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Find the mean deviation about the mean for the data in Exercises 1 and 2.

Ex 15.1 Class 11 Maths Question 1.
4, 7, 8, 9, 10, 12, 13, 17
Solution:
Mean of the given data is

Ex 15.1 Class 11 Maths Question 2.
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
Mean of the given data is

Ex 15.1 Class 11 Maths Question 3.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Solution:
Arranging the data in ascending order, we have
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Here n = 12 (which is even)
So median is the average of 6th and 7th observations

Ex 15.1 Class 11 Maths Question 4.
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Solution:
Arranging the data in ascending order, we have 36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Here n = 10 (which is even)
So median is the average of 5th and 6th observations

Find the mean deviation about the mean for the data in Exercises 5 and 6.

Ex 15.1 Class 11 Maths Question 5.

Solution:

Ex 15.1 Class 11 Maths Question 6.

Solution:

Find the mean deviation about the median for the data in Exercises 7 and 8.

Ex 15.1 Class 11 Maths Question 7.

Solution:

Ex 15.1 Class 11 Maths Question 8.

Solution:

Find the mean deviation about the mean for the data in Exercises 9 and 10.

Ex 15.1 Class 11 Maths Question 9.

Solution:

Ex 15.1 Class 11 Maths Question 10.

Solution:

Ex 15.1 Class 11 Maths Question 11.
Find the mean deviation about median for the following data:

Solution:

Ex 15.1 Class 11 Maths Question 12.
Calculate the mean deviation about median age for the age distribution of 100 persons given below:


[Hint: Convert the given data into continuous frequency distribution by subtracting 0.5 from lower limit and adding 0.5 to the upper limit of each class interval]
Solution:

We hope the NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1, drop a comment below and we will get back to you at the earliest.

Find the mean and variance for each of the data in Exercises 1 to 5.

Ex 15.2 Class 11 Maths Question 1.
6, 7, 10, 12, 13, 4, 8, 12
Solution:
Here x_i = 6, 7, 10, 12, 13, 4, 8, 12
∴ Σx_i = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72
n = 8

Ex 15.2 Class 11 Maths Question 2.
First n natural numbers
Solution:
Here x_i = 1, 2, 3, 4, ……….n

Ex 15.2 Class 11 Maths Question 3.
First 10 multiples of 3
Solution:
Here x_i = 3, 6, 9, 12, 15, 18, 21, 27, 30,
Σx_i = 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30 = 165
n = 10

Ex 15.2 Class 11 Maths Question 4.

Solution:

Ex 15.2 Class 11 Maths Question 5.

Solution:

Ex 15.2 Class 11 Maths Question 6.
Find the mean and standard deviation using short-cut method

Solution:

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

Ex 15.2 Class 11 Maths Question 7.

Solution:

Ex 15.2 Class 11 Maths Question 8.

Solution:

Ex 15.2 Class 11 Maths Question 9.
Find the mean, variance and standard deviation using short-cut method.

Solution:

Ex 15.2 Class 11 Maths Question 10.
The diameters of circles (in mm) drawn in a design are given below:

Calculate the standard deviation and mean diameter of the circles.
[Hint: First make the data continuous by making the classes as 32.5 – 36.5, 36.5 – 40.5, 40.5 – 44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]
Solution:

We hope the NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.2, drop a comment below and we will get back to you at the earliest.

Ex 15.3 Class 11 Maths Question 1.
From the data given below state which group is more variable, A or B?

Solution:
For Group A :

Ex 15.3 Class 11 Maths Question 2.
From the prices of shares X and Y below, find out which is more stable in value:

Solution:

Ex 15.3 Class 11 Maths Question 3.
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, AorB, shows greater variability in individual wages?
Solution:
(i) Firm A :
Number of wage earners (n₁) = 586
Mean of monthly wages () = Rs.5253
∴ Total monthly wages = 5253 x 586
= Rs. 3078258
Firm B :
Number of wage earners (n₂) = 648
Mean of monthly wages () = Rs.5253
∴ Total monthly wages = 5253 x 648
= Rs. 3403944
Hence, Firm B pays larger amount as monthly wages.

(ii) Since both the firms have same mean of monthly wages, so the firm with greater variance will have more variability in individual wages. Thus firm B will have more variability in individual wages.

Ex 15.3 Class 11 Maths Question 4.
The following is the record of goals scored by team A in a football session:

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Solution:
For team A:

Ex 15.3 Class 11 Maths Question 5.
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

Which is more varying, the length or weight?
Solution:

We hope the NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.3, drop a comment below and we will get back to you at the earliest.

