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Class 11th Chapter – 5 Complex Numbers and Quadratic Equations | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter -5 Complex Numbers and Quadratic Equations NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Complex Numbers and Quadratic Equations solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Complex Numbers and Quadratic Equations NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 5 Complex Numbers and Quadratic Equations| NCERT MATHS SOLUTION |

Chapter 5 Complex Numbers and Quadratic Equations EX 5.1 NCERT Solutions

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.
Ex 5.1 Class 11 Maths Question 1.
$\left( 5i \right) \left( -\frac { 3 }{ 5 } i \right)$
Solution.
$\left( 5i \right) \left( -\frac { 3 }{ 5 } i \right)$
= -3i² = -3(-1) [∵ i² = -1]
= 3 = 3 + 0i

Ex 5.1 Class 11 Maths Question 2.
i⁹+ i¹⁹
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 1

Ex 5.1 Class 11 Maths Question 3.
i^-39
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 2

Ex 5.1 Class 11 Maths Question 4.
3(7 + i7) + i(7 + i7)
Solution.
3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i²
= 21 + (21 + 7)i + (-1)7 = 21 – 7 + 28i
= 14 + 28i.

Ex 5.1 Class 11 Maths Question 5.
(1 – i) – (- 1 +i6)
Solution.
(1 – i) – (-1 + i6) = 1 – i + 1 – 6i
= (1 +1) – i(1 + 6)
= 2 – 7i

Ex 5.1 Class 11 Maths Question 6.
$\left( \frac { 1 }{ 5 } +i\frac { 2 }{ 5 } \right) -\left( 4+i\frac { 5 }{ 2 } \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 3

Ex 5.1 Class 11 Maths Question 7.
$\left[ \left( \frac { 1 }{ 3 } +i\frac { 7 }{ 3 } \right) +\left( 4+i\frac { 1 }{ 3 } \right) \right] -\left( -\frac { 4 }{ 3 } +i \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 4

Ex 5.1 Class 11 Maths Question 8.
(1 -i)⁴
Solution.
(1 -i)⁴ = [(1 – i)²]² = [1 – 2i + i²]²
= [1 – 2i + (-1)]²
= (-2i)² = 4i² = 4(-1) = – 4
= – 4 + 0i

Ex 5.1 Class 11 Maths Question 9.
${ \left( \frac { 1 }{ 3 } +3i \right) }^{ 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 5

Ex 5.1 Class 11 Maths Question 10.
${ \left( -2-\frac { 1 }{ 3 } i \right) }^{ 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 6

Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.
Ex 5.1 Class 11 Maths Question 11.
4 – 3i
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 7

Ex 5.1 Class 11 Maths Question 12.
$\sqrt { 5 } +3i$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 8

Ex 5.1 Class 11 Maths Question 13.
-i
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 9

Ex 5.1 Class 11 Maths Question 14.
Express the following expression in the form of a + ib:
$\frac { \left( 3+i\sqrt { 5 } \right) \left( 3-i\sqrt { 5 } \right) }{ \left( \sqrt { 3 } +\sqrt { 2 } i \right) -\left( \sqrt { 3 } -i\sqrt { 2 } \right) }$
Solution.
We have, $\frac { \left( 3+i\sqrt { 5 } \right) \left( 3-i\sqrt { 5 } \right) }{ \left( \sqrt { 3 } +\sqrt { 2 } i \right) -\left( \sqrt { 3 } -i\sqrt { 2 } \right) }$
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 10

We hope the NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1, drop a comment below and we will get back to you at the earliest.

Chapter 5 Complex Numbers and Quadratic Equations EX 5.2 NCERT Solutions

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.
Ex 5.2 Class 11 Maths Question 1.
$z=-1-i\sqrt { 3 }$
Solution.
We have, $z=-1-i\sqrt { 3 }$
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 1

Ex 5.2 Class 11 Maths Question 2.
$z=-\sqrt { 3 } +i$
Solution.
We have, $z=-\sqrt { 3 } +i$
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 2

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
Ex 5.2 Class 11 Maths Question 3.
1 – i
Solution.
We have, z = 1 – i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 3

Ex 5.2 Class 11 Maths Question 4.
-1 + i
Solution.
We have, z = -1 + i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 4

Ex 5.2 Class 11 Maths Question 5.
-1 – i
Solution.
We have, z = -1 – i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 5

Ex 5.2 Class 11 Maths Question 6.
-3
Solution.
We have, z = -3, i.e., z = -3 + 0i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 6

Ex 5.2 Class 11 Maths Question 7.
$\sqrt { 3 } +i$
Solution.
We have, $z=\sqrt { 3 } +i$
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 7

Ex 5.2 Class 11 Maths Question 8.
i
Solution.
We have, z = i, i.e., z = 0 + 1.i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 8

We hope the NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2, drop a comment below and we will get back to you at the earliest.

Chapter 5 Complex Numbers and Quadratic Equations EX 5.3 NCERT Solutions

Solve each of the following equations:
Ex 5.3 Class 11 Maths Question 1.
x² + 3 = 0
Solution.
We have, x² + 3 = 0 ⇒ x² = -3
⇒ $x=\pm \sqrt { -3 }$ ⇒ x = $\pm \sqrt { 3 } i$

Ex 5.3 Class 11 Maths Question 2.
2x² + x + 1 = 0
Solution.
We have, 2x² + x + 1 = 0
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 1

Ex 5.3 Class 11 Maths Question 3.
x² + 3x + 9 = 0
Solution.
We have, x² + 3x + 9 = 0
Comparing the given equation with the general form ax² + bx + c = 0,we get a = 1, b = 3, c = 9
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 2

Ex 5.3 Class 11 Maths Question 4.
-x² + x – 2 = 0
Solution.
We have, -x² + x – 2 = 0
Comparing the given equation with the general form ax² + bx + c = 0,we get
a = 1, b = 1, c = -2
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 3

Ex 5.3 Class 11 Maths Question 5.
x² + 3x + 5 = 0
Solution.
We have, x² + 3x + 5 = 0
Comparing the given equation with the general form ax² + bx + c = 0, we get a = 1, b = 3, c = 5.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 4

Ex 5.3 Class 11 Maths Question 6.
x² – x + 2 = 0
Solution.
We have, x² – x + 2 = 0
Comparing the given equation with the general form ax² + bx + c = 0, we get a = 1, b = -1, c = 2.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 5

Ex 5.3 Class 11 Maths Question 7.
$\sqrt { 2 } { x }^{ 2 }+x+\sqrt { 2 } =0$
Solution.
We have, $\sqrt { 2 } { x }^{ 2 }+x+\sqrt { 2 } =0$
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 6

Ex 5.3 Class 11 Maths Question 8.
$\sqrt { 3 } { x }^{ 2 }+\sqrt { 2 } x+3\sqrt { 3 } =0$
Solution.
We have, $\sqrt { 3 } { x }^{ 2 }+\sqrt { 2 } x+3\sqrt { 3 } =0$
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 7

Ex 5.3 Class 11 Maths Question 9.
${ x }^{ 2 }+x+\frac { 1 }{ \sqrt { 2 } } =0$
Solution.
We have, ${ x }^{ 2 }+x+\frac { 1 }{ \sqrt { 2 } } =0$
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 8

Ex 5.3 Class 11 Maths Question 10.
${ x }^{ 2 }+\frac { x }{ \sqrt { 2 } } +1=0$
Solution.
We have, ${ x }^{ 2 }+\frac { x }{ \sqrt { 2 } } +1=0$
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 9

We hope the NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3, drop a comment below and we will get back to you at the earliest.

