 3x³ + x² – 20x + 12

Hence, g(x) = 3x – 2 is a factor of the given polynomial p(x). Question 9

Using factor theorem, show that:

(x – ) is a factor of Solution 9

f(x) = 

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0. 

Here, 

= 14 – 8 – 6

= 14 – 14 = 0

Question 10

Using factor theorem, show that:

(x + ) is a factor of Solution 10

f(x) = 

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here, 

Question 11

Show that (p – 1) is a factor of (p¹⁰ – 1) and also of (p¹¹ – 1).Solution 11

Let q(p) = (p¹⁰ – 1) and f(p) = (p¹¹ – 1)

By the factor theorem, (p – 1) will be a factor of q(p) and f(p) if q(1) and f(1) = 0.

Now, q(p) = p¹⁰ – 1

⇒ q(1) = 1¹⁰ – 1 = 1 – 1 = 0

Hence, (p – 1) is a factor of p¹⁰ – 1.

And, f(p) = p¹¹ – 1

⇒ f(1) = 1¹¹ – 1 = 1 – 1 = 0

Hence, (p – 1) is also a factor of p¹¹ – 1. Question 12

Find the value of k for which (x – 1) is a factor of (2x³+ 9x² + x + k).Solution 12

f(x) = (2x³ + 9x² + x + k)

x – 1 = 0  x = 1

 f(1) = 2  1³ + 9  1² + 1 + k

= 2 + 9 + 1 + k

= 12 + k

Given that (x – 1) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(1) = 0.

f(1) = 12 + k = 0

k = -12.
Question 13

Find the value of a for which (x – 4) is a factor of (2x³ – 3x² – 18x + a).Solution 13

f(x) = (2x³ – 3x² – 18x + a)

x – 4 = 0  x = 4

 f(4) = 2(4)³ – 3(4)² – 18  4 + a

= 128 – 48 – 72 + a

= 128 – 120 + a

= 8 + a

Given that (x – 4) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(4) = 0.

 f(4) = 8 + a = 0

 a = -8
Question 14

Find the value of a for which (x + 1) is a factor of (ax³ + x² – 2x + 4a – 9).Solution 14

Let p(x) = ax³ + x² – 2x + 4a – 9

It is given that (x + 1) is a factor of p(x).

⇒ p(-1) = 0

⇒ a(-1)³ + (-1)² – 2(-1) + 4a – 9 = 0

⇒ -a + 1 + 2 + 4a – 9 = 0

⇒ 3a – 6 = 0

⇒ 3a = 6

⇒ a = 2 Question 15

Find the value of a for which (x + 2a) is a factor of (x⁵ – 4a²x³ + 2x + 2a + 3).Solution 15

Let p(x) = x⁵ – 4a²x³ + 2x + 2a + 3

It is given that (x + 2a) is a factor of p(x).

⇒ p(-2a) = 0

⇒ (-2a)⁵ – 4a²(-2a)³ + 2(-2a) + 2a + 3 = 0

⇒ -32a⁵ – 4a²(-8a³) – 4a + 2a + 3 = 0

⇒ -32a⁵ + 32a⁵ -2a + 3 = 0

⇒ 2a = 3

Question 16

Find the value of m for which (2x – 1) is a factor of (8x⁴ + 4x³ – 16x² + 10x + m).Solution 16

Let p(x) = 8x⁴ + 4x³ – 16x² + 10x + m

It is given that (2x – 1) is a factor of p(x).

Question 17

Find the value of a for which the polynomial (x⁴ – x³ – 11x² – x + a) is divisible by (x + 3).Solution 17

Let p(x) = x⁴ – x³ – 11x² – x + a

It is given that p(x) is divisible by (x + 3).

⇒ (x + 3) is a factor of p(x).

⇒ p(-3) = 0

⇒ (-3)⁴ – (-3)³ – 11(-3)² – (-3) + a = 0

⇒ 81 + 27 – 99 + 3 + a = 0

⇒ 12 + a = 0

⇒ a = -12 Question 18

Without actual division, show that (x³ – 3x² – 13x + 15) is exactly divisible by (x² + 2x – 3).Solution 18

Let f(x) = x³ – 3x² – 13x + 15

Now, x² + 2x – 3 = x² + 3x – x – 3

= x (x + 3) – 1 (x + 3)

= (x + 3) (x – 1)

Thus, f(x) will be exactly divisible by x² + 2x – 3 = (x + 3) (x – 1) if (x + 3) and (x – 1) are both factors of f(x), so by factor theorem, we should have f(-3) = 0 and f(1) = 0.

Now, f(-3) = (-3)³ – 3 (-3)² – 13 (-3) + 15

= -27 – 3  9 + 39 + 15

= -27 – 27 + 39 + 15

= -54 + 54 = 0

And, f(1) = 1³ – 3  1² – 13  1 + 15

= 1 – 3 – 13 + 15

= 16 – 16 = 0

 f(-3) = 0 and f(1) = 0

So, x² + 2x – 3 divides f(x) exactly.Question 19

If (x³ + ax² + bx + 6) has (x – 2) as a factor and leaves a remainder 3 when divided by (x – 3), find the values of a and b.Solution 19

Letf(x) = (x³ + ax² + bx + 6)

Now, by remainder theorem, f(x) when divided by (x – 3) will leave a remainder as f(3).

So, f(3) = 3³ + a 3² + b 3 + 6 = 3

27 + 9a + 3b + 6 = 3

9 a + 3b + 33 = 3

9a + 3b = 3 – 33

9a + 3b = -30

3a + b = -10(i)

Given that (x – 2) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(2) = 0.

 f(2) =  2³ + a 2² + b 2 + 6 = 0

                       8 + 4a+ 2b + 6 = 0

                               4a + 2b = -14

                                   2a + b = -7(ii)

Subtracting (ii) from (i), we get,

a = -3

Substituting the value of a = -3 in (i), we get,

3(-3) + b = -10

-9 + b = -10

b = -10 + 9

b = -1

a = -3 and b = -1.Question 20

Find the values of a and b so that the polynomial (x³ – 10x² + ax + b) is exactly divisible by (x – 1) as well as (x – 2).Solution 20

Let f(x) = (x³ – 10x² + ax + b), then by factor theorem

(x – 1) and (x – 2) will be factors of f(x) if f(1) = 0 and f(2) = 0.

f(1) = 1³ – 10 1² + a 1 + b = 0

1 – 10 + a + b = 0

a + b = 9(i)

Andf(2) = 2³ – 10 2² + a 2 + b = 0

8 – 40 + 2a + b = 0

2a + b = 32(ii)

Subtracting (i) from (ii), we get

a = 23

Substituting the value of a = 23 in (i), we get

23 + b = 9

b = 9 – 23

b = -14

a = 23 and b = -14.Question 21

Find the values of a and b so that the polynomial (x⁴ + ax³ – 7x² – 8x + b) is exactly divisible by (x + 2) as well as (x + 3).Solution 21

Letf(x)= (x⁴ + ax³ – 7x² – 8x + b)

Now, x + 2 = 0 x = -2 and x + 3 = 0 x = -3

By factor theorem, (x + 2) and (x + 3) will be factors of f(x) if f(-2) = 0 and f(-3) = 0

f(-2) = (-2)⁴ + a (-2)³ – 7 (-2)² – 8 (-2) + b = 0

16 – 8a – 28 + 16 + b = 0

-8a + b = -4

8a – b = 4(i)

And, f(-3) = (-3)⁴ + a (-3)³ – 7 (-3)² – 8 (-3) + b = 0

81 – 27a – 63 + 24 + b = 0

-27a + b = -42

27a – b = 42(ii)

Subtracting (i) from (ii), we get,

19a = 38

So, a = 2

Substituting the value of a = 2 in (i), we get

8 2 – b = 4

16 – b = 4

-b = -16 + 4

-b = -12

b = 12

a = 2 and b = 12.Question 22

If both (x – 2) and  are factors of px² + 5x + r, prove that p = r.Solution 22

Let f(x) = px² + 5x + r

Now, (x – 2) is a factor of f(x).