Ex 14.1 Class 11 Maths Question 1.
Which of the following sentences are statements? Give reasons for your answer.
(i) There are 35 days in a month.
(ii) Mathematics is difficult.
(iii) The sum of 5 and 7 is greater than 10.
(iv) The square of a number is an even number.
(v) The sides of a quadrilateral have equal length.
(vi) Answer this question.
(vii) The product of (-1) and 8 is 8.
(viii) The sum of all interior angles of a triangle is 180°.
(ix) Today is a windy day.
(x) All real numbers are complex numbers.
Solution:
(i) This sentence is false since the maximum number of days in a month can never exceed 31. Therefore, this sentence is a statement.
(ii) This sentence is subjective in the sense that for those who hate mathematics, it is difficult but for others it may not be. This means that this sentence is not always true. Hence it is not a statement.
(iii) This sentence is true as sum of 5 and 7 is greater than 10. Hence it is a statement.
(iv) This sentence is subjective in the sense that it depends on the number that is being squared. Hence it is not a statement.
(v) This sentence is sometimes true and sometimes false since sides in squares and rhombuses have equal length whereas rectangles and trapeziums have unequal length. Hence it is not a statement.
(vi) This sentence is an order and so, it is not a statement.
(vii) This sentence is false as product of (-1) and 8 is -8. So, it is a statement.
(viii) This sentence is true and therefore it is a statement.
(ix) It is not clear from the context which day is referred. Therefore, it is not a statement.
(x) All real numbers can be written in the form of complex numbers. So, this sentence is true and it is a statement.

Ex 14.1 Class 11 Maths Question 2.
Give three examples of sentences which are not statements. Give reasons for the answers.
Solution:
(i) Who are you?
This sentence is an interrogative sentence. Hence, it is not a statement.
(ii) May God bless you!
This sentence is an exclamatory sentence. Hence, it is not a statement.
(iii) How are you?
This sentence is an interrogative sentence. Hence, it is not a statement.

We hope the NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.1, drop a comment below and we will get back to you at the earliest.

 Ex 14.2 Class 11 Maths Question 1.
Write the negation of the following statements:
(i) Chennai is the capital of Tamil Nadu.
(ii)  is not a complex number.
(iii) All triangles are not equilateral triangle.
(iv) The number 2 is greater than 7.
(v) Every natural number is an integer.
Solution:
(i) Negation of statement is: Chennai is not the capital of Tamil Nadu.
(ii) Negation of statement is:  is a complex number.
(iii) Negation of statement is: All triangles are equilateral triangles.
(iv) Negation of statement is: The number 2 is not greater than 7.
(v) Negation of statement is: Every natural number is not an integer.

Ex 14.2 Class 11 Maths Question 2.
Are the following pairs of statements negations of each other:
(i) The number x is not a rational number.
The number x is not an irrational number.
(ii) The number x is a rational number.
The number x is an irrational number.
Solution:
(i) Let p: The number x is not a rational number.
q: The number x is not an irrational number.
Now, ~p: The number x is a rational number. ~q: The number x is an irrational number.
∴ ~p = q and ~q = p
Thus, p and q are negations of each other.

(ii) Let p: The number x is a rational number.
q: The number x is an irrational number.
Now, ~p: The number x is not a rational number.
~q: The number x is not an irrational number.
∴ ~p = q and ~q = p
Thus, p and q are negations of each other.

Ex 14.2 Class 11 Maths Question 3.
Find the component statements of the following compound statements and check whether they are true or false.
(i) Number 3 is prime or it is odd.
(ii) All integers are positive or negative.
(iii) 100 is divisible by 3,11 and 5.
Solution:
(i) The component statements are:
p: Number 3 is prime
q: Number 3 is odd.
Both the component statements p and q are true.

(ii) The component statements are:
p: All integers are positive.
q: All integers are negative.
Both the component statements p and q are false.

(iii) The component statements are:
p: 100 is divisible by 3.
q: 100 is divisible by 11.
r: 100 is divisible by 5.
The component statements p and q are false whereas r is true.

We hope the NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.2, drop a comment below and we will get back to you at the earliest.

 Ex 14.3 Class 11 Maths Question 1.
For each of the following compound statements first, identify the connecting words and then break it into component statements.
(i) All rational numbers are real and all real numbers are not complex.
(ii) Square of an integer is positive or negative.
(iii) The sand heats up quickly in the Sun and does not cool down fast at night.
(iv) x = 2 and x = 3 are the roots of the equation 3x² – x – 10 = 0.
Solution:
(i) The compound statement has the connecting word ‘and’. Component statements are
p: All rational numbers are real.
q: All real numbers are not complex.

(ii) The compound statement has the connecting word ‘or’. Component statements are:
p: Square of an integer is positive.
q: Square of an integer is negative.

(iii) The compound statement has the connecting word ‘and’. Component statements are:
p: The sand heats up quickly in the sun.
q: The sand does not cool down fast at night.

(iv) The compound statement has the connecting word ‘and’. Component statements are:
p: x- 2 is a root of the equation 3x² – x – 10 = 0.
q: x = 3 is a root of the equation 3x² – x – 10 = 0.

Ex 14.3 Class 11 Maths Question 2.
Identify the quantifier in the following statements and write the negation of the statements.
(i) There exists a number which is equal to its square.
(ii) For every real number x, x is less than x + 1.
(iii) There exists a capital for every state in India.
Solution:
(i) Here the quantifier is ‘there exists’.
The negation of statement is: There does not exist a number which is equal to its square.
(ii) Here the quantifier is ‘for every’
The negation of statement is: For at least one real number x, x is not less than x + 1.
(iii) Here the quantifier is ‘there exists’
The negation of statement is: There exists a state in India which does not have a capital.