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Class 11th Chapter -4 Principle of Mathematical Induction | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter – 4 Principle of Mathemarical Induction NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Principle of Mathemarical Induction solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Principle of Mathemarical Induction NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 4 Principle of Mathemarical Induction| NCERT MATHS SOLUTION |

Chapter 4 Principle of Mathemarical Induction EX 4.1 NCERT Solutions

Prove the following by using the principle of mathematical induction for aline n ∈ N :
Ex 4.1 Class 11 Maths Question 1.
$1+{ 3 }^{ 2 }+{ 3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ 3 }^{ n }=frac { left( { 3 }^{ n }-1 right) }{ 2 }$
Solution.
Let the given statement be P(n) i.e.,
P(n) : $1+{ 3 }^{ 2 }+{ 3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ 3 }^{ n }=frac { left( { 3 }^{ n }-1 right) }{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 1

Ex 4.1 Class 11 Maths Question 2.
${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ n }^{ 3 }={ left( frac { nleft( n+1 right) }{ 2 } right) }^{ 2 }$
Solution.
Let the given statement be P(n) i.e.,
P(n) : ${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ n }^{ 3 }={ left( frac { nleft( n+1 right) }{ 2 } right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 3

Ex 4.1 Class 11 Maths Question 3.
$1+frac { 1 }{ left( 1+2 right) } +frac { 1 }{ left( 1+2+3 right) } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 1+2+3+cdot cdot cdot cdot cdot cdot cdot cdot +n right) } =frac { 2 }{ left( n+1 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1+frac { 1 }{ left( 1+2 right) } +frac { 1 }{ left( 1+2+3 right) } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 1+2+3+.cdot cdot cdot cdot cdot cdot cdot cdot +n right) } =frac { 2 }{ left( n+1 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 5

Ex 4.1 Class 11 Maths Question 4.
$1.2.3+2.3.4+cdot cdot cdot cdot cdot cdot cdot cdot +nleft( n+1 right) left( n+2 right) =frac { nleft( n+1 right) left( n+2 right) left( n+3 right) }{ 4 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.2.3+2.3.4+cdot cdot cdot cdot cdot cdot cdot cdot +nleft( n+1 right) left( n+2 right) =frac { nleft( n+1 right) left( n+2 right) left( n+3 right) }{ 4 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 7

Ex 4.1 Class 11 Maths Question 5.
$1.3+{ 2.3 }^{ 2 }+{ 3.3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ n.3 }^{ n }=frac { left( 2n-1 right) { 3 }^{ n+1 }+3 }{ 4 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.3+{ 2.3 }^{ 2 }+{ 3.3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ n.3 }^{ n }=frac { left( 2n-1 right) { 3 }^{ n+1 }+3 }{ 4 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 9

Ex 4.1 Class 11 Maths Question 6.
$1.2+2.3+3.4+cdot cdot cdot cdot cdot cdot cdot cdot +n.left( n+1 right) =left[ frac { nleft( n+1 right) left( n+2 right) }{ 3 } right]$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.2+2.3+3.4+cdot cdot cdot cdot cdot cdot cdot cdot +n.left( n+1 right) =left[ frac { nleft( n+1 right) left( n+2 right) }{ 3 } right]$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 11

Ex 4.1 Class 11 Maths Question 7.
$1.3+3.5+5.7+cdot cdot cdot cdot cdot cdot cdot cdot +left( 2n-1 right) left( 2n+1 right) =frac { nleft( { 4n }^{ 2 }+6n-1 right) }{ 3 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.3+3.5+5.7+cdot cdot cdot cdot cdot cdot cdot cdot +left( 2n-1 right) left( 2n+1 right) =frac { nleft( { 4n }^{ 2 }+6n-1 right) }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 13

Ex 4.1 Class 11 Maths Question 8.
$1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +n.{ 2 }^{ n }=left( n-1 right) { 2 }^{ n+1 }+2$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +n.{ 2 }^{ n }=left( n-1 right) { 2 }^{ n+1 }+2$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 16

Ex 4.1 Class 11 Maths Question 9
$frac { 1 }{ 2 } +frac { 1 }{ 4 } +frac { 1 }{ 8 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ { 2 }^{ n } } =1-frac { 1 }{ { 2 }^{ n } }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 2 } +frac { 1 }{ 4 } +frac { 1 }{ 8 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ { 2 }^{ n } } =1-frac { 1 }{ { 2 }^{ n } }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 17

Ex 4.1 Class 11 Maths Question 10.
$frac { 1 }{ 2.5 } +frac { 1 }{ 5.8 } +frac { 1 }{ 8.11 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 3n-1 right) left( 3n+2 right) } =frac { n }{ left( 6n+4 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 2.5 } +frac { 1 }{ 5.8 } +frac { 1 }{ 8.11 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 3n-1 right) left( 3n+2 right) } =frac { n }{ left( 6n+4 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 19

Ex 4.1 Class 11 Maths Question 11.
$frac { 1 }{ 1.2.3 } +frac { 1 }{ 2.3.4 } +frac { 1 }{ 3.4.5 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ nleft( n+1 right) left( n+2 right) } =frac { nleft( n+3 right) }{ 4left( n+1 right) left( n+2 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 1.2.3 } +frac { 1 }{ 2.3.4 } +frac { 1 }{ 3.4.5 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ nleft( n+1 right) left( n+2 right) } =frac { nleft( n+3 right) }{ 4left( n+1 right) left( n+2 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 21

Ex 4.1 Class 11 Maths Question 12.
$a+ar+{ ar }^{ 2 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ ar }^{ n-1 }=frac { aleft( { r }^{ n }-1 right) }{ r-1 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $a+ar+{ ar }^{ 2 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ ar }^{ n-1 }=frac { aleft( { r }^{ n }-1 right) }{ r-1 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 24

Ex 4.1 Class 11 Maths Question 13.
$left( 1+frac { 3 }{ 1 } right) left( 1+frac { 5 }{ 4 } right) left( 1+frac { 7 }{ 9 } right) cdot cdot cdot cdot cdot cdot cdot cdot left( 1+frac { left( 2n+1 right) }{ { n }^{ 2 } } right) ={ left( n+1 right) }^{ 2 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $left( 1+frac { 3 }{ 1 } right) left( 1+frac { 5 }{ 4 } right) left( 1+frac { 7 }{ 9 } right) cdot cdot cdot cdot cdot cdot cdot cdot left( 1+frac { left( 2n+1 right) }{ { n }^{ 2 } } right) ={ left( n+1 right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 26

Ex 4.1 Class 11 Maths Question 14.
$left( 1+frac { 1 }{ 1 } right) left( 1+frac { 1 }{ 2 } right) left( 1+frac { 1 }{ 3 } right) cdot cdot cdot cdot cdot cdot cdot cdot left( 1+frac { 1 }{ n } right) =left( n+1 right)$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $left( 1+frac { 1 }{ 1 } right) left( 1+frac { 1 }{ 2 } right) left( 1+frac { 1 }{ 3 } right) cdot cdot cdot cdot cdot cdot cdot cdot left( 1+frac { 1 }{ n } right) =left( n+1 right)$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 28