⇒ f(2) = 0

⇒ p(2)² + 5(2) + r = 0

⇒ 4p + 10 + r = 0

⇒ 4p + r = -10

Also,  is a factor of f(x).

From (i) and (ii), we have

4p + r = p + 4r

⇒ 4p – p = 4r – r

⇒ 3p = 3r

⇒ p = rQuestion 23

Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.Solution 23

Let f(x) = 2x⁴ – 5x³ + 2x² – x + 2

and g(x) = x² – 3x + 2

= x² – 2x – x + 2

= x(x – 2) – 1(x – 2)

= (x – 2)(x – 1)

Clearly, (x – 2) and (x – 1) are factors of g(x).

In order to prove that f(x) is exactly divisible by g(x), it is sufficient to prove that f(x) is exactly divisible by (x – 2) and (x – 1).

Thus, we will show that (x – 2) and (x – 1) are factors of f(x).

Now,

f(2) = 2(2)⁴ – 5(2)³ + 2(2)² – 2 + 2 = 32 – 40 + 8 = 0 and

f(1) = 2(1)⁴ – 5(1)³ + 2(1)² – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0

Therefore, (x – 2) and (x – 1) are factors of f(x).

⇒ g(x) = (x – 2)(x – 1) is a factor of f(x).

Hence, f(x) is exactly divisible by g(x). Question 24

What must be added to 2x⁴ – 5x³ + 2x² – x – 3 so that the result is exactly divisible by (x – 2)?Solution 24

Let the required number to be added be k.

Then, p(x) = 2x⁴ – 5x³ + 2x² – x – 3 + k and g(x) = x – 2

Thus, we have

p(2) = 0

⇒ 2(2)⁴ – 5(2)³ + 2(2)² – 2 – 3 + k = 0

⇒ 32 – 40 + 8 – 5 + k = 0

⇒ k – 5 = 0

⇒ k = 5

Hence, the required number to be added is 5. Question 25

What must be subtracted from (x⁴ + 2x³ – 2x² + 4x + 6) so that the result is exactly divisible by (x² + 2x – 3)?Solution 25

Let p(x) = x⁴ + 2x³ – 2x² + 4x + 6 and q(x) = x² + 2x – 3.

When p(x) is divided by q(x), the remainder is a linear expression in x.

So, let r(x) = ax + b be subtracted from p(x) so that p(x) – r(x) is divided by q(x).

Let f(x) = p(x) – r(x) = p(x) – (ax + b)

= (x⁴ + 2x³ – 2x² + 4x + 6) – (ax + b)

= x⁴ + 2x³ – 2x² + (4 – a)x + 6 – b

We have,

q(x) = x² + 2x – 3

= x² + 3x – x – 3

= x(x + 3) – 1(x + 3)

= (x + 3)(x – 1)

Clearly, (x + 3) and (x – 1) are factors of q(x).

Therefore, f(x) will be divisible by q(x) if (x + 3) and (x – 1) are factors of f(x).

i.e., f(-3) = 0 and f(1) = 0

Consider, f(-3) = 0

⇒ (-3)⁴ + 2(-3)³ – 2(-3)² + (4 – a)(-3) + 6 – b = 0

⇒ 81 – 54 – 18 – 12 + 3a + 6 – b = 0

⇒ 3 + 3a – b = 0

⇒ 3a – b = -3 ….(i)

And, f(1) = 0

⇒ (1)⁴ + 2(1)³ – 2(1)² + (4 – a)(1) + 6 – b = 0

⇒ 1 + 2 – 2 + 4 – a + 6 – b = 0

⇒ 11 – a – b = 0

⇒ -a – b = -11 ….(ii)

Subtracting (ii) from (i), we get

4a = 8

⇒ a = 2

Substituting a = 2 in (i), we get

3(2) – b = -3

⇒ 6 – b = -3

⇒ b = 9

Putting the values of a and b in r(x) = ax + b, we get

r(x) = 2x + 9

Hence, p(x) is divisible by q(x), if r(x) = 2x + 9 is subtracted from it. Question 26

Use factor theorem to prove that (x + a) is a factor of (xⁿ + aⁿ) for any odd positive integer n.Solution 26

Let f(x) = xⁿ + aⁿ

In order to prove that (x + a) is a factor of f(x) for any odd positive integer n, it is sufficient to show that f(-a) = 0.

Now,

f(-a) = (-a)ⁿ + aⁿ

= (-1)ⁿ aⁿ + aⁿ

= [(-1)ⁿ + 1] aⁿ

= [-1 + 1] aⁿ …[n is odd ⇒ (-1)ⁿ = -1]

= 0 × aⁿ

= 0

Hence, (x + a) is a factor of xⁿ + aⁿ for any odd positive integer n. 

Exercise Ex. 2C

Question 1

By actual division, find the quotient and the remainder when (x⁴ + 1) is divided by (x – 1).

Verify that remainder = f(1).Solution 1

Quotient = x³ + x² + x + 1

Remainder = 2

Verification:

f(x) = x⁴ + 1

Then, f(1) = 1⁴ + 1 = 1 + 1 = 2 = Remainder Question 2

Verify the division algorithm for the polynomials

p(x) = 2x⁴ – 6x³ + 2x² – x + 2 and g(x) = x + 2.Solution 2

Question 3

Using remainder theorem, find the remainder when:

(x³ – 6x² + 9x + 3) is divided by (x – 1)Solution 3

f(x) = x³ – 6x² + 9x + 3

Now, x – 1 = 0  x = 1

By the remainder theorem, we know that when f(x) is divided by (x – 1) the remainder is f(1).

Now, f(1) = 1³ – 6  1² + 9  1 + 3

= 1 – 6 + 9 + 3

= 13 – 6 = 7

 The required remainder is 7.Question 4

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 2x³ – 7x² + 9x – 13, g(x) = x – 3.Solution 4

x – 3 = 0

⇒ x = 3

By the remainder theorem, we know that when p(x) = 2x³ – 7x² + 9x – 13 is divided by g(x) = x – 3, the remainder is g(3).

Now,

g(3) = 2(3)³ – 7(3)² + 9(3) – 13 = 54 – 63 + 27 – 13 = 5

Hence, the required remainder is 5.Question 5

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 3x⁴ – 6x² – 8x – 2, g(x) = x – 2.Solution 5

x – 2 = 0

⇒ x = 2

By the remainder theorem, we know that when p(x) = 3x⁴ – 6x² – 8x – 2 is divided by g(x) = x – 2, the remainder is g(2).

Now,

g(2) = 3(2)⁴ – 6(2)² – 8(2) – 2 = 48 – 24 – 16 – 2 = 6

Hence, the required remainder is 6. Question 6

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 2x³ – 9x² + x + 15, g(x) = 2x – 3.Solution 6

2x – 3 = 0

⇒ x =  

By the remainder theorem, we know that when p(x) = 2x³ – 9x² + x + 15 is divided by g(x) = 2x – 3, the remainder is g.

Now,

Hence, the required remainder is 3. Question 7

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = x³ – 2x² – 8x – 1, g(x) = x + 1.Solution 7

x + 1 = 0

⇒ x = -1

By the remainder theorem, we know that when p(x) = x³ – 2x² – 8x – 1 is divided by g(x) = x + 1, the remainder is g(-1).

Now,

g(-1) = (-1)³ – 2(-1)² – 8(-1) – 1 = -1 – 2 + 8 – 1 = 4

Hence, the required remainder is 4. Question 8

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 2x³ + x²– 15x – 12, g(x) = x + 2.Solution 8

x + 2 = 0

⇒ x = -2

By the remainder theorem, we know that when p(x) = 2x³ + x² – 15x – 12 is divided by g(x) = x + 2, the remainder is g(-2).