Ex 14.3 Class 11 Maths Question 3.
Check whether the following pair of statements is negation of each other. Give reasons for your answer.
(i) x + y = y + x is true for every real numbers x and y.
(ii) There exists real numbers x and y for which x + y = y + x.
Solution:
Let p: x + y = y + x is true for every real numbers x and y.
q: There exists real numbers x and y for which
x+y=y + x.
Now, ~p: There exists real numbers x and y for which x + y ≠ y + x.
Thus, ~p ≠ q.

Ex 14.3 Class 11 Maths Question 4.
State whether the “Or” used in the following statements is “exclusive” or “inclusive”. Give reasons for your answer.
(i) Sunrises or Moon sets.
(ii) To apply for a driving license, you should have a ration card or a passport.
(iii) All integers are positive or negative.
Solution:
(i) This statement makes use of exclusive “or”. Since when sun rises, moon does not set during day-time.
(ii) This statement makes use of inclusive ‘or’. Since you can apply for a driving license even if you have a ration card as well as a passport.
(iii) This statement makes use of exclusive ‘or’. Since a integer is either positive or negative, it cannot be both.

We hope the NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.3, drop a comment below and we will get back to you at the earliest.

Ex 14.4 Class 11 Maths Question 1.
Rewrite the following statement with “if-then” in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd.
Solution:
(i) A natural number is odd implies that its square is odd.
(ii) A natural number is odd only if its square is odd.
(iii) For a natural number to be odd it is necessary that its square is odd.
(iv) For the square of a natural number to be odd, it is sufficient that the number is odd.
(v) If the square of a natural number is not odd, then the natural number is not odd.

Ex 14.4 Class 11 Maths Question 2.
Write the contrapositive and converse of the following statements.
(i) If x is a prime number, then x is odd.
(ii) If the two lines are parallel, then they do not intersect in the same plane.
(iii) Something is cold implies that it has low temperature.
(iv) You cannot comprehend geometry if you do not know how to reason deductively.
(v) x is an even number implies that x is divisible by 4.
Solution:
(i) The contra positive of given statement is:
If a number x is not odd, then x is not a prime number.
The converse of given statement is:
If x is an odd number, then x is a prime number.

(ii) The contra positive of given statement is:
If two lines intersect in the same plane, then they are not parallel.
The converse of given statement is:
If two lines do not intersect in the same plane, then they are parallel.

(iii) The contra positive of given statement is:
If something is not at low temperature, then it is not cold.
The converse of given statement is:
If something is at low temperature, then it is cold.

(iv) The contra positive of given statement is:
If you know how to reason deductively, then you can comprehend geometry.
The converse of given statement is:
If you do not know how to reason deductively, then you cannot comprehend geometry.

(v) The contra positive of given statement is:
If x is not divisible by 4, then x is not an even number.
The converse of given statement is:
If x is divisible by 4, then x is an even number.

Ex 14.4 Class 11 Maths Question 3.
Write each of the following statements in the form “if-then”
(i) You get a job implies that your credentials are good.
(ii) The Banana trees will bloom if it stays warm for a month.
(iii) A quadrilateral is a parallelogram if its diagonals bisect each other.
(iv) To get an A+ in the class, it is necessary that you do all the exercises of the book.
Solution:
(i) If you get a job, then your credentials are good.
(ii) If the banana tree stays warm for a month, then it will bloom.
(iii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
(iv) If you get A+ in the class, then you do all the exercises in the book.

Ex 14.4 Class 11 Maths Question 4.
Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other.
(a) If you live in Delhi, then you have winter clothes.
(i) If you do not have winter clothes, then you do not live in Delhi.
(ii) If you have winter clothes, then you live in Delhi.

(b) If a quadrilateral is a parallelogram, then its diagonals bisect each other.
(i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.
(ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Solution:
(a)
(i) contrapositive
(ii) converse

(b)
(i) contrapositive
(ii) converse

We hope the NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.4 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.4, drop a comment below and we will get back to you at the earliest.

Ex 14.5 Class 11 Maths Question 1.
Show that the statement
p: “If x is a real number such that x³ + 4x = 0, then x is 0″ is true by
(i) direct method,
(ii) method of contradiction,
(iii) method of contrapositive.
Solution:
The given compound statement is of the form “if p then q”
p: x ϵ R such that x³ + 4x = 0
q: x = 0
(i) Direct method:
We assume that p is true, then
x ϵ R such that x³ + 4x = 0
⇒ x ϵ R such that x(x² + 4) = 0
⇒ x ϵ R such that x = 0 or x² + 4 = 0
⇒ x = 0 => q is true.
So, when p is true, q is true.
Thus, the given compound statement is true.

(ii) Method of contradiction :
We assume that p is true and q is false, then
x ϵ R such that x³ + 4x = 0
⇒ x ϵ R such that x(x² + 4) = 0
⇒ x ϵ R such that x = 0 or x² + 4 = 0
⇒ x = 0.
which is a contradiction. So, our assumption that x ≠ 0 is false. Thus, the given compound statement is true.

(iii) Method of contrapositive: We assume that q is false, then x ≠ 0
x ϵ R such that x³ + 4x = 0
⇒ x ϵ R such that x = 0 or x² + 4 = 0
∴ statement q is false, so x ≠ 0. So, we have,
x ϵ R such that x² = -2
Which is not true for any x ϵ R.
⇒ p is false
So, when q is false, p is false.
Thus, the given compound statement is true.