Ex 4.1 Class 11 Maths Question 15.
${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ left( 2n-1 right) }^{ 2 }=frac { nleft( 2n-1 right) left( 2n+1 right) }{ 3 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : ${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ left( 2n-1 right) }^{ 2 }=frac { nleft( 2n-1 right) left( 2n+1 right) }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 30

Ex 4.1 Class 11 Maths Question 16.
$frac { 1 }{ 1.4 } +frac { 1 }{ 4.7 } +frac { 1 }{ 7.10 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 3n-2 right) left( 3n+1 right) } =frac { n }{ left( 3n+1 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 1.4 } +frac { 1 }{ 4.7 } +frac { 1 }{ 7.10 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 3n-2 right) left( 3n+1 right) } =frac { n }{ left( 3n+1 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 32

Ex 4.1 Class 11 Maths Question 17.
$frac { 1 }{ 3.5 } +frac { 1 }{ 5.7 } +frac { 1 }{ 7.9 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 2n+1 right) left( 2n+3 right) } =frac { n }{ 3left( 2n+3 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 3.5 } +frac { 1 }{ 5.7 } +frac { 1 }{ 7.9 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 2n+1 right) left( 2n+3 right) } =frac { n }{ 3left( 2n+3 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 34

Ex 4.1 Class 11 Maths Question 18.
$1+2+3+cdot cdot cdot cdot cdot cdot cdot cdot +n<frac { 1 }{ 8 } { left( 2n+1 right) }^{ 2 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1+2+3+cdot cdot cdot cdot cdot cdot cdot cdot +n<frac { 1 }{ 8 } { left( 2n+1 right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 36

Ex 4.1 Class 11 Maths Question 19.
n(n+1 )(n + 5) is a multiple of 3.
Solution.
Let the given statement be P(n), i.e.,
P(n): n(n + l)(n + 5) is a multiple of 3.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 38

Ex 4.1 Class 11 Maths Question 20.
${ 10 }^{ 2n-1 }+1$ is divisible by 11.
Solution.
Let the given statement be P(n), i.e.,
P(n): ${ 10 }^{ 2n-1 }+1$ is divisible by 11
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 40

Ex 4.1 Class 11 Maths Question 21.
${ x }^{ 2n }-{ y }^{ 2n }$ is divisible by x + y.
Solution.
Let the given statement be P(n), i.e.,
P(n): ${ x }^{ 2n }-{ y }^{ 2n }$ is divisible by x + y.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 41

Ex 4.1 Class 11 Maths Question 22.
${ 3 }^{ 2n+2 }-8n-9$ is divisible by 8.
Solution.
Let the given statement be P(n), i.e.,
P(n): ${ 3 }^{ 2n+2 }-8n-9$ is divisible by 8.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 42

Ex 4.1 Class 11 Maths Question 23.
${ 41 }^{ n }-{ 14 }^{ n }$ is a multiple of 27.
Solution.
Let the given statement be P(n), i.e.,
P(n): ${ 41 }^{ n }-{ 14 }^{ n }$ is a multiple of 27.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 43

Ex 4.1 Class 11 Maths Question 24.
$left( 2n+7 right) <{ left( n+3 right) }^{ 2 }$
Solution.
Let the given statement be P(n), i.e.,
P(n): $left( 2n+7 right) <{ left( n+3 right) }^{ 2 }$
First we prove that the statement is true for n = 1.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 44

We hope the NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction, drop a comment below and we will get back to you at the earliest.

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Class 11th Chapter -3 Trigonometric Functions | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter -3 Trigonometric Fuctions NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Trigonometric Fuctions solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Trigonometric Fuctions NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 3 Trigonometric Fuctions | NCERT MATHS SOLUTION |

Chapter 3 Trigonometric Fuctions EX 3.1 NCERT Solutions

Ex 3.1 Class 11 Maths Question 1.
Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) -47°30′
(iii) 240°
(iv) 520°
Solution.
We have, 180° = π Radians
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 1
Ex 3.1 Class 11 Maths Question 2.
Find the degree measures corresponding to the following radian measures $\left( Use\quad \pi =\frac { 22 }{ 7 } \right)$
(i) $\frac { 11 }{ 16 }$
(ii) -4
(iii) $\frac { 5\pi }{ 3 }$
(iv) $\frac { 7\pi }{ 6 }$
Solution.
We have π Radians = 180°
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 2

Ex 3.1 Class 11 Maths Question 3.
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second ?
Solution.
Number of revolutions made by wheel in one minute = 360
As we know that, 1 Revolution = 27 π Radians
∴ 360 Revolutions = 720 π Radians
∴ In 1 minute wheel can make = 720 π Radians
⇒ In 60 seconds wheel can make = 720 π Radians
⇒ In 1 second wheel can make
$\frac { 720\pi }{ 3 } Radians=12\pi \quad Radians$

Ex 3.1 Class 11 Maths Question 4.
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm $\left( Use\quad \pi =\frac { 22 }{ 7 } \right)$
Solution.
Let O be the centre and AB be the arc length of the circle.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 3

Ex 3.1 Class 11 Maths Question 5.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of Xhe chord.
Solution.
Let AB be the minor arc of the chord.
AB = 20 cm, OA = OB = 20 cm
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 4

Ex 3.1 Class 11 Maths Question 6.
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their, radii.
Solution.
Let r₁ r₂ and θ₁, θ₂ be the radii and angles subtended at the centre of two circles respectively.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 5

Ex 3.1 Class 11 Maths Question 7.
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 6

We hope the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, drop a comment below and we will get back to you at the earliest.

Chapter 3 Trigonometric Fuctions EX 3.2 NCERT Solutions

Find the values of other five trigonometric functions in Exercises 1 to 5.
Ex 3.2 Class 11 Maths Question 1.
$\cos { x } =\frac { -1 }{ 2 }$ , x lies in third quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 1

Ex 3.2 Class 11 Maths Question 2.
$\sin { x } =\frac { 3 }{ 5 }$ , x lies in second quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 2

Ex 3.2 Class 11 Maths Question 3.
$\cot { x= } \frac { 3 }{ 4 }$ , xlies in third quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 3

Ex 3.2 Class 11 Maths Question 4.
$\sec { x } =\frac { 13 }{ 5 }$ , x lies in fourth quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 4

Ex 3.2 Class 11 Maths Question 5.
$\tan { x } =-\frac { 5 }{ 12 }$ , x lies in second quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 4

Find the values of the trigonometric functions in Exercises 6 to 10.
Ex 3.2 Class 11 Maths Question 6.
sin 765°
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 5

Ex 3.2 Class 11 Maths Question 7.
cosec (-1410°)
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 6

Ex 3.2 Class 11 Maths Question 8.
$tan\quad \frac { 19\pi }{ 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 7

Ex 3.2 Class 11 Maths Question 9.
$sin\left( -\frac { 11\pi }{ 3 } \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 8

Ex 3.2 Class 11 Maths Question 10.
$cot\left( -\frac { 15\pi }{ 4 } \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 9

We hope the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2, drop a comment below and we will get back to you at the earliest.