Now,

g(-2) = 2(-2)³ + (-2)² – 15(-2) – 12 = -16 + 4 + 30 – 12 = 6

Hence, the required remainder is 6. Question 9

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 6x³ + 13x² + 3, g(x) = 3x + 2.Solution 9

3x + 2 = 0

⇒ x =  

By the remainder theorem, we know that when p(x) = 6x³ + 13x² + 3 is divided by g(x) = 3x + 2, the remainder is g.

Now,

Hence, the required remainder is 7. Question 10

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = x³ – 6x² + 2x – 4, g(x) = .Solution 10

By the remainder theorem, we know that when p(x) = x³ – 6x² + 2x – 4 is divided by g(x) = , the remainder is g.

Now,

Hence, the required remainder is . Question 11

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 2x³ + 3x² – 11x – 3, g(x) = .Solution 11

By the remainder theorem, we know that when p(x) = 2x³ + 3x² – 11x – 3 is divided by g(x) = , the remainder is g.

Now,

Hence, the required remainder is 3. Question 12

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = x³ – ax² + 6x – a, g(x) = x – a.Solution 12

x – a = 0

⇒ x = a

By the remainder theorem, we know that when p(x) = x³ – ax² + 6x – a is divided by g(x) = x – a, the remainder is g(a).

Now,

g(a) = (a)³ – a(a)² + 6(a) – a = a³– a³+ 6a – a = 5a

Hence, the required remainder is 5a. Question 13

The polynomial (2x³ + x² – ax + 2) and (2x³ – 3x² – 3x + a) when divided by (x – 2) leave the same remainder. Find the value of a.Solution 13

Let p(x) = 2x³ + x² – ax + 2 and q(x) = 2x³ – 3x² – 3x + a be the given polynomials.

The remainders when p(x) and q(x) are divided by (x – 2) are p(2) and q(2) respectively.

By the given condition, we have

p(2) = q(2)

⇒ 2(2)³ + (2)² – a(2) + 2 = 2(2)³ – 3(2)² – 3(2) + a

⇒ 16 + 4 – 2a + 2 = 16 – 12 – 6 + a

⇒ 22 – 2a = -2 + a

⇒ a + 2a = 22 + 2

⇒ 3a = 24

⇒ a = 8 Question 14

The polynomial f(x) = x⁴ – 2x³ + 3x² – ax + b when divided by (x – 1) and (x + 1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when f(x) is divided by (x – 2).Solution 14

Letf(x) = (x⁴ – 2x³ + 3x² – ax + b)

_{From the given information,}

f(1) = 1⁴ – 2(1)³ + 3(1)² – a 1 + b = 5

 1 – 2 + 3 – a + b = 5

 2 – a + b = 5(i)

And,

f(-1) = (-1)⁴ – 2(-1)³ + 3(-1)² – a(-1) + b = 19

 1 + 2 + 3 + a + b = 19

 6 + a + b = 19(ii)

Adding (i) and (ii), we get

8 + 2b = 24

2b= 24 – 8 = 16

b = 

Substituting the value of b = 8 in (i), we get

2 – a + 8 = 5

-a + 10 = 5

-a = -10 + 5

-a = -5

a = 5

a = 5 and b = 8

f(x) = x⁴ – 2x³ + 3x² – ax + b

= x⁴ – 2x³ + 3x² – 5x + 8

f(2) = (2)⁴ – 2(2)³ + 3(2)² – 5 2 + 8

= 16 – 16 + 12 – 10 + 8

= 20 – 10 = 10

The required remainder is 10.Question 15

If p(x) = x³ – 5x² + 4x – 3 and g(x) = x – 2, show that p(x) is not a multiple of g(x).Solution 15

The polynomial p(x) will be a multiple of g(x) if g(x) divides p(x) completely.

i.e. when p(x) is divided by g(x), it does not leave any remainder.

Now, x – 2 = 0 ⇒ x = 2

Also,

p(2) = (2)³ – 5(2)² + 4(2) – 3 = 8 – 20 + 8 – 3 = -7 ≠ 0

Thus, p(x) is not a multiple of g(x).Question 16

If p(x) = 2x³ – 11x² – 4x + 5 and g(x) = 2x + 1, show that g(x) is not a factor of p(x).Solution 16

The polynomial g(x) will be a factor of p(x) if g(x) divides p(x) completely.

i.e. when p(x) is divided by g(x), it does not leave any remainder.

Now, 2x + 1 = 0 ⇒ x = 

Also,

Thus, g(x) is not a factor of p(x). 

Exercise VSAQ

Question 1

What can you say about the sum of a rational number and an irrational number?Solution 1

The sum of a rational number and an irrational number is irrational.

Example: 5 +  is irrational. Question 2

Solve .Solution 2

Question 3

The number  will terminate after how many decimal places?Solution 3

Thus, the given number will terminate after 3 decimal places. Question 4

Find the value of (1296)^0.17× (1296)^0.08. Solution 4

(1296)^0.17× (1296)^0.08

Question 5

Simplify .Solution 5

Question 6

Find an irrational number between 5 and 6.Solution 6

An irrational number between 5 and 6 =  Question 7

Find the value of .Solution 7

Question 8

Rationalise Solution 8

Question 9

Solve for x: .Solution 9

Question 10

Simplify (32)^1/5 + (-7)⁰ + (64)^1/2.Solution 10

Question 11

Evaluate .Solution 11

Question 12

Simplify .Solution 12

Question 13

If a = 1, b = 2 then find the value of (a^b + b^a)^-1.Solution 13

Given, a = 1 and b = 2

Question 14

Simplify .Solution 14

Question 15

Give an example of two irrational numbers whose sum as well as product is rational.Solution 15

Question 16

Is the product of a rational and irrational numbers always irrational? Give an example.Solution 16

Yes, the product of a rational and irrational numbers is always irrational.

For example,

Question 17

Give an example of a number x such that x² is an irrational number and x³ is a rational number.Solution 17

Question 18

Write the reciprocal of ().Solution 18

The reciprocal of ()

Question 19

Solution 19

Question 20

Simplify Solution 20

Question 21

If 10^x = 64, find the value of .Solution 21

Question 22

Evaluate Solution 22

Question 23

Simplify .Solution 23

Exercise MCQ

Question 1

Which of the following is a rational number?

(a) 

(b) π

(c) 

(d) 0Solution 1

Correct option: (d)

0 can be written as  where p and q are integers and q ≠ 0.Question 2

A rational number between -3 and 3 is

(a) 0

(b) -4.3

(c) -3.4

(d) 1.101100110001….Solution 2

Correct option: (a)

On a number line, 0 is a rational number that lies between -3 and 3. Question 3

Two rational numbers between  are

(a) 

(b) 

(c) 

(d) Solution 3

Correct option: (c)

Two rational numbers between   Question 4

Every point on number line represents

(a) a rational number

(b) a natural number

(c) an irrational number

(d) a unique numberSolution 4

Correct option: (d)

Every point on number line represents a unique number. Question 5

Which of the following is a rational number?

Solution 5

Question 6

Every rational number is

(a) a natural number

(b) a whole number

(c) an integer

(d)a real numberSolution 6

Question 7

Between any two rational numbers there

(a) is no rational number

(b) is exactly one rational number

(c) are infinitely many rational numbers

(d)is no irrational numberSolution 7

Question 8

The decimal representation of a rational number is

(a) always terminating

(b) either terminating or repeating

(c) either terminating or non-repeating

(d)neither terminating nor repeatingSolution 8

Question 9

The decimal representation of an irrational number is

(a) always terminating

(b) either terminating or repeating

(c) either terminating or non-repeating

(d)neither terminating nor repeatingSolution 9

Question 10

The decimal expansion that a rational number cannot have is

(a) 0.25

(b) 

(c) 

(d) 0.5030030003….Solution 10

Correct option: (d)

The decimal expansion of a rational number is either terminating or non-terminating recurring.