Ex 14.5 Class 11 Maths Question 2.
Show that the statement” For any real numbers a and b, a² = b² implies that a = b” is not true by giving a counter-example.
Solution:
The given compound statement is of the form “if p then q”
We assume that p is true, then a, b ⍷ R such that a² = b²
Let us take a = -3 and b = 3
Now, a² = b², but a ≠ b
So, when p is true, q is false.
Thus, the given compound statement is not true.

Ex 14.5 Class 11 Maths Question 3.
Show that the following statement is true by the method of contrapositive.
p: If x is an integer and x² is even, then x is also even.
Solution:
The given compound statement is of the form “if p then q”
p: x ϵ Z and x² is even.
q: x is an even integer.
We assume that q is false, then x is not an even integer.
⇒ x is an odd integer.
⇒ x² is an odd integer.
⇒ p is false
So, when q is false, p is false.
Thus, the given compound statement is true.

Ex 14.5 Class 11 Maths Question 4.
By giving a counter example, show that the following statements are not true.
(i) p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.
(ii) q: The equation x² – 1 = 0 does not have a root lying between 0 and 2.
Solution:
(i) Since the triangle is an obtuse angled triangle then 0 > 90°.
Let 0 = 100°
Also, all the angles of the triangle are equal.
∴ Sum of all angles of the triangle is 300°, which is not possible.
Thus, the given compound statement is not true,

(ii) We see that x = 1 is a root of the equation x² – 1 = 0, which lies between 0 and 2. Thus, the given compound statement is not true.

Ex 14.5 Class 11 Maths Question 5.
Which of the following statements are true and which are false? In each case give a valid reason for saying so.
(i) p. Each radius of a circle is a chord of the circle.
(ii) q: The center of a circle bisects each chord of the circle.
(iii) r. Circle is a particular case of an ellipse.
(iv) s: If x and y are integers such that x > y, then -x < -y.
(v) t.  is a rational number.
Solution:
(i) A chord of a circle is a line whose two endpoints lie on the circle and all the points on the line lie inside the circle. So, the radius of a circle is not a chord of the circle.Thus, the given statement is false.
(ii) The center of a circle bisects chord of circle when the chord is diameter of circle. When the chord is other than diameter then center of circle does not lie on the chord. Thus, the given statement is false.
(iii) In the equation of an ellipse if we put a = b, then we get an equation of circle.
Thus, the given statement is true.
(iv) It is given that x, y ϵ Z such that x > y. Multiplying both sides by negative sign, we have
x, y ϵ Z such that -x < -y.
Thus, the given statement is true.
(v) Since  cannot be expressed in the form , where a and b are integers and b ≠ 0. Thus, the given statement is false.

We hope the NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5, drop a comment below and we will get back to you at the earliest.

Evaluate the following limits in Exercises 1 to 22.

Ex 13.1 Class 11 Maths Question 1.

Solution:

Ex 13.1 Class 11 Maths Question 2.

Solution:

Ex 13.1 Class 11 Maths Question 3.

Solution:

Ex 13.1 Class 11 Maths Question 4.

Solution:

Ex 13.1 Class 11 Maths Question 5.

Solution:

Ex 13.1 Class 11 Maths Question 6.

Solution:

Ex 13.1 Class 11 Maths Question 7.

Solution:

Ex 13.1 Class 11 Maths Question 8.

Solution:

Ex 13.1 Class 11 Maths Question 9.

Solution:

Ex 13.1 Class 11 Maths Question 10.

Solution:

Ex 13.1 Class 11 Maths Question 11.

Solution:

Ex 13.1 Class 11 Maths Question 12.

Solution:

Ex 13.1 Class 11 Maths Question 13.

Solution:

Ex 13.1 Class 11 Maths Question 14.

Solution:

Ex 13.1 Class 11 Maths Question 15.

Solution:

Ex 13.1 Class 11 Maths Question 16.

Solution:

Ex 13.1 Class 11 Maths Question 17.

Solution:

Ex 13.1 Class 11 Maths Question 18.

Solution:

Ex 13.1 Class 11 Maths Question 19.

Solution:

Ex 13.1 Class 11 Maths Question 20.

Solution:

Ex 13.1 Class 11 Maths Question 21.

Solution:

Ex 13.1 Class 11 Maths Question 22.

Solution:

Ex 13.1 Class 11 Maths Question 23.

Solution:

Ex 13.1 Class 11 Maths Question 24.

Solution:

Ex 13.1 Class 11 Maths Question 25.

Solution:

Ex 13.1 Class 11 Maths Question 26.

Solution:

Ex 13.1 Class 11 Maths Question 27.

Solution:

Ex 13.1 Class 11 Maths Question 28.

Solution:

Adding (ii) and (iii), we get 2b = 8 ⇒ b = 4
Subtituting the value of b in (iii), we get
4 – a = 4 ⇒ a = 0
Thus a = 0 and b = 4.

Ex 13.1 Class 11 Maths Question 29.

Solution:

Ex 13.1 Class 11 Maths Question 30.