Chapter 3 Trigonometric Fuctions EX 3.3 NCERT Solutions

Ex 3.3 Class 11 Maths Question 1.
Prove that: ${ sin }^{ 2 }\frac { \pi }{ 6 } +{ cos }^{ 2 }\frac { \pi }{ 3 } -{ tan }^{ 2 }\frac { \pi }{ 4 } =-\frac { 1 }{ 2 }$
Solution.
L.H.S. = ${ sin }^{ 2 }\frac { \pi }{ 6 } +{ cos }^{ 2 }\frac { \pi }{ 3 } -{ tan }^{ 2 }\frac { \pi }{ 4 } =-\frac { 1 }{ 2 }$
$=\left[ { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( 1 \right) }^{ 2 } \right] =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } -1=\frac { -1 }{ 2 } =\quad R.H.S.$

Ex 3.3 Class 11 Maths Question 2.
$2{ sin }^{ 2 }\frac { \pi }{ 6 } +{ cosec }^{ 2 }\frac { 7\pi }{ 6 } { cos }^{ 2 }\frac { \pi }{ 3 } =\frac { 3 }{ 2 }$
Solution.
L.H.S. = $2{ sin }^{ 2 }\frac { \pi }{ 6 } +{ cosec }^{ 2 }\frac { 7\pi }{ 6 } { cos }^{ 2 }\frac { \pi }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 1

Ex 3.3 Class 11 Maths Question 3.
${ cot }^{ 2 }\frac { \pi }{ 6 } +cosec\frac { 5\pi }{ 6 } +3{ tan }^{ 2 }\frac { \pi }{ 6 } =6$
Solution.
L.H.S. = ${ cot }^{ 2 }\frac { \pi }{ 6 } +cosec\frac { 5\pi }{ 6 } +3{ tan }^{ 2 }\frac { \pi }{ 6 }$
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 2

Ex 3.3 Class 11 Maths Question 4.
$2{ sin }^{ 2 }\frac { 3\pi }{ 4 } +2{ cos }^{ 2 }\frac { \pi }{ 4 } +2{ sec }^{ 2 }\frac { \pi }{ 3 } =10$
Solution.
L.H.S. = $2{ sin }^{ 2 }\frac { 3\pi }{ 4 } +2{ cos }^{ 2 }\frac { \pi }{ 4 } +2{ sec }^{ 2 }\frac { \pi }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 3

Ex 3.3 Class 11 Maths Question 5.
Find the value of:
(i) sin 75°
(ii) tan 15°
Solution.
(i) sin (75°) = sin (30° + 45°)
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 4

(ii) tan 15° = tan (45° – 30°)
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 5

Prove the following:
Ex 3.3 Class 11 Maths Question 6.
$cos\left( \frac { \pi }{ 4 } -x \right) cos\left( \frac { \pi }{ 4 } -y \right) -sin\left( \frac { \pi }{ 4 } -x \right) sin\left( \frac { \pi }{ 4 } -y \right)$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 6

Ex 3.3 Class 11 Maths Question 7.
$\frac { tan\left( \frac { \pi }{ 4 } +x \right) }{ tan\left( \frac { \pi }{ 4 } -x \right) } ={ \left( \frac { 1+tan\quad x }{ 1-tan\quad x } \right) }^{ 2 }$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 7

Ex 3.3 Class 11 Maths Question 8.
$\frac { cos\left( \pi +x \right) cos\left( -x \right) }{ sin\left( \pi -x \right) cos\left( \frac { \pi }{ 2 } +x \right) } ={ cot }^{ 2 }x$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 8

Ex 3.3 Class 11 Maths Question 9.
$cos\left( \frac { 3\pi }{ 2 } +x \right) cos\left( 2\pi +x \right) \left[ cot\left( \frac { 3\pi }{ 2 } -x \right) +cot\left( 2\pi +x \right) \right] =1$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 9

Ex 3.3 Class 11 Maths Question 10.
sin(n +1 )x sin(n + 2)x + cos(n +1 )x cos(n + 2)x = cosx
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 10

Ex 3.3 Class 11 Maths Question 11.
$cos\left( \frac { 3\pi }{ 4 } +x \right) -cos\left( \frac { 3\pi }{ 4 } -x \right) =-\sqrt { 2 } sinx$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 11

Ex 3.3 Class 11 Maths Question 12.
sin²6x – sin²4x= sin²x sin10x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 12

Ex 3.3 Class 11 Maths Question 13.
cos²2x – cos²6x = sin 4x sin 8x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 13

Ex 3.3 Class 11 Maths Question 14.
sin2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 14

Ex 3.3 Class 11 Maths Question 15.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 15

Ex 3.3 Class 11 Maths Question 16.
$\frac { cos9x-cos5x }{ sin17x-sin3x } =-\frac { sin2x }{ cos10x }$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 16

Ex 3.3 Class 11 Maths Question 17.
$\frac { sin5x+sin3x }{ cos5x+cos3x } =tan4x$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 17

Ex 3.3 Class 11 Maths Question 18.
$\frac { sinx-siny }{ cosx+cosy } =tan\left( \frac { x-y }{ 2 } \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 18

Ex 3.3 Class 11 Maths Question 19.
$\frac { sinx+sin3x }{ cosx+cos3x } =tan2x$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 19

Ex 3.3 Class 11 Maths Question 20.
$\frac { sinx-sin3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } =2sinx$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 20

Ex 3.3 Class 11 Maths Question 21.
$\frac { cos4x+cos3x+cos2x }{ sin4x+sin3x+sin2x } =cot3x$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 21

Ex 3.3 Class 11 Maths Question 22.
cot x cot 2x – cot 2x cot 3x – cot3x cotx = 1
Solution.
We know that 3x = 2x + x.
Therefore,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 22

Ex 3.3 Class 11 Maths Question 23.
$tan4x=\frac { 4tanx\left( 1-{ tan }^{ 2 }x \right) }{ 1-6{ tan }^{ 2 }x+{ tan }^{ 4 }x }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 23

Ex 3.3 Class 11 Maths Question 24.
cos 4x = 1 – 8 sin²x cos²x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 24

Ex 3.3 Class 11 Maths Question 25.
cos 6x = 32 cos6 x – 48 cos⁴x + 18 cos² x -1
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 25

We hope the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3, drop a comment below and we will get back to you at the earliest.

Chapter 3 Trigonometric Fuctions EX 3.4 NCERT Solutions

Find the principal and general solutions of the following equations:
Ex 3.4 Class 11 Maths Question 1.
$tanx=\sqrt { 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 1

Ex 3.4 Class 11 Maths Question 2.
sec x = 2
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 2

Ex 3.4 Class 11 Maths Question 3.
$cotx=-\sqrt { 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 3

Ex 3.4 Class 11 Maths Question 4.
cosec x = -2
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 5

Find the general solution for each of the following equations:
Ex 3.4 Class 11 Maths Question 5.
cos 4x = cos 2x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 6

Ex 3.4 Class 11 Maths Question 6.
cos 3x + cos x – cos 2x=0
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 7

Ex 3.4 Class 11 Maths Question 7.
sin 2 x + cos x = 0
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 8

Ex 3.4 Class 11 Maths Question 8.
sec²2x = 1 – tan 2x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 9

Ex 3.4 Class 11 Maths Question 9.
sin x + sin 3x + sin 5x = 0
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 10

We hope the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4, drop a comment below and we will get back to you at the earliest.