The decimal expansion of 0.5030030003…. is non-terminating non-recurring, which is not a property of a rational number. Question 11

Which of the following is an irrational number?

(a) 3.14

(b) 3.141414….

(c) 3.14444…..

(d) 3.141141114….Solution 11

Correct option: (d)

The decimal expansion of an irrational number is non-terminating non-recurring.

Hence, 3.141141114….. is an irrational number. Question 12

A rational number equivalent to  is

(a) 

(b) 

(c) 

(d) Solution 12

Correct option: (d)

Question 13

Choose the rational number which does not lie between 

(a) 

(b) 

(c) 

(d) Solution 13

Correct option: (b)

Given two rational numbers are negative and  is a positive rational number.

So, it does not lie between  Question 14

Π is

(a) a rational number

(b) an integer

(c) an irrational number

(d) a whole numberSolution 14

Correct option: (c)

Π = 3.14159265359…….., which is non-terminating non-recurring.

Hence, it is an irrational number.Question 15

The decimal expansion of  is

(a) finite decimal

(b) 1.4121

(c) nonterminating recurring

(d) nonterminating, nonrecurringSolution 15

Correct option: (d)

The decimal expansion of , which is non-terminating, non-recurring. Question 16

Which of the following is an irrational number?

(a) 

(b) 

(c) 0.3799

(d) Solution 16

Correct option: (a)

The decimal expansion of , which is non-terminating, non-recurring.

Hence, it is an irrational number. Question 17

Hoe many digits are there in the repeating block of digits in the decimal expansion of 

(a) 16

(b) 6

(c) 26

(d) 7Solution 17

Correct option: (b)

Question 18

Which of the following numbers is irrational?

(a) 

(b) 

(c) 

(d) Solution 18

Correct option: (c)

The decimal expansion of , which is non-terminating, non-recurring.

Hence, it is an irrational number. Question 19

The product of two irrational numbers is

(a) always irrational

(b) always rational

(c) always an integer

(d)sometimes rational and sometimes irrationalSolution 19

Question 20

Which of the following is a true statement?

(a) The sum of two irrational numbers is an irrational number

(b) The product of two irrational numbers is an irrational number

(c) Every real number is always rational

(d) Every real number is either rational or irrationalSolution 20

Question 21

Which of the following is a true statement?

(a) 

(b) 

(c) 

(d) Solution 21

Question 22

A rational number lying between  is

(a) 

(b) 

(c) 1.6

(d) 1.9Solution 22

Correct option: (c)

Question 23

Which of the following is a rational number?

(a) 

(b) 0.101001000100001…

(c) π 

(d) 0.853853853…Solution 23

Correct option: (d)

The decimal expansion of a rational number is either terminating or non-terminating recurring.

Hence, 0.853853853… is a rational number. Question 24

The product of a nonzero rational number with an irrational number is always a/an

(a) irrational number

(b) rational number

(c) whole number

(d) natural numberSolution 24

Correct option: (a)

The product of a non-zero rational number with an irrational number is always an irrational number. Question 25

The value of , where p and q are integers and q ≠ 0, is

(a) 

(b) 

(c) 

(d) Solution 25

Correct option: (b)

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

An irrational number between 5 and 6 is

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

The sum of 

(a) 

(b) 

(c) 

(d) Solution 33

Correct option: (b)

Let x = 

i.e. x = 0.3333…. ….(i)

⇒ 10x = 3.3333…. ….(ii)

On subtracting (i) from (ii), we get

9x = 3

Let y = 

i.e. y = 0.4444…. ….(i)

⇒ 10y = 4.4444…. ….(ii)

On subtracting (i) from (ii), we get

9y = 4

Question 34

The value of 

(a) 

(b) 

(c) 

(d) Solution 34

Correct option: (c)

Let x = 

i.e. x = 2.4545….  ….(i)

⇒ 100x = 245.4545…….  ….(ii)

On subtracting (i) from (ii), we get

99x = 243

Let y =  

i.e. y = 0.3636….  ….(iii)

⇒ 100y = 36.3636….  ….(iv)

On subtracting (iii) from (iv), we get

99y = 36

Question 35

Which of the following is the value of ?

(a) -4

(b) 4

(c) 

(d) Solution 35

Correct option: (b)

Question 36

 when simplified is

(a) positive and irrational

(b) positive and rational

(c) negative and irrational

(d) negative and rationalSolution 36

Correct option: (b)

Which is positive and rational number. Question 37

 when simplified is

(a) positive and irrational

(b) positive and rational

(c) negative and irrational

(d) negative and rationalSolution 37

Correct option: (b)

Which is positive and rational number. Question 38

When  is divided by , the quotient is

(a) 

(b) 

(c) 

(d) Solution 38

Correct option: (c)

Question 39

The value of  is

(a) 10

(b) 

(c) 

(d) Solution 39

Correct option: (a)

Question 40

The value of  is

(a) 

(b) 

(c) 

(d) Solution 40

Correct option: (b)

Question 41

= ?

(a) 

(b) 

(c) 

(d) None of theseSolution 41

Correct option: (b)

Question 42

=?

(a) 

(b) 2

(c) 4

(d) 8Solution 42

Correct option: (b)

Question 43

(125)^-1/3 = ?

(a) 5

(b) -5

(c) 

(d) Solution 43

Correct option: (c)

Question 44

The value of 7^1/2⋅ 8^1/2 is

(a) (28)^1/2

(b) (56)^1/2

(c) (14)^1/2

(d) (42)^1/2Solution 44

Correct option: (b)

Question 45

After simplification,  is

(a) 13^2/15

(b) 13^8/15

(c) 13^1/3

(d) 13^-2/15Solution 45

Correct option: (d)

Question 46

The value of is

(a) 

(b) 

(c) 8

(d) Solution 46

Correct option: (a)

Question 47

The value of is

(a) 0

(b) 2

(c) 

(d) Solution 47

Correct option: (b)

Question 48

The value of (243)^1/5 is

(a) 3

(b) -3

(c) 5

(d) Solution 48

Correct option: (a)

Question 49

9³ + (-3)³ – 6³ = ?

(a) 432

(b) 270

(c) 486

(d) 540Solution 49

Correct option: (c)

9³ + (-3)³ – 6³ = 729 – 27 – 216 = 486 Question 50

Simplified value of  is

(a) 0

(b) 1

(c) 4

(d) 16Solution 50

Correct option: (b)

Question 51

The value of is

(a) 2^-1/6

(b) 2^-6

(c) 2^1/6

(d) 2⁶Solution 51

Correct option: (c)

Question 52

Simplified value of (25)^1/3× 5^1/3 is

(a) 25

(b) 3

(c) 1

(d) 5Solution 52

Correct option: (d)

Question 53

The value of is

(a) 3

(b) -3

(c) 9

(d) Solution 53

Correct option: (a)

Question 54

There is a number x such that x² is irrational but x⁴ is rational. Then, x can be

(a) 

(b) 

(c) 

(d) Solution 54

Correct option: (d)

Question 55

If  then value of p is

(a) 

(b) 

(c) 

(d) Solution 55

Correct option: (b)

Question 56

The value of is

(a) 

(b) 

(c) 

(d) Solution 56

Correct option: (b)

Question 57

The value of x^p-q⋅ x^{q – r}⋅ x^{r – p} is equal to

(a) 0

(b) 1

(c) x

(d) x^pqrSolution 57

Correct option: (b)

x^p-q⋅ x^{q – r}⋅ x^{r – p}

= x^{p – q + q – r + r – p}

= x⁰

= 1 Question 58

The value of  is

(a) -1

(b) 0

(c) 1

(d) 2Solution 58

Correct option: (c)

Question 59

= ?