Solution:

Ex 13.1 Class 11 Maths Question 31.

Solution:

Ex 13.1 Class 11 Maths Question 32.

Solution:

We hope the NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.1, drop a comment below and we will get back to you at the earliest.

Ex 13.2 Class 11 Maths Question 1.
Find the derivative of x² – ² at x = 10.
Solution:
let f(x) = x² – 2
Differentiating (i) with respect to x, we get
f'(x) = 2x
At x = 10, f'(10) = 2(10) = 20.

Ex 13.2 Class 11 Maths Question 2.
Find the derivative of 99x at x = 10.
Solution:
let f(x) = 99x
Differentiating (i) with respect to x, we get
f'(x) = 90
At x = 100, f'(100) = 99.

Ex 13.2 Class 11 Maths Question 3.
Find the derivative of x at x = 10.
Solution:
let f(x) = x
Differentiating (i) with respect to x, we get
f'(x) = 1
At x = 1, f'(1) = 1.

Ex 13.2 Class 11 Maths Question 4.
Find the derivative of the following functions from first principle.
(i) x³ – 27
(ii) (x – 1)(x – 2)
(iii) 
(iv) 
Solution:

Ex 13.2 Class 11 Maths Question 5.
For the function

Prove that f'(1) = 100f'(0)
Solution:
We have

Ex 13.2 Class 11 Maths Question 6.
Find the derivative of xⁿ + ax^n-1 + a²x^n-2+
…. + a^n-1x + aⁿ for some fixed real number a.
Solution:
Let f(x) = xⁿ + ax^n-1 + a²x^n-2+
…. + a^n-1x + aⁿ
Differentiating (i) with respect to x, we get
f'(x) = nx^n-1 + (n – 1)ax^n-2 + …… + a^n-1

Ex 13.2 Class 11 Maths Question 7.
For some constants a and b, find the derivative of
(i) (x – a)(x – b)
(ii) (ax² + b)²
(iii) 
Solution:
(i) Let f(x) = (x – a)(x – b) ….(1)
Differentiating (1) with respect to x, we get
f'(x) = (x – a)(x – b)’ + (x – a)’ (x – b)
⇒ f'(x) = (x – a) + (x – b) = 2x – a – b

Ex 13.2 Class 11 Maths Question 8.
Find the derivative  for some constant a.
Solution:
Let f(x) =  ….(i), where a is a constant.
Differentiating (i) with respect to x, we get

Ex 13.2 Class 11 Maths Question 9.
Find the derivative of

Solution:
(i) Let f(x) =  …(1)
Differentiating (i) with respect to x, we get
f'(x) = 2·1 – 0 ⇒ f'(x) = 2.
(ii) Let f(x) = 5x³ + 3x – 1)(x – 1)

Ex 13.2 Class 11 Maths Question 10.
Find the derivative of cos x from first principle.
Solution:
Let f(x) = cos x

Ex 13.2 Class 11 Maths Question 11.
Find the derivative of the following functions:
(i) sin x cos x
(ii) secx
(iii) 5 secx + 4 cosx
(iv) cosecx
(v) 3 cotx + 5 cosecx
(vi) 5sinx – 6 cosx + 7
(vii) 2 tanx – 7 secx.
Solution:
(i) Let f(x) = sin x cos x … (1)
Differentiating (1) with respect to x, we get

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Ex 12.1 Class 11 Maths Question 1.
A point is on the x-axis. What are its y-coordinate and z-coordinate?
Solution:
The coordinates of any point on the x-axis will be (x, 0, 0). Thus y-coordinate and z-coordinate of the point are zero.

Ex 12.1 Class 11 Maths Question 2.
A point is in the XZ-plane. What can you say about its y-coordinate?
Solution:
The coordinates of any point in XZ-plane will be (x, 0, z). Thus y-coordinate of the point is zero.

Ex 12.1 Class 11 Maths Question 3.
Name the octants in which the following points lie:
(1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (2, -4, -7)
Solution:
Point (1, 2, 3) lies in Octant I.
Point (4, -2, 3) lies in Octant IV.
Point (4, -2, -5) lies in Octant VIII.
Point (4, 2, -5) lies in Octant V.
Point (- 4, 2, -5) lies in Octant VI.
Point (- 4, 2, 5) lies in Octant II.
Point (- 3, -1, 6) lies in Octant III.
Point (2, – 4, -7) lies in Octant VIII.

Ex 12.1 Class 11 Maths Question 4.
Fill in the blanks:
(i) The x-axis and y-axis taken together determine a plane known as ______
(ii) The coordinates of points in the XY-plane are of the form _______
(iii) Coordinate planes divide the space into ______ octants.
Solution:
(i) XY-plane
(ii) (x, y, 0)
(iii) Eight

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Ex 12.2 Class 11 Maths Question 1.
Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3)
Solution:
(i) The distance PQ between the points P(2, 3, 5) and Q(4, 3, 1) is

= 

(ii) The distance PQ between the points P(-3, 7, 2) and Q(2, 4, -1) is

(iii) The distance PQ between the points P(-1, 3, -4) and Q(1, -3, 4) is

(iv) The distance PQ between the points P(2, -1, 3) and Q(-2, 1, 3) is

Ex 12.2 Class 11 Maths Question 2.
Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Solution:
Let A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.