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Class 11th Chapter -2 Relations and Functions | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter 2 Relations and functions NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Relations and functions solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Relations and functions NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 1 Relations and Functions | NCERT MATHS SOLUTION |

Chapter 1 - Relations and functions EX 2.1 NCERT Solutions

Ex 2.1 Class 11 Maths Question 1.
If $\left( \frac { x }{ 3 } +1,y-\frac { 2 }{ 3 } \right) =\left( \frac { 5 }{ 3 } ,\frac { 1 }{ 3 } \right)$ , find the values of x and y.
Solution.
Since the ordered pairs are equal. So, the corresponding elements are equal
∴ $\frac { x }{ 3 } +1=\frac { 5 }{ 3 }$ and $y-\frac { 2 }{ 3 } =\frac { 1 }{ 3 }$
⇒ $\frac { x }{ 3 } =\frac { 5 }{ 3 } -1$ and $y=\frac { 1 }{ 3 } +\frac { 2 }{ 3 }$ ⇒ x = 2 and y = 1.

Ex 2.1 Class 11 Maths Question 2.
If the set A has 3 elements and the set B {3, 4, 5}, then find the number of elements in (A x B).
Solution.
According to question, n(A) = 3 and n(B) = 3.
∴ n(A x B) = n(A) x n(B) = 3 x 3 = 9
∴ There are total 9 elements in (A x B).

Ex 2.1 Class 11 Maths Question 3.
If G = {7, 8} and H = {5, 4, 2}, find G x H and H x G.
Solution.
We have G = {7, 8} and H = {5, 4, 2} Then, by the definition of the cartesian product, we have
G x H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H x G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.

Ex 2.1 Class 11 Maths Question 4.
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m}, then P x Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then Ax B is a non-empty set of ordered pairs (x, y) such
that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A x (B ∩φ) = φ
Solution.
(i) False, if P = {m, n} and Q = {n, m}
Then P x Q = {(m, n), (m, m), (n, n), (n, m)}.
(ii) True, by the definition of cartesian product.
(iii) True, We have A = {1, 2} and B = {3, 4}
Now, B ∩ φ = φ ∴ A x (B ∩ φ) = A x φ = φ.

Ex 2.1 Class 11 Maths Question 5.
If A = {-1, 1},find A x A x A.
Solution.
A = {-1, 1}
Then, A x A = {-1, 1} x {-1, 1} = {(-1, -1), (-1,1),(1,-1), (1,1)}
A x A x A = ((-1,-1),(-1,1),(1,-1),(1,1)} x {-1,1}
= {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1,1), (1, -1, -1), (1, -1,1), (1,1,-1), (1,1,1)}

Ex 2.1 Class 11 Maths Question 6.
If A x B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution.
Given, A x B = {(a, x), (a, y), (b, x), (b, y)}
If {p, q) ∈ A x B, then p ∈ A and q ∈ B
∴ A = {a, b} and B = {x, y}.

Ex 2.1 Class 11 Maths Question 7.
Let A = {1, 2}, B = (1, 2, 3, 4), C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A x (B ∩ C = (A x B) ∩ (AxC)
(ii) A x C is a subset of B x D.
Solution.
Given, A = {1, 2}, B ={1, 2, 3, 4}, C = {5, 6}, D = (5, 6, 7, 8}
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1 1

Ex 2.1 Class 11 Maths Question 8.
Let A = {1, 2} and B = {3, 4}. Write 4 x B. How many subsets will 4 x B have? List them.
Solution.
Given, A = {1, 2} and B = {3, 4}
Then, A x B = {(1, 3), (1,4), (2, 3), (2, 4)}
i. e., A x B has 4 elements. So, it has 2⁴ i.e. 16 subsets.
The subsets of A x B are as follows :
φ, {(1, 3)1, ((1, 4)), {(2, 3)|, {(2, 4)}, {(1, 3), (1,4)}, {(1,3), (2,3)},{(1,3), (2,4)), ((1,4), (2,3)},
{(1, 4), (2, 4)},{(2, 3), (2, 4)},{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4), (2, 3), (2,4)},{(1, 3), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2,4)}.

Ex 2.1 Class 11 Maths Question 9.
Let A and B be two sets such that n (A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A x B, find A and B, where x, y and z are distinct elements.
Solution.
Given, n(A) = 3 and n(B) = 2
Now (x, 1) ∈ A x B ⇒ x ∈ A and 1 ∈ B,
(y, 2) ∈ A x B ⇒ y ∈ A and 2 ∈ B
(z, 1) ∈ A x B ⇒z ∈ A and 1 ∈ B
∴ x, y, z ∈ A and 1, 2 ∈ B
Hence, A = {x, y, z} and B = {1, 2}.

Ex 2.1 Class 11 Maths Question 10.
The Cartesian product 4×4 has 9 elements among which are found (-1, 0) and (0, 1). Find the set 4 and the remaining elements of 4 x 4.
Solution.
Since, we have n(A x A) = 9
⇒ n(A) x n(A) = 9 [ ∵ n (A x B) = n(A) x n(B)]
⇒ (n(A))² = 9 ⇒ n(A) = 3
Also, given (-1, 0) ∈ A x A ⇒ -1, 0 ∈ A ,
and (0,1) ∈ A x A ⇒ 0, 1 ∈ A
∴ -1, 0,1 ∈ A
Hence, A = {-1, 0, 1} (∵ n(A) = 3)
and the remaining elements of A x A are (-1, -1), (-1,1), (0, -1), (0,0), (1, -1), (1,0), (1,1).

We hope the NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1, drop a comment below and we will get back to you at the earliest.

Chapter 1 - Relations and functions EX 2.2 NCERT Solutions

Ex 2.2 Class 11 Maths Question 1.
Let 4 = {1,2,3, ,14}. Define a relation R from A to A by R = {(x,y): 3x – y = 0, where x, y ∈ 4}.
Write down its domain, codomain and range.
Solution.
We have A = (1, 2, 3,……..,14)
Given relation R = {(x, y) : 3x – y = 0, where x, y ∈ A}
= {(x, y): y = 3x, where x, y ∈ A)
= {(x, 3x), where x, 3x ∈ A}
= {(1, 3), (2, 6), (3, 9), (4,12)}
[∵ 1 ≤ 3x ≤ 14, ∴ $\frac { 1 }{ 3 } \le x\le \frac { 14 }{ 3 }$ ⇒ x = 1, 2, 3, 4 ]
Domain of R = {1, 2, 3, 4}
Codomain of R = {1, 2,……, 14}
Range of R = {3, 6, 9, 12}.

Ex 2.2 Class 11 Maths Question 2.
Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Solution.
Given relation R = {(x, y): y = x + 5, x < 4 and x, y ∈ N)
= {(x, y): y = x + 5, x ∈ (1, 2, 3) & y ∈ N}
= {(x, x + 5): x = 1, 2, 3}
Thus, R = {(1, 6), (2, 7), (3, 8)}.
Domain of R = {1, 2, 3}, Range of R = {6, 7, 8}.

Ex 2.2 Class 11 Maths Question 3.
A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ 4, y ∈B}. Write R in roster form.
Solution.
We have, A = {1, 2, 3, 5} and B = {4, 6, 9} R = {(x, y): difference between x and y is odd;
x ∈ A, y ∈B}
= {(x, y): y – x = odd; x ∈ A, y ∈ B}
Hence R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

Ex 2.2 Class 11 Maths Question 4.
The figure shows a relationship between the sets P and Q. Write this relation
(i) in set-builder form
(ii) roster form.
What is its domain and range?
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2 1
Solution.
(i) Its set builder form is
R = {(x, y): x – y = 2; x ∈ P, y ∈ Q}
i.e., R = {(x, y): y = x – 2 for x = 5, 6, 7)}

(ii) Roster form is R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7} = P,
Range of R = {3, 4, 5} = Q.