(a) 2

(b) 

(c) 

(d) Solution 59

Correct option: (a)

Question 60

If  then x = ?

(a) 1

(b) 2

(c) 3

(d) 4Solution 60

Correct option: (d)

Question 61

If (3³)² = 9^x then 5^x = ?

(a) 1

(b) 5

(c) 25

(d) 125Solution 61

Correct option: (d)

(3³)² = 9^x

⇒ (3²)³ = (3²)^x

⇒ x = 3

Then 5^x = 5³ = 125 Question 62

On simplification, the expression  equals

(a) 

(b) 

(c) 

(d) Solution 62

Correct option: (b)

Question 63

The simplest rationalisation factor of  is

(a) 

(b) 

(c) 

(d) Solution 63

Correct option: (d)

Thus, the simplest rationalisation factr of   Question 64

The simplest rationalisation factor of  is

(a) 

(b) 

(c) 

(d) Solution 64

Correct option: (b)

The simplest rationalisation factor of  is  Question 65

The rationalisation factor of  is

(a) 

(b) 

(c) 

(d) Solution 65

Correct option: (d)

Question 66

Rationalisation of the denominator of  gives

(a) 

(b) 

(c) 

(d) Solution 66

Correct option: (d)

Question 67

(a) 

(b) 2

(c) 4

(d) Solution 67

Correct option: (c)

Question 68

(a) 

(b) 

(c) 

(d) None of theseSolution 68

Correct option: (c)

Question 69

(a) 

(b) 14

(c) 49

(d) 48Solution 69

Correct option: (b)

Question 70

(a) 0.075

(b) 0.75

(c) 0.705

(d) 7.05Solution 70

Correct option: (c)

Question 71

(a) 0.375

(b) 0.378

(c) 0.441

(d) None of theseSolution 71

Correct option: (b)

Question 72

The value of  is

(a) 

(b) 

(c) 

(d) Solution 72

Correct option: (d)

Question 73

The value of  is

(a) 

(b) 

(c) 

(d) Solution 73

Correct option: (c)

Question 74

(a) 0.207

(b) 2.414

(c) 0.414

(d) 0.621Solution 74

Correct option: (c)

Question 75

= ?

(a) 34

(b) 56

(c) 28

(d) 63Solution 75

Correct option: (a)

Question 76

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
	A rational number between two rational numbers p and q is .

The correct answer is: (a)/(b)/(c)/(d).Solution 76

Question 77

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
	Square root of a positive integer which is not a perfect square is an irrational number.

The correct answer is: (a)/(b)/(c)/(d).Solution 77

Question 78

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
e is an irrational number.	Π is an irrational number.

The correct answer is: (a)/(b)/(c)/(d).Solution 78

Question 79

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
	The sum of a rational number and an irrational number is an irrational number.

The correct answer is: (a)/(b)/(c)/(d).Solution 79

Question 80

Match the following columns:

Column I	Column II
	(p) 14(q) 6(r) a rational number(s) an irrational number

The correct answer is:

(a)-…….,

(b)-…….,

(c)-…….,

(d)-…….,Solution 80

Question 81

Match the following columns:

Column I	Column II

The correct answer is:

(a)-…….,

(b)-…….,

(c)-…….,

(d)-…….,Solution 81

Exercise Ex. 1B

Question 1(i)

Without actual division, find which of the following rationals are terminating decimals.

Solution 1(i)

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

_{Since, 80 has prime factors 2 and 5,}  is a terminating decimal.Question 1(ii)

Without actual division, find which of the following rationals are terminating decimals.

Solution 1(ii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since, 24 has prime factors 2 and 3 and 3 is different from 2 and 5,

 is not a terminating decimal.Question 1(iii)

Without actual division, find which of the following rationals are terminating decimals.

Solution 1(iii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since 12 has prime factors 2 and 3  and 3 is different from 2 and 5,

 is not a terminating decimal.Question 1(iv)

Without actual division, find which of the following rational numbers are terminating decimals.

Solution 1(iv)

Since the denominator of a given rational number is not of the form 2^m × 2ⁿ, where m and n are whole numbers, it has non-terminating decimal. Question 2(i)

Write each of the following in decimal form and say what kind of decimal expansion each has.

Solution 2(i)

Hence, it has terminating decimal expansion. Question 2(ii)

Write each of the following in decimal form and say what kind of decimal expansion each has.

Solution 2(ii)

Hence, it has terminating decimal expansion. Question 2(iii)

Write each of the following in decimal form and say what kind of decimal expansion each has.

Solution 2(iii)

Hence, it has non-terminating recurring decimal expansion. Question 2(iv)

Write each of the following in decimal form and say what kind of decimal expansion each has.

Solution 2(iv)

Hence, it has non-terminating recurring decimal expansion. Question 2(v)

Write each of the following in decimal form and say what kind of decimal expansion each has.

Solution 2(v)

Hence, it has non-terminating recurring decimal expansion. Question 2(vi)

Write each of the following in decimal form and say what kind of decimal expansion each has.

Solution 2(vi)

Hence, it has terminating decimal expansion. Question 2(vii)

Write each of the following in decimal form and say what kind of decimal expansion each has.

Solution 2(vii)

Hence, it has terminating decimal expansion. Question 2(viii)

Write each of the following in decimal form and say what kind of decimal expansion each has.

Solution 2(viii)

Hence, it has non-terminating recurring decimal expansion. Question 3(i)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(i)

Let x = 

i.e. x = 0.2222…. ….(i)

⇒ 10x = 2.2222…. ….(ii)

On subtracting (i) from (ii), we get

9x = 2

Question 3(ii)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(ii)

Let x = 

i.e. x = 0.5353….  ….(i)

⇒ 100x = 53.535353…. ….(ii)

On subtracting (i) from (ii), we get

99x = 53

Question 3(iii)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(iii)

Let x = 

i.e. x = 2.9393….  ….(i)

⇒ 100x = 293.939……. ….(ii)

On subtracting (i) from (ii), we get

99x = 291

Question 3(iv)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(iv)

Let x = 

i.e. x = 18.4848….  ….(i)

⇒ 100x = 1848.4848……. ….(ii)

On subtracting (i) from (ii), we get

99x = 1830

Question 3(v)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(v)

Let x = 

i.e. x = 0.235235..…   ….(i)

⇒ 1000x = 235.235235……. ….(ii)

On subtracting (i) from (ii), we get

999x = 235

Question 3(vi)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(vi)

Let x = 

i.e. x = 0.003232..…   

⇒ 100x = 0.323232……. ….(i)

⇒ 10000x = 32.3232…. ….(ii)

On subtracting (i) from (ii), we get

9900x = 32

Question 3(vii)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(vii)

Let x = 

i.e. x = 1.3232323..… ….(i)   

⇒ 100x = 132.323232……. ….(ii)

On subtracting (i) from (ii), we get

99x = 131

Question 3(viii)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(viii)

Let x = 

i.e. x = 0.3178178..… 

⇒ 10x = 3.178178…… ….(i)   

⇒ 10000x = 3178.178……. ….(ii)

On subtracting (i) from (ii), we get

9990x = 3175

Question 3(ix)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(ix)

Let x = 

i.e. x = 32.123535..… 

⇒ 100x = 3212.3535…… ….(i)   

⇒ 10000x = 321235.3535……. ….(ii)

On subtracting (i) from (ii), we get

9900x = 318023

Question 3(x)

Express each of the following decimals in the form , where p, q are integers and q ≠ 0.