Now AC = AB + BC
Thus, points A, B and C are collinear.

Ex 12.2 Class 11 Maths Question 3.
Verify the following:
(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
(iii) (-1, 2, 1), (1, -2, 5), (4, -7,8) and (2, -3,4) are the vertices of a parallelogram.
Solution:
(i) Let A(0, 7, -10), B(l, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC. Then

Now, AB = BC
Thus, ABC is an isosceles triangle.

(ii) Let A(0, 7,10), B(-l, 6, 6) and C(-A, 9, 6) be three vertices of triangle ABC. Then

Now, AC² = AB² + BC²
Thus, ABC is a right angled triangle.

(iii) Let A(-1, 2, 1), B(1, -2, 5) and C(4, -7, 8) and D(2, -3,4) be four vertices of quadrilateral ABCD. Then

Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.

Ex 12.2 Class 11 Maths Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Solution:
Let A(x, y, z) be any point which is equidistant from points B(1, 2, 3) and C(3, 2, -1).

Ex 12.2 Class 11 Maths Question 5.
Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(-4,0,0) is equal to 10.
Solution:
Let P(x, y, z) be any point.

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Ex 12.3 Class 11 Maths Question 1.
Find the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) in the ratio
(i) 2 : 3 internally,
(ii) 2 : 3 enternally
Solution:
(i) Let P(x, y, z) be any point which divides the line segment joining the points A(-2, 3, 5) and B(1, -4, 6) in the ratio 2 : 3 internally.

(ii) Let P(x, y, z) be any point which divides the line segment joining the points 71 (-2, 3, 5) and B(1, -4, 6) in the ratio 2 : 3 externally. Then

Ex 12.3 Class 11 Maths Question 2.
Given that P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear. Find the ratio in which Q divides PR.
Solution:
Let Q(5, 4, -6) divides the line segment joining the points P(3, 2, -4) and R(9, 8, -10) in the ratio k : 1 internally.

Ex 12.3 Class 11 Maths Question 3.
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).
Solution:
Let the line segment joining the points A(-2, 4, 7) and B(3, -5, 8) be divided by the YZ -plane at a point C in the ratio k : 1.

Ex 12.3 Class 11 Maths Question 4.
Using section formula, show that the points A(2, -3, 4), B(-1, 2, 1) and C are collinear.
Solution:
Let the points A(2, -3, 4), B(-l, 2,1) and C be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of P are 
Let us examine whether for some value of k, the point P coincides with point C.

AB internally in the ratio 2:1. Hence A, B, C are collinear.

Ex 12.3 Class 11 Maths Question 5.
Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6).
Solution:
Let R and S be two points which trisect the line segment PQ.

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In each of the following Exercises 1 to 5, find the equation of the circle with

Ex 11.1 Class 11 Maths Question 1.
centre (0, 2) and radius 2
Solution:
Here h = 0,k = 2 and r = 2
The equation of circle is,
(x-h)² + (y- k)² = r²
∴ (x – 0)² + (y – 2)² = (2)²
⇒ x² + y² + 4 – 4y = 4
⇒ x² + y² – 4y = 0

Ex 11.1 Class 11 Maths Question 2.
centre (-2,3) and radius 4
Solution:
Here h=-2,k = 3 and r = 4
The equation of circle is,
(x – h)² + (y – k)² = r²
∴(x + 2)² + (y – 3)² = (4)²
⇒ x² + 4 + 4x + y² + 9 – 6y = 16
⇒ x² + y² + 4x – 6y – 3 = 0

Ex 11.1 Class 11 Maths Question 3.
centre  and radius 
Solution:
here h = , k =  and r = 
The equation of circle is,

Ex 11.1 Class 11 Maths Question 4.
centre (1, 1) and radius 
Solution:
Here h = l, k=l and r = 
The equation of circle is,
(x – h)² + (y – k)² = r²
∴ (x – 1)² + (y – 1)² = ²
⇒ x² + 1 – 2x + y² +1 – 2y = 2
⇒ x² + y² – 2x – 2y = 0

Ex 11.1 Class 11 Maths Question 5.
centre (-a, -b) and radius .
Solution:
Here h=-a, k = -b and r = 
The equation of circle is, (x – h)² + (y – k)² = r²
∴ (x + a)² + (y + b)² = 
⇒ x² + a² + 2ax + y² + b² + 2by = a² -b²
⇒ x² + y² + 2ax + 2 by + 2b² = 0

In each of the following exercises 6 to 9, find the centre and radius of the circles.

Ex 11.1 Class 11 Maths Question 6.
(x + 5)² + (y – 3)² = 36
Solution:
The given equation of circle is,
(x + 5)² + (y – 3)² = 36
⇒ (x + 5)² + (y – 3)² = (6)²
Comparing it with (x – h)2² + (y – k)² = r², we get
h = -5, k = 3 and r = 6.
Thus the co-ordinates of the centre are (-5, 3) and radius is 6.