Ex 2.2 Class 11 Maths Question 5.
Let A = {1,2,3,4,6}. Let R be the relation on A defined by {(a, b): a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R?.
Solution.
Given A = {1, 2, 3, 4, 6}
Given relation is R = {(a, b):a,b ∈ A, b is exactly divisible by a}
(i) Roster form of R = {(1,1), (1,2), (1,3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6) (4, 4), (6, 6)}.
(ii) Domain of R = {1, 2, 3, 4, 6} = A.
(iii) Range of R = {1, 2, 3, 4, 6} = A.

Ex 2.2 Class 11 Maths Question 6.
Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Solution.
Given relation is R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5)}
= {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5,10)}
∴Domain of R = {0,1, 2, 3, 4, 5} and
Range of R = {5, 6, 7, 8, 9, 10}.

Ex 2.2 Class 11 Maths Question 7.
Write the relation R = {(x, x³): x is a prime number less than 10} in roster form.
Solution.
Given relation is R = {(x, x³): x is a prime number less than 10)
= {(x, x³): x ∈ {2, 3, 5, 7}}
= {(2, 23), (3, 33), (5, 53), (7, 73)}
= {(2, 8), (3, 27), (5, 125), (7, 343)}.

Ex 2.2 Class 11 Maths Question 8.
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Solution.
Given A = {x, y, z} and B = {1, 2}
∴ n(A) = 3 & n(B) = 2
Since n(A x B) = n(A) x n(B)
∴ n(A x B) = 3 x 2 = 6
Number of relations from A to B is equal to the number of subsets of A x B.
Since A x B contains 6 elements.
⇒ Number of subsets of A x B = 2⁶ = 64
So, there are 64 relations from A to B.

Ex 2.2 Class 11 Maths Question 9.
Let R be the relation on Z defined by R = {{a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution.
Given relation is R = {(a, b): a, b ∈ Z, a – b is an integer}
If a, b ∈ Z, then a- b ∈ Z ⇒ Every ordered pair of integers is contained in R.
R = {(a, b) :a,b ∈ Z}
So, Range of R = Domain of R = Z.

We hope the NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2, drop a comment below and we will get back to you at the earliest.

Chapter 1 - Relations and functions EX 2.3 NCERT Solutions

Ex 2.3 Class 11 Maths Question 1.
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {{2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) {(1, 3), (1, 5), (2, 5)}.
Solution.
(i) We have a relation R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} Since 2, 5, 8, 11, 14, 17 are the elements of domain of R having their unique images.
∴ The given relation is a function.
Hence domain = {2, 5, 8, 11, 14, 17) and Range = {1}.

(ii) We have a relation
R = {(2,1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, 14 are the elements of domain of R having their unique images.
∴ The given relation is a function.
Hence domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7}.

(iii) We have a relation R = {(1, 3), (1, 5), (2, 5)}
Since the distinct ordered pairs (1, 3) and (1, 5) have same first element i.e., 1 does not have a unique image under R.
∴ It is not a function.

Ex 2.3 Class 11 Maths Question 2.
Find the domain and range of the following real functions:
(i) f(x) = $-\left| x \right|$
(ii) f(x) = $\sqrt { 9-{ x }^{ 2 } }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 1

Ex 2.3 Class 11 Maths Question 3.
A function f is defined by f (x) = 2x – 5. Write down the values of
(i) f (0)
(ii) f (7)
(iii) f (-3)
Solution.
We are given f (x) = 2x – 5
(i) f (0) = 2(0) – 5 = 0- 5 = -5
(ii) f (7) = 2(7) – 5 = 14- 5 = 9
(iii) f (-3) = 2(-3) – 5 = -6 – 5 = -11.

Ex 2.3 Class 11 Maths Question 4.
The function T which maps temperature in degree Celsius into temperature in degree by
$t(C)=\frac { 9C }{ 5 } +32$
Find
(i) t (0)
(ii) t (28)
(iii) t (-10)
(iv) The value of C, when t (C = 212
Solution.
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 2
Ex 2.3 Class 11 Maths Question 5.
Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x>0.
(ii) f(x)=x²+ 2, x is a real number.
(iii) f (x) = x, x is a real number.
Solution.
(i) Given f (x) = 2 – 3x, x ∈ R, x > 0
∵ x > 0 ⇒ -3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f (x) < 2
∴ The range of f (x) is (-2).

(ii) Given f (x) = x² + 2, x is a real number
We know x²≥ 0 ⇒ x² + 2 ≥ 0 + 2
⇒ x² + 2 > 2 ∴ f (x) ≥ 2
∴ The range of f (x) is [2, ∞).

(iii) Given f (x) = x, x is a real number.
Let y =f (x) = x ⇒ y = x
∴ Range of f (x) = Domain of f (x)
∴ Range of f (x) is R.

We hope the NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3, drop a comment below and we will get back to you at the earliest.

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NCERT MCQ CLASS 9 CHAPTER – 8 | QUADRILATERALS | EDUGROWN

NCERT MCQ ON QUADRILATERALS

Question 1.

Three angles of a quadrilateral are 75°, 90°and 75°, the fourth angle is
(a) 90°
(b) 95°
(c) 105°
(d) 120°

Answer: (d) 120°

Question 2.

A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is
(a) 55°
(b) 50°
(c) 40°
(d) 25°

Answer: (b) 50°

Question 3.

ABCD is a rhombus such that ∠ACB = 40°, then ∠ADB is
(a) 40°
(b) 45°
(c) 50°
(d) 60°

Answer: (c) 50°

Question 4.

If angles A, B, C and D of a quadrilateral ABCD, taken in order, are in the ratio 3 : 7 : 6 : 4, then ABCD is a
(a) rhombus
(b) parallelogram
(c) trapezium
(d) kite.

Answer: (c) trapezium

Question 5.

The diagonals AC and BD of a || gm ABCD intersect each other at the point O. If ∠DAC = 32° and ∠AOB = 70°, then ∠DBC is equal to
(a) 24°
(b) 86°
(c) 38°
(d) 32°

Answer: (c) 38°

Question 6.

ABCD is a rhombus such that ∠ABC = 40°, then ∠ADC is equal to
(a) 40°
(b) 45°
(c) 50°
(d) 20°

Answer: (a) 40°

Question 7.

In the following figure, ABCD and AEFG are two parallelograms. If ∠C = 60°, then ∠GFE is
MCQ Questions for Class 9 Maths Chapter 8 Quadrilaterals with Answers
(a) 30°
(b) 60°
(c) 90°
(d) 120°

Answer: (b) 60°

Question 8.

The bisectors of any two adjacent angles of a || gm intersect at
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer: (d) 90°

Question 9.

If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of a parallelogram is
(a) 176°
(b) 68°
(c) 112°
(d) 102°

Answer: (c) 112°

Question 10.

If the diagonal of a rhombus are 18 cm and 24 cm respectively, then its side is equal to
(a) 16 cm
(b) 15 cm
(c) 20 cm
(d) 17 cm

Answer: (b) 15 cm

Question 11.

In the given figure, ABCD is a parallelogram. Find the value of x.
MCQ Questions for Class 9 Maths Chapter 8 Quadrilaterals with Answers
(a) 25°
(b) 60°
(c) 75°
(d) 45°

Answer: (d) 45°

Question 12 .