Solution 3(x)

Let x = 

i.e. x = 0.40777..… 

⇒ 100x = 40.777…… ….(i)   

⇒ 1000x = 407.777……. ….(ii)

On subtracting (i) from (ii), we get

900x = 367

Question 4

Express as a fraction in simplest form.Solution 4

Let x = 

i.e. x = 2.3636….  ….(i)

⇒ 100x = 236.3636……. ….(ii)

On subtracting (i) from (ii), we get

99x = 234

Let y =  

i.e. y = 0.2323….  ….(iii)

⇒ 100y = 23.2323…. ….(iv)

On subtracting (iii) from (iv), we get

99y = 23

Question 5

Express in the form of  Solution 5

Let x =  

i.e. x = 0.3838….  ….(i)

⇒ 100x = 38.3838….  ….(ii)

On subtracting (i) from (ii), we get

99x = 38

Let y = 

i.e. y = 1.2727….  ….(iii)

⇒ 100y = 127.2727…….  ….(iv)

On subtracting (iii) from (iv), we get

99y = 126

Question 9(v)

Without actual division, find which of the following rationals are terminating decimals.

Solution 9(v)

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

Since 125 has prime factor 5 only

   is a terminating decimal.

Exercise Ex. 1C

Question 1

What are irrational numbers? How do they differ from rational numbers? Give examples.Solution 1

Irrational number: A number which cannot be expressed either as a terminating decimal or a repeating decimal is known as irrational number. Rather irrational numbers cannot be expressed in the fraction form, 

For example, 0.101001000100001 is neither a terminating nor a repeating decimal and so is an irrational number.

Also, etc. are examples of irrational numbers.Question 2(iii)

Classify the following numbers as rational or irrational. Give reasons to support you answer.

Solution 2(iii)

We know that, if n is a not a perfect square, then  is an irrational number.

Here,  is a not a perfect square number.

So, is irrational.Question 2(v)

Classify the following numbers as rational or irrational. Give reasons to support you answer.

Solution 2(v)

 is the product of a rational number  and an irrational number .

Theorem: The product of a non-zero rational number and an irrational number _{is an irrational number}.

Thus, by the above theorem,  is an irrational number.

So,  is an irrational number.Question 2(i)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

Solution 2(i)

Since quotient of a rational and an irrational is irrational, the given number is irrational.  Question 2(ii)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

Solution 2(ii)

Question 2(iv)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

Solution 2(iv)

Question 2(vi)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

4.1276Solution 2(vi)

The given number 4.1276 has terminating decimal expansion.

Hence, it is a rational number. Question 2(vii)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

Solution 2(vii)

Since the given number has non-terminating recurring decimal expansion, it is a rational number. Question 2(viii)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

1.232332333….Solution 2(viii)

The given number 1.232332333…. has non-terminating and non-recurring decimal expansion.

Hence, it is an irrational number. Question 2(ix)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

3.040040004…..Solution 2(ix)

The given number 3.040040004….. has non-terminating and non-recurring decimal expansion.

Hence, it is an irrational number. Question 2(x)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

2.356565656…..Solution 2(x)

The given number 2.356565656….. has non-terminating recurring decimal expansion.

Hence, it is a rational number. Question 2(xi)

Classify the following numbers as rational or irrational. Give reasons to support your answer.

6.834834….Solution 2(xi)

The given number 6.834834…. has non-terminating recurring decimal expansion.

Hence, it is a rational number. Question 3

Let x be a rational number and y be an irrational number. Is x + y necessarily an irrational number? Give an example in support of your answer.Solution 3

We know that the sum of a rational and an irrational is irrational.

Hence, if x is rational and y is irrational, then x + y is necessarily an irrational number.

For example,

Question 4

Let a be a rational number and b be an irrational number. Is ab necessarily an irrational number? Justify your answer with an example.Solution 4

We know that the product of a rational and an irrational is irrational.

Hence, if a is rational and b is irrational, then ab is necessarily an irrational number.

For example,

Question 5

Is the product of two irrationals always irrational? Justify your answer.Solution 5

No, the product of two irrationals need not be an irrational.

For example,

Question 6

Give an example of two irrational numbers whose

(i) difference is an irrational number.

(ii) difference is a rational number.

(iii) sum is an irrational number.

(iv) sum is an rational number.

(v) product is an irrational number.

(vi) product is a rational number.

(vii) quotient is an irrational number.

(viii) quotient is a rational number. Solution 6

(i) Difference is an irrational number:

(ii) Difference is a rational number:

(iii) Sum is an irrational number:

(iv) Sum is an rational number:

(v) Product is an irrational number:

(vi) Product is a rational number:

(vii) Quotient is an irrational number:

(viii) Quotient is a rational number:

Question 7

Examine whether the following numbers are rational or irrational.

Solution 7

Question 8

Insert a rational and an irrational number between 2 and 2.5Solution 8

Rational number between 2 and 2.5 = 

Irrational number between 2 and 2.5 = Question 9

How many irrational numbers lie between? Find any three irrational numbers lying between .Solution 9

There are infinite irrational numbers between.

We have

Hence, three irrational numbers lying between  are as follows:

1.5010010001……., 1.6010010001…… and 1.7010010001……. Question 10

Find two rational and two irrational numbers between 0.5 and 0.55.Solution 10

Since 0.5 < 0.55

Let x = 0.5, y = 0.55 and y = 2

Two irrational numbers between 0.5 and 0.55 are 0.5151151115……. and 0.5353553555…. Question 11

Find three different irrational numbers between the rational numbers .Solution 11

Thus, three different irrational numbers between the rational numbers  are as follows:

0.727227222….., 0.757557555….. and 0.808008000….. Question 12

Find two rational numbers of the form between the numbers 0.2121121112… and  0.2020020002……Solution 12

Let a and b be two rational numbers between the numbers 0.2121121112… and 0.2020020002……

Now, 0.2020020002…… <0.2121121112…

Then, 0.2020020002…… < a < b < 0.2121121112…

Question 13

Find two irrational numbers between 0.16 and 0.17.Solution 13

Two irrational numbers between 0.16 and 0.17 are as follows:

0.1611161111611111611111…… and 0.169669666……. Question 14(i)

State in each case, whether the given statement is true or false.

The sum of two rational numbers is rational.Solution 14(i)

TrueQuestion 14(ii)

State in each case, whether the given statement is true or false.

The sum of two irrational numbers is irrational.Solution 14(ii)

FalseQuestion 14(iii)

State in each case, whether the given statement is true or false.

The product of two rational numbers is rational.Solution 14(iii)

TrueQuestion 14(iv)

State in each case, whether the given statement is true or false.

The product of two irrational numbers is irrational.Solution 14(iv)

FalseQuestion 14(v)

State in each case, whether the given statement is true or false.

The sum of a rational number and an irrational number is irrational.Solution 14(v)

TrueQuestion 14(vi)

State in each case, whether the given statement is true or false.

The product of a nonzero rational number and an irrational number is a rational number.Solution 14(vi)

FalseQuestion 14(vii)

State in each case, whether the given statement is true or false.

Every real number is rational.Solution 14(vii)

FalseQuestion 14(viii)

State in each case, whether the given statement is true or false.

Every real number is either rational or irrational.Solution 14(viii)

TrueQuestion 14(ix)

State in each case, whether the given statement is true or false.

is irrational and is rational.Solution 14(ix)

True

Exercise Ex. 1D

Question 1(i)

Add:

Solution 1(i)

We have:

Question 1(ii)

Add:

Solution 1(ii)

We have:

Question 1(iii)

Add:

Solution 1(iii)

Question 2(i)

Multiply:

Solution 2(i)

Question 2(ii)

Multiply:

Solution 2(ii)

Question 2(iii)

Multiply:

Solution 2(iii)

Question 2(iv)

Multiply:

Solution 2(iv)

Question 2(v)

Multiply:

Solution 2(v)

Question 2(vi)

Multiply:

Solution 2(vi)

Question 3(i)

Divide:

Solution 3(i)

Question 3(ii)

Divide:

Solution 3(ii)

Question 3(iii)

Divide:

Solution 3(iii)

Question 4(iii)

Simplify:

Solution 4(iii)

Question 4(iv)

Simplify:

Solution 4(iv)

Question 4(vi)

Simplify:

Solution 4(vi)

Question 4(i)

Simplify

Solution 4(i)

= 9 – 11

= -2  Question 4(ii)

Simplify

Solution 4(ii)

= 9 – 5

= 4 Question 4(v)

Simplify

Solution 4(v)

Question 5

Simplify

Solution 5

Question 6(i)

Examine whether the following numbers are rational or irrational:

Solution 6(i)

Thus, the given number is rational. Question 6(ii)

Examine whether the following numbers are rational or irrational:

Solution 6(ii)

Clearly, the given number is irrational. Question 6(iii)

Examine whether the following numbers are rational or irrational:

Solution 6(iii)

Thus, the given number is rational. Question 6(iv)

Examine whether the following numbers are rational or irrational:

Solution 6(iv)

Thus, the given number is irrational. Question 7

On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by .