Ex 11.1 Class 11 Maths Question 7.
x² + y² – 4x – 8y – 45 = 0
Solution:
The given equation of circle is
x² + y² – 4x – 8y – 45 = 0
∴ (x² – 4x) + (y² – 8y) = 45
⇒ [x² – 4x + (2)²] + [y² – 8y + (4)²] = 45 + (2)² + (4)²
⇒ (x – 2)² + (y – 4)² = 45 + 4 + 16
⇒ (x – 2)² + (y – 4)² = 65
⇒ (x – 2)² + (y – 4)²= 
Comparing it with (x – h)² + (y – k)² = r², we
have h = 2,k = 4 and r = .
Thus co-ordinates of the centre are (2, 4) and radius is .

Ex 11.1 Class 11 Maths Question 8.
x² + y² – 8x + 10y – 12 = 0
Solution:
The given equation of circle is,
x² + y² – 8x + 10y -12 = 0
∴ (x² – 8x) + (y² + 10y) = 12
⇒ [x² – 8x + (4)²] + [y² + 10y + (5)²] = 12 + (4)² + (5)²
⇒ (x – 4)² + (y + 5)² = 12 + 16 + 25
⇒ (x – 4)² + (y + 5)² = 53
⇒ (x – 4)² + (y + 5)² = 
Comparing it with (x – h)² + (y – k)² = r², we have h = 4, k = -5 and r = 
Thus co-ordinates of the centre are (4, -5) and radius is .

Ex 11.1 Class 11 Maths Question 9.
2x² + 2y² – x = 0
Solution:
The given equation of circle is,
2x² + 2y² – x = 0

Ex 11.1 Class 11 Maths Question 10.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
The equation of the circle is,
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (4, 1)
∴ (4 – h)² + (1 – k)² = r²
⇒ 16 + h² – 8h + 1 + k² – 2k = r²
⇒ h²+ k² – 8h – 2k + 17 = r² …. (ii)
Also, the circle passes through point (6, 5)
∴ (6 – h² + (5 – k)² = r²
⇒ 36 + h² -12h + 25 + k² – 10k = r²
⇒ h² + k² – 12h – 10kk + 61 = r² …. (iii)
From (ii) and (iii), we have h² + k² – 8h – 2k +17
= h² + k²– 12h – 10k + 61
⇒ 4h + 8k = 44 => h + 2k = ll ….(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
∴ 4h + k = 16 …(v)
Solving (iv) and (v), we get h = 3 and k = 4.
Putting value of h and k in (ii), we get
(3)² + (4)² – 8 x 3 – 2 x 4 + 17 = r²
∴ r² = 10
Thus required equation of circle is
(x – 3)² + (y – 4)² = 10
⇒ x² + 9 – 6x + y² +16 – 8y = 10
⇒ x² + y² – 6x – 8y +15 = 0.

Ex 11.1 Class 11 Maths Question 11.
Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre is on the line x – 3y – 11 = 0.
Solution:
The equation of the circle is,
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (2, 3)
∴ (2 – h)² + (3 – k)² = r²
⇒ 4 + h² – 4h + 9 + k² – 6k = r²
⇒ h²+ k² – 4h – 6k + 13 = r² ….(ii)
Also, the circle passes through point (-1, 1)
∴ (-1 – h)² + (1 – k)² = r²
⇒ 1 + h² + 2h + 1 + k² – 2k = r²
⇒ h² + k² + 2h – 2k + 2 = r² ….(iii)
From (ii) and (iii), we have
h² + k² – 4h – 6k + 13 = h² + k² + 2h – 2k + 2
⇒ -6h – 4k = -11 ⇒ 6h + 4k = 11 …(iv)
Since the centre (h, k) of the circle lies on the line x – 3y-11 = 0.
∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v)
Solving (iv) and (v), we get
h =  and k = 
Putting these values of h and k in (ii), we get

⇒  ⇒ 
Thus required equation of circle is
⇒ 
⇒ 
⇒ 4x² + 49 – 28x + 4y² + 25 + 20y = 130
⇒ 4x² + 4y² – 28x + 20y – 56 = 0
⇒ 4(x² + y² – 7x + 5y -14) = 0
⇒ x² + y² – 7x + 5y -14 = 0.

Ex 11.1 Class 11 Maths Question 12.
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution:
Since the centre of the circle lies on x-axis, the co-ordinates of centre are (h, 0). Now the circle passes through the point (2, 3).
∴ Radius of circle

Ex 11.1 Class 11 Maths Question 13.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.
Solution:
Let the circle makes intercepts a with x-axis and b with y-axis.
∴ OA = a and OB = b
So the co-ordinates of A are (a, 0) and B are (0,b)
Now, the circle passes through three points 0(0, 0), A(a, 0) and B(0, b).

Ex 11.1 Class 11 Maths Question 14.
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Solution:
The equation of circle is
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2)
∴ radius of circle

Ex 11.1 Class 11 Maths Question 15.
Does the point (-2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?
Solution:
The equation of given circle is x² + y² = 25
⇒ (x – 0)² + (y – 0)² = (5)²
Comparing it with (x – h)² + (y – k)² = r², we
get
h = 0,k = 0, and r = 5
Now, distance of the point (-2.5, 3.5) from the centre (0, 0)

 = 4.3 < 5.
Thus the point (-2.5, 3.5) lies inside the circle.