1. The quadrilateral whose all its sides are equal and angles are equal to 90 degrees, it is called:
a. Rectangle
b. Square
c. Kite
d. Parallelogram

Answer: (b) square

Question 13.

The sum of all the angles of a quadrilateral is equal to:
a. 180°
b. 270°
c. 360°
d. 90°

Answer: (c) 360°

Question 14.

A trapezium has:
a. One pair of opposite sides parallel
b. Two pair of opposite sides parallel to each other
c. All its sides are equal
d. All angles are equal

Answer: (a) One pair of opposite sides parallel

Question 15.

A rhombus can be a:
a. Parallelogram
b. Trapezium
c. Kite
d. Square

Answer: (d) Square

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NCERT MCQ CLASS 9 CHAPTER – 7 | TRIANGLES | EDUGROWN

NCERT MCQ ON TRIANGLES

Question 1.

In given figure, the measure of ∠BAC is
MCQ Questions for Class 9 Maths Chapter 7 Triangles with Answers 1
(a) 60°
(b) 50°
(c) 70°
(d) 80°

Answer: (b) 50°

Question 2.

In figure if AE || DC and AB = AC, find the value of ∠ABD.
MCQ Questions for Class 9 Maths Chapter 7 Triangles with Answers 2
(a) 130°
(b) 110°
(c) 120°
(d) 70°

Answer: (a) 130°

Question 3.

In figure X is a point in the interior of square ABCD, AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then find BY.
MCQ Questions for Class 9 Maths Chapter 7 Triangles with Answers 3
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm

Answer: (c) 7 cm

Question 4.

In figure AB ⊥ AE, BC ⊥ AB, CE = DE and ∠AED = 120°, then find ∠ECD.
MCQ Questions for Class 9 Maths Chapter 7 Triangles with Answers 4
(a) 80°
(b) 70°
(c) 85°
(d) 60°

Answer: (d) 60°

Question 5.

In ΔABC, ∠C = ∠A and BC = 4 cm and AC = 5 cm, then find length of AB.
(a) 5 cm
(b) 3 cm
(c) 4 cm
(d) 2.5 cm

Answer: (c) 4 cm

Question 6.
In ΔABC, AB = AC and ∠B = 50°, then find ∠C.
(a) 50°
(b) 40°
(c) 80°
(d) 120°

Answer: (a) 50°

Question 7.

In figure, D is the mid-point of side BC of a ΔABC and ∠ABD = 50°. If AD = BD = CD, then find the measure of ∠ACD.
MCQ Questions for Class 9 Maths Chapter 7 Triangles with Answers 5
(a) 30°
(b) 70°
(c) 80°
(d) 40°

Answer: (d) 40°

Question 8.

In figure, ABC is an isosceles triangle whose side AC is produce to E and through C, CD is drawn parallel to BA. Find the value of x.
MCQ Questions for Class 9 Maths Chapter 7 Triangles with Answers 6
(a) 52°
(b) 156°
(c) 76°
(d) 104°

Answer: (d) 104°

Question 9.

In figure AB ⊥ BE and EF ⊥ BE. If BC = DE and AB = EF, then ΔABD is congruent to
MCQ Questions for Class 9 Maths Chapter 7 Triangles with Answers 7
(a) ΔEFC
(b) ΔECF
(c) ΔDEF
(d) ΔFEC

Answer: (d) ΔFEC

Question 10.

D is a point on the side BC of a ΔABC such that AD bisects ∠BAC. Then
(a) BD = CD
(b) BA > BD
(c) BD > BA
(d) CD > CA

Answer: (b) BA > BD

Question 11.

It is given that ΔABC = ΔFDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then which of the following is true?
(а) DF = 5 cm, ∠F = 60°
(b) DF = 5 cm, ∠E = 60°
(c) DE = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40°
MCQ Questions for Class 9 Maths Chapter 7 Triangles with Answers

Answer: (b) DF = 5 cm, ∠E = 60°

Question 12.

Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm

Answer: (d) 3.4 cm

Question 13.

In ΔPQR, if ∠R > ∠Q, then
(a) QR > PR
(b) PQ > PR
(c) PQ < PR
(d) QR < PR

Answer: (b) PQ > PR

Question 14.

In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles

Answer: (a) isosceles but not congruent

Question 15.

In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if
(a) BC = EF
(b) AC = DE
(c) AC = EF
(d) BC = DE

Answer: (b) AC = DE

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NCERT MCQ CLASS 9 CHAPTER 6 | LINES AND ANGLES |

NCERT MCQ FOR LINES AND ANGLES

Question 1.

In ΔABC, ∠A = 50° and the external bisectors of ∠B and ∠C meet at O as shown in figure. The measure of ∠BOC is
MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers 1
(a) 40°
(b) 65°
(c) 115°
(d) 140°

Answer: (b) 65°

Question 2.

In figure the value of x is
MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers 2
(a) 120°
(b) 130°
(c) 110°
(d) 100°

Answer: (b) 130°

Question 3.

An exterior angle of a triangle is 80° and the interior opposite angles are in the ratio 1 : 3, measure of interior opposite angles are
(a) 30°, 90°
(b) 40°, 120°
(c) 20°, 60°
(d) 30°, 60°

Answer: (c) 20°, 60°

Question 4.

In ΔABC, the bisectors of ∠ABC and ∠BCA intersect each other at O. The measure of ∠BOC is
(a) 90° + ∠A
(b) 90° + ∠A2
(c) 180 – ∠A
(d) 90° – ∠A2

Answer: (b) 90° + ∠A2

Question 5.

In figure if ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = k right angles, then find value of k.
MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers 3
(а) 2
(b) 3
(c) 4
(d) 5

Answer: (c) 4

Question 6.

In the given figure, the measure of ∠ABC is
MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers 4
(a) 80°
(b) 20°
(c) 100°
(d) 60°

Answer: (a) 80°

Question 7.

The angle of a triangle are in the ratio 5 : 3 : 7, the triangle is
(а) an acute-angled triangle
(b) an obtuse angled triangle
(c) an right angled triangle
(d) an isosceles triangle.

Answer: (а) an acute-angled triangle

Question 8.

In figure l₁ || l₂, the value of x is
MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers 5
(a) 80°
(b) 100°
(c) 110°
(d) 70°

Answer: (a) 80°

Question 9.

If one angle of triangle is equal to the sum of the other two, then the triangle is
(a) an isosceles triangle
(b) an obtuse-angled triangle
(c) an equilateral triangle
(d) a right triangle

Answer: (d) a right triangle

Question 10.

One of the angles of a triangle is 75°. If the difference of other two is 35°, then the largest angle of other two angles has a measure
(a) 80°
(b) 75°
(c) 70°
(d) 135°

Answer: (c) 70°

Question 11 .

If AB = x + 3, BC = 2x and AC = 4x – 5, then for what value of ‘x’, B lies on AC?

a) 2

b) 3

c) 5

d) 8

Answer: (d) 8

Question 12 .

In the given figure, if AOB is a line then the measure of ∠BOC, ∠COD and ∠DOA respectively are:

a) 36^o, 54^o, 90^o

b) 36^o, 90^o, 54^o

c) 90^o, 36^o, 54^o

d) 90^o, 54^o, 36^o

Answer: (a) 36^o, 54^o, 90^o

Question 13 .