(i) Find the number of chocolates distributed by her.

(ii) Write the moral values depicted here by Reema.Solution 7

(i) Number of chocolates distributed by Reema

(ii) Loving, helping and caring attitude towards poor and needy children.Question 8(i)

Simplify

Solution 8(i)

Question 8(ii)

Simplify

Solution 8(ii)

Question 8(iii)

Simplify

Solution 8(iii)

Exercise Ex. 1G

Question 1(iii)

Simplify:

Solution 1(iii)

Question 1(i)

Simplify

Solution 1(i)

Question 1(ii)

Simplify

Solution 1(ii)

Question 1(iv)

Simplify

Solution 1(iv)

Question 2(i)

Simplify:

Solution 2(i)

Question 2(ii)

Simplify:

Solution 2(ii)

Question 2(iii)

Simplify:

Solution 2(iii)

Question 3(i)

Simplify:

Solution 3(i)

Question 3(ii)

Simplify:

Solution 3(ii)

Question 3(iii)

Simplify:

Solution 3(iii)

Question 4(i)

Simplify:

Solution 4(i)

Question 4(ii)

Simplify:

Solution 4(ii)

Question 4(iii)

Simplify:

Solution 4(iii)

Question 5(i)

Evaluate:

Solution 5(i)

Question 5(ii)

Evaluate:

Solution 5(ii)

Question 5(iii)

Evaluate:

Solution 5(iii)

Question 5(iv)

Evaluate:

Solution 5(iv)

Question 5(v)

Evaluate:

Solution 5(v)

Question 5(vi)

Evaluate:

Solution 5(vi)

Question 6(i)

If a = 2, b = 3, find the value of (a^b + b^a)^-1Solution 6(i)

Given, a = 2 and b = 3

Question 6(ii)

If a = 2, b = 3, find the value of (a^a + b^b)^-1Solution 6(ii)

Given, a = 2 and b = 3

Question 7(i)

Simplify

Solution 7(i)

Question 7(ii)

Simplify

(14641)^0.25Solution 7(ii)

(14641)^0.25

Question 7(iii)

Simplify

Solution 7(iii)

Question 7(iv)

Simplify

Solution 7(iv)

Question 8(i)

Evaluate

Solution 8(i)

Question 8(ii)

Evaluate

Solution 8(ii)

Question 8(iii)

Evaluate

Solution 8(iii)

Question 8(iv)

Evaluate

Solution 8(iv)

Question 9(i)

Evaluate

Solution 9(i)

Question 9(ii)

Evaluate

Solution 9(ii)

Question 9(iii)

Evaluate

Solution 9(iii)

Question 9(iv)

Evaluate

Solution 9(iv)

Question 10(i)

Prove that

Solution 10(i)

Question 10(ii)

Prove that

Solution 10(ii)

Question 10(iii)

Prove that

Solution 10(iii)

Question 11

Simplify and express the result in the exponential form of x.Solution 11

Question 12

Simplify the product Solution 12

Question 13(i)

Simplify

Solution 13(i)

Question 13(ii)

Simplify

Solution 13(ii)

Question 13(iii)

Simplify

Solution 13(iii)

Question 14(i)

Find the value of x in each of the following.

Solution 14(i)

Question 14(ii)

Find the value of x in each of the following.

Solution 14(ii)

Question 14(iii)

Find the value of x in each of the following.

Solution 14(iii)

Question 14(iv)

Find the value of x in each of the following.

5^{x – 3}× 3^{2x – 8} = 225Solution 14(iv)

5^{x – 3} × 3^{2x – 8} = 225

⇒ 5^{x – 3}× 3^{2x – 8} = 5² × 3²

⇒ x – 3 = 2 and 2x – 8 = 2

⇒ x = 5 and 2x = 10

⇒ x = 5 Question 14(v)

Find the value of x in each of the following.

Solution 14(v)

Question 15(i)

Prove that

Solution 15(i)

Question 15(ii)

Prove that

Solution 15(ii)

Question 15(iii)

Prove that

Solution 15(iii)

Question 15(iv)

Prove that

Solution 15(iv)

Question 16

If x is a positive real number and exponents are rational numbers, simplify

Solution 16

Question 17

If prove that m – n = 1.Solution 17

Question 18

Write the following in ascending order of magnitude.

Solution 18

Exercise Ex. 1A

Question 1

Is zero a rational number? Justify.Solution 1

A number which can be expressed as , where ‘a’ and ‘b’ both are integers and b ≠ 0, is called a rational number.

Since, 0 can be expressed as , it is a rational number.Question 2(i)

Represent each of the following rational numbers on the number line:

(i) Solution 2(i)

(i) 

Question 2(ii)

Represent each of the following rational numbers on the number line:

(ii) Solution 2(ii)

(ii) 

Question 2(iii)

Represent each of the following rational numbers on the number line:

Solution 2(iii)

Question 2(iv)

Represent each of the following rational numbers on the number line:

(iv) 1.3Solution 2(iv)

(iv) 1.3

Question 2(v)

Represent each of the following rational numbers on the number line:

(v) -2.4Solution 2(v)

(v) -2.4

Question 3(i)

Find a rational number lying between

Solution 3(i)

Question 3(ii)

Find a rational number lying between

1.3 and 1.4Solution 3(ii)

Question 3(iii)

Find a rational number lying between

-1 and Solution 3(iii)

Question 3(iv)

Find a rational number lying between

Solution 3(iv)

Question 3(v)

Find a rational number between

Solution 3(v)

Question 4

Find three rational numbers lying between 

How many rational numbers can be determined between these two numbers?Solution 4

Infinite rational numbers can be determined between given two rational numbers.Question 5

Find four rational numbers between Solution 5

We have

We know that 9 < 10 < 11 < 12 < 13 < 14 < 15

Question 6

Find six rational numbers between 2 and 3.Solution 6

2 and 3 can be represented asrespectively.

Now six rational numbers between 2 and 3 are 

. Question 7

Find five rational numbers between Solution 7

Question 8

Insert 16 rational numbers between 2.1 and 2.2.Solution 8

Let x = 2.1 and y = 2.2

Then, x < y because 2.1 < 2.2

Or we can say that, 

Or, 

That is, we have, 

We know that,

Therefore, we can have,

Therefore, 16 rational numbers between, 2.1 and 2.2 are:

So, 16 rational numbers between 2.1 and 2.2 are:

2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175, 2.18

Question 9(i)

State whether the given statement is true or false. Give reasons. for your answer.

Every natural number is a whole number.Solution 9(i)

True. Since the collection of natural number is a sub collection of whole numbers, and every element of natural numbers is an element of whole numbersQuestion 9(ii)

Write, whether the given statement is true or false. Give reasons.

Every whole number is a natural number.Solution 9(ii)

False. Since 0 is whole number but it is not a natural number.Question 9(iii)

State whether the following statements are true or false. Give reasons for your answer.

Every integer is a whole number.Solution 9(iii)

False, integers include negative of natural numbers as well, which are clearly not whole numbers. For example -1 is an integer but not a whole number.Question 9(iv)

Write, whether the given statement is true or false. Give reasons.