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In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

Ex 11.2 Class 11 Maths Question 1.
y²= 12x
Solution:
The given equation of parabola is y² = 12x which is of the form y² = 4ax.
∴ 4a = 12 ⇒ a = 3
∴ Coordinates of focus are (3, 0)
Axis of parabola is y = 0
Equation of the directrix is x = -3 ⇒ x + 3 = 0
Length of latus rectum = 4 x 3 = 12.

Ex 11.2 Class 11 Maths Question 2.
x² = 6y
Solution:
The given equation of parabola is x² = 6y which is of the form x² = 4ay.

Ex 11.2 Class 11 Maths Question 3.
y² = – 8x
Solution:
The given equation of parabola is
y² = -8x, which is of the form y² = – 4ax.
∴ 4a = 8 ⇒ a = 2
∴ Coordinates of focus are (-2, 0)
Axis of parabola is y = 0
Equation of the directrix is x = 2 ⇒ x – 2 = 0
Length of latus rectum = 4 x 2 = 8.

Ex 11.2 Class 11 Maths Question 4.
x² = -16y
Solution:
The given equation of parabola is
x² = -16y, which is of the form x² = -4ay.
∴ 4a = 16 ⇒ a = 4
∴ Coordinates of focus are (0, -4)
Axis of parabola is x = 0
Equation of the directrix is y = 4 ⇒ y – 4 = 0
Length of latus rectum = 4 x 4 = 16.

Check the Parabola Calculator to solve Parabola Equation.

Ex 11.2 Class 11 Maths Question 5.
y²= 10x
Solution:
The given equation of parabola is y² = 10x, which is of the form y² = 4ax.

Ex 11.2 Class 11 Maths Question 6.
x² = -9y
Solution:
The given equation of parabola is
x² = -9y, which is of the form x² = -4ay.

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

Ex 11.2 Class 11 Maths Question 7.
Focus (6, 0); directrix x = -6
Solution:
We are given that the focus (6, 0) lies on the x-axis, therefore x-axis is the axis of parabola. Also, the directrix is x = -6 i.e. x = -a and focus (6, 0) i.e. (a, 0). The equation of parabola is of the form y² = 4ax.
The required equation of parabola is
y² = 4 x 6x ⇒ y² = 24x.

Ex 11.2 Class 11 Maths Question 8.
Focus (0, -3); directri xy=3
Solution:
We are given that the focus (0, -3) lies on the y-axis, therefore y-axis is the axis of parabola. Also the directrix is y = 3 i.e. y = a and focus (0, -3) i.e. (0, -a). The equation of parabola is of the form x² = -4ay.
The required equation of parabola is
x² = – 4 x 3y ⇒ x² = -12y.

Ex 11.2 Class 11 Maths Question 9.
Vertex (0, 0); focus (3, 0)
Solution:
Since the vertex of the parabola is at (0, 0) and focus is at (3, 0)
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y² = 4ax
The required equation of the parabola is
y² = 4 x 3x ⇒ y² = 12x.

Ex 11.2 Class 11 Maths Question 10.
Vertex (0, 0); focus (-2, 0)
Solution:
Since the vertex of the parabola is at (0, 0) and focus is at (-2, 0).
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y² = – 4ax
The required equation of the parabola is
y² = – 4 x 2x ⇒ y² = -8x.

Ex 11.2 Class 11 Maths Question 11.
Vertex (0, 0), passing through (2, 3) and axis is along x-axis.
Solution:
Since the vertex of the parabola is at (0, 0) and the axis is along x-axis.
∴ The equation of the parabola is of the form y² = 4ax
Since the parabola passes through point (2, 3)

Ex 11.2 Class 11 Maths Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.
Solution:
Since the vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
∴ The equation of the parabola is of the form x² = 4ay
Since the parabola passes through point (5, 2)

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.2, drop a comment below and we will get back to you at the earliest.

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

Ex 11.3 Class 11 Maths Question 1.

Solution:
Given equation of ellipse of 
Clearly, 36 > 16
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 2.

Solution:
Given equation of ellipse is 
Clearly, 25 > 4
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 3.

Solution:
Given equation of ellipse is 
Clearly, 16 > 9
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 4.

Solution:
Given equation of ellipse is 
Clearly, 100 > 25
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 5.

Solution:
Given equation of ellipse is 
Clearly, 49 > 36
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 6.

Solution:
Given equation of ellipse is 
Clearly, 400 > 100
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 7.
36x² + 4y² = 144
Solution:
Given equation of ellipse is 36x² + 4y² = 144

Ex 11.3 Class 11 Maths Question 8.
16x² + y² = 16
Solution:
Given equation of ellipse is16x² + y² = 16

Ex 11.3 Class 11 Maths Question 9.
4x² + 9y² = 36
Solution:
Given equation of ellipse is4x² + 9y² = 36

In each 0f the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

Ex 11.3 Class 11 Maths Question 10.
Vertices (±5, 0), foci (±4,0)
Solution:
Clearly, The foci (±4, o) lie on x-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 11.
Vertices (0, ±13), foci (0, ±5)
Solution:
Clearly, The foci (0, ±5) lie on y-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 12.
Vertices (±6, 0), foci (±4,0)
Solution:
Clearly, The foci (±4, 0) lie