In the figure, if x, y and z are exterior angles of ΔABC then x + y + z is:

CBSE Class 9th MCQs on Maths Chapter 6 Lines and Angles

a) 90^o

b) 180^o

c) 270^o

d) 360^o

Answer: (d) 360^o

Question 14 .

In the given figure, PQ || RS and ∠ACS = 127°, ∠BAC is:
MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers 10
a) 50^o
b) 53^o
c) 77^o
d) 107^o

Answer c) 77^o

Question 15.

In figure, AB || ED, the value of x is:
MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers 11
a) 26^o
b) 36^o
c) 54^o
d) 62^o

Answer a) 26^o

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NCERT MCQ CLASS 9 CHAPTER 5 | INTRODUCTION TO ELUCID’S GEOMETRY |

NCERT MCQ FOR INTRODUCTION TO ELUCID’S GEOMETRY

Question 1.

The number of dimensions, a solid has:
(a) 1
(b) 2
(c) 3
(d) 0

Answer: (c) 3

Question 2.

The total number of propositions in the elements are:
(a) 465
(b) 460
(c) 13
(d) 55

Answer: (a) 465

Question 3.

In Indus valley civilisation (about 3000 BC), the bricks used for construction work were having dimensions in the ratio
(a) 1 : 3 : 4
(b) 4 : 2 : 1
(c) 4 : 4 : 1
(d) 4 : 3 : 2

Answer: (b) 4 : 2 : 1

Question 4.

The things which are double of same thing are
(a) equal
(b) halves of same thing
(c) unequal
(d) double of the same thing

Answer: (a) equal

Question 5.

Which of the following statements is incorrect?
(а) A line segment has definite length.
(b) Three lines are concurrent if and only if they have a common point.
(c) Two lines drawn in a plane always intersect at a point.
(d) One and only one line can be drawn passing through a given point and parallel to a given line.

Answer: (c) Two lines drawn in a plane always intersect at a point.

Question 6.

Select the wrong statement:
(а) Only one line can pass through a single point.
(b) Only one line can pass through two distinct points.
(c) A terminated line can be produced indefinitely on both the sides.
(d) If two circles are equal, then their radii are equal.

Answer: (а) Only one line can pass through a single point.

Question 7.

Which one of the following statements is true?
(a) Only one line can pass through a single point.
(b) There are an infinite number lines which pass through two distinct points.
(c) Two distinct lines cannot have more than one point in common.
(d) If two circles are equal, then their radii are not equal.

Answer: (c) Two distinct lines cannot have more than one point in common.

Question 8.

If the point P lies in between M and N and C is mid point of MP, then:
(a) MC + PN = MN
(b) MP + CP – MN
(c) MC + CN = MN
(d) CP + CN = MN

Answer: (c) MC + CN = MN

Question 9.

‘Lines are parallel if they do not intersect’ is stated in the form of:
(a) an axiom
(b) a definition
(c) a postulate
(d) a proof

Answer: (b) a definition

Question 10.

Two planes intersect each other to form a:
(a) plane
(b) point
(c) straight line
(d) angle

Answer: (c) straight line

Question 11.

The number of lines that can pass through a given point is
(a) two
(b) none
(c) only one
(d) infinitely many

Answer: (d) infinitely many

Question 12.

How many lines do pass through two distinct points?
(a) 1
(b) 2
(c) 3
(d) 4

Answer: (a) 1

Question 13.

The number of line segments determined by three collinear points is:
(a) two
(b) three
(c) only one
(d) four

Answer: (b) three

Question 14.

Number of dimension(s) a surface has:
(a) 0
(b) 1
(c) 2
(d) 3

Answer: (c) 2

Question 15.

Euclid stated that things which are equal to the same thing are equal to one another in the form of:
(a) an axiom
(b) a definition
(c) a postulate
(d) a proof

Answer: (a) an axiom

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NCERT MCQ CLASS-9 CHAPTER 1 | NUMBER SYSTEM | EDUGROWN

NCERT MCQ ON NUMBER SYSTEM

Question 1.

A rational number lying between √2 and √3 is
(a) √2+√32/2
(b) √6
(c) 1.6
(d) 1.9

Answer: (c) 1.6

Question 2.

Find the value of 54250√3
(a) 925
(b) 35
(c) 27125
(d) 2√35

Answer: (b) 35

Question 3.

Simplified value of (16)^–14 × 16√4 is
(a) 16
(b) 4
(c) 1
(d) 0

Answer: (c) 1

Question 4.

Find the value of 64−2√4.
(a) 18
(b) 12
(c) 8
(d) 164

Answer: (a) 18

Question 5.

Find the value of 216√3 – 125√3
(a) 1
(b) -1
(c) 91√3
(d) 6/5

Answer: (a) 1

Question 6.

When 1515√ is divided by 3√3 find the quotient.
(a) 5√3
(b) 3√5
(c) 5√5
(d) 3√3

Answer: (c) 5√5

Question 7.

If √3 = 1.732 and √2 = 1.414, find the value of 1√3−√2
(a) 0.318
(b) 3.146
(c) 13.146
(d) √1.732 – √ 1.414

Answer: (b) 3.146

Question 8.

Which of the following numbers is an irrational number?
(a) 16√ 4
(b) (3 – √3) (3 + √3)
(c) √5 + 3
(d) – √ 25

Answer: (c) √5 + 3

Question 9.

The decimal expansion of √2 is
(а) finite decimal
(b) 1.4121
(c) non-terminating recurring
(d) non-terminating non-recurring

Answer: (d) non-terminating non-recurring

Question 10.

If x = √75 and 5x = p√7, then the value of p is
(а) 5√7
(b) 257
(c) 725
(d) √75

Answer: (b) 257

Question 11.

If x = 2+√3, then x + 1/x =
(a) 5
(b) 4
(c) -4
(d) -5

Answer: (b) 4

Question 12.

Every rational number is:
(a) Whole number
(b) Natural number
(c) Integer
(d) Real number

Answer: (d) Real number

Question 13.

3√6 + 4√6 is equal to:
(a) 6√6
(b) 7√6
(c) 4√12
(d) 7√12

Answer: (b) 7√6

Question 14.

Expression of 2.2323… in the form of a/ b is ________.
(a) 221/99
(b) 75/31
(c) 7/99
(d) 223/99

Answer: (a) 221/99

Question 15.

Which of the following lies between 0 and -1?
(a) 0
(b) -3
(c) -2/3
(d) 4/3

Answer: (c) -2/3

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Class 11 Political Science video lectures by Richa Sharma Mam-EduGrown

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Class 11th Chapter – 5 Complex Numbers and Quadratic Equations | NCERT Maths Solution | NCERT Solution | Edugrown

Class 11th Chapter 5 Complex Numbers and Quadratic Equations| NCERT MATHS SOLUTION |

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Class 11th Chapter 1 Relations and Functions | NCERT MATHS SOLUTION |

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NCERT MCQ CLASS 9 CHAPTER – 7 | TRIANGLES | EDUGROWN

NCERT MCQ CLASS 9 CHAPTER 6 | LINES AND ANGLES |

NCERT MCQ CLASS 9 CHAPTER 5 | INTRODUCTION TO ELUCID’S GEOMETRY |

NCERT MCQ CLASS-9 CHAPTER 1 | NUMBER SYSTEM | EDUGROWN

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