Ever integer is a rational number.Solution 9(iv)

True. Every integer can be represented in a fraction form with denominator 1.Question 9(v)

State whether the following statements are true or false. Give reasons for your answer.

Every rational number is an integer.Solution 9(v)

False, integers are counting numbers on both sides of the number line i.e. they are both positive and negative while rational numbers are of the form . Hence, Every rational number is not an integer but every integer is a rational number.Question 9(vi)

Write, whether the given statement is true or false. Give reasons.

Every rational number is a whole number.Solution 9(vi)

False. Since division of whole numbers is not closed under division, the value of , may not be a whole number.

Exercise Ex. 1E

Question 1

Represent on the number line.Solution 1

Draw a number line as shown.

On the number line, take point O corresponding to zero.

Now take point A on number line such that OA = 2 units.

Draw perpendicular AZ at A on the number line and cut-off arc AB = 1 unit.

By Pythagoras Theorem,

OB² = OA² + AB² = 2² + 1² = 4 + 1 = 5

⇒ OB = 

Taking O as centre and OB =  as radius draw an arc cutting real line at C.

Clearly, OC = OB = 

Hence, C represents  on the number line.Question 2

Locate on the number line. Solution 2

Draw a number line as shown.

On the number line, take point O corresponding to zero.

Now take point A on number line such that OA = 1 unit.

Draw perpendicular AZ at A on the number line and cut-off arc AB = 1 unit.

By Pythagoras Theorem,

OB² = OA² + AB² = 1² + 1² = 1 + 1 = 2

⇒ OB = 

Taking O as centre and OB =  as radius draw an arc cutting real line at C.

Clearly, OC = OB = 

Thus, C represents  on the number line.

Now, draw perpendicular CY at C on the number line and cut-off arc CE = 1 unit.

By Pythagoras Theorem,

OE² = OC² + CE² = ² + 1² = 2 + 1 = 3

⇒ OE = 

Taking O as centre and OE =  as radius draw an arc cutting real line at D.

Clearly, OD = OE = 

Hence, D represents  on the number line. Question 3

Locate on the number line.Solution 3

Draw a number line as shown.

On the number line, take point O corresponding to zero.

Now take point A on number line such that OA = 3 units.

Draw perpendicular AZ at A on the number line and cut-off arc AB = 1 unit.

By Pythagoras Theorem,

OB² = OA² + AB² = 3² + 1² = 9 + 1 = 10

⇒ OB = 

Taking O as centre and OB =  as radius draw an arc cutting real line at C.

Clearly, OC = OB = 

Hence, C represents  on the number line. Question 4

Locate on the number line. Solution 4

Draw a number line as shown.

On the number line, take point O corresponding to zero.

Now take point A on number line such that OA = 2 units.

Draw perpendicular AZ at A on the number line and cut-off arc AB = 2 units.

By Pythagoras Theorem,

OB² = OA² + AB² = 2² + 2² = 4 + 4 = 8

⇒ OB = 

Taking O as centre and OB =  as radius draw an arc cutting real line at C.

Clearly, OC = OB = 

Hence, C represents  on the number line. Question 5

Represent  geometrically on the number line.Solution 5

Draw a line segment AB = 4.7 units and extend it to C such that BC = 1 unit.

Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle.

Now, draw BD ⊥ AC, intersecting the semicircle at D.

Then, BD =  units.

With B as centre and BD as radius, draw an arc, meeting AC produced at E.

Then, BE = BD =  units. Question 6

Represent on the number line.Solution 6

Draw a line segment OB = 10.5 units and extend it to C such that BC = 1 unit.

Find the midpoint D of OC.

With D as centre and DO as radius, draw a semicircle.

Now, draw BE ⊥ AC, intersecting the semicircle at E.

Then, BE =  units.

With B as centre and BE as radius, draw an arc, meeting AC produced at F.

Then, BF = BE =  units.Question 7

Represent geometrically on the number line.Solution 7

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit.

Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle.

Now, draw BD AC, intersecting the semicircle at D.

Then, BD = units.

With D as centre and BD as radius, draw an arc, meeting AC produced at E.

Then, BE = BD = units.Question 8

Represent  on the number line.Solution 8

Draw a line segment OB = 9.5 units and extend it to C such that BC = 1 unit.

Find the midpoint D of OC.

With D as centre and DO as radius, draw a semicircle.

Now, draw BE ⊥ AC, intersecting the semicircle at E.

Then, BE =  units.

With B as centre and BE as radius, draw an arc, meeting AC produced at F.

Then, BF = BE =  units.

Extend BF to G such that FG = 1 unit.

Then, BG =  

Question 9

Visualize the representation of 3.765 on the number line using successive magnification.Solution 9

Question 10

Visualize the representation of  on the number line up to 4 decimal places.Solution 10

Exercise Ex. 1F

Question 1

Write the rationalising factor of the denominator in . Solution 1

The rationalising factor of the denominator in  is  Question 2(i)

Rationalise the denominator of following:

Solution 2(i)

On multiplying the numerator and denominator of the given number by , we get

Question 2(ii)

Rationalise the denominator of following:

Solution 2(ii)

On multiplying the numerator and denominator of the given number by , we get

Question 2(iii)

Rationalise the denominator of following:

Solution 2(iii)

Question 2(iv)

Rationalise the denominator of following:

Solution 2(iv)

Question 2(v)

Rationalise the denominator of following:

Solution 2(v)

Question 2(vi)

Rationalise the denominator of each of the following.

Solution 2(vi)

Question 2(vii)

Rationalise the denominator of each of the following.

Solution 2(vii)

Question 2(viii)

Rationalise the denominator of each of the following.

Solution 2(viii)

Question 2(ix)

Rationalise the denominator of each of the following.

Solution 2(ix)

Question 3(i)

find the value to three places of decimals, of each of the following.

Solution 3(i)

Question 3(ii)

find the value to three places of decimals, of each of the following.

Solution 3(ii)

Question 3(iii)

find the value to three places of decimals, of each of the following.

Solution 3(iii)

Question 4(i)

Find rational numbers a and b such that

Solution 4(i)

Question 4(ii)

Find rational numbers a and b such that

Solution 4(ii)

Question 4(iii)

Find rational numbers a and b such that

Solution 4(iii)

Question 4(iv)

Find rational numbers a and b such that

Solution 4(iv)

Question 5(i)

find to three places of decimals, the value of each of the following.

Solution 5(i)

Question 5(ii)

find to three places of decimals, the value of each of the following.

Solution 5(ii)

Question 5(iii)

find to three places of decimals, the value of each of the following.

Solution 5(iii)

Question 5(iv)

find to three places of decimals, the value of each of the following.

Solution 5(iv)

Question 5(v)

find to three places of decimals, the value of each of the following.

Solution 5(v)

Question 5(vi)

find to three places of decimals, the value of each of the following.

Solution 5(vi)

Question 6(i)

Simplify by rationalising the denominator.

Solution 6(i)

Question 6(ii)

Simplify by rationalising the denominator.

Solution 6(ii)

Question 7(i)

Simplify: Solution 7(i)

Question 7(ii)

Simplify

Solution 7(ii)

Question 7(iii)

Simplify

Solution 7(iii)

Question 7(iv)

Simplify

Solution 7(iv)

Question 8(i)

Prove that

Solution 8(i)

Question 8(ii)

Prove that

Solution 8(ii)

Question 9

Find the values of a and b if

Solution 9

*Back answer incorrect Question 10

Simplify

Solution 10

Question 11

Solution 11

Thus, the given number is rational. Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

*Question modified Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

.Solution 21

Question 22(i)

Rationalise the denominator of each of the following.

Solution 22(i)

Question 22(ii)

Rationalise the denominator of each of the following.

Solution 22(ii)

Question 22(iii)

Rationalise the denominator of each of the following.

Solution 22(iii)

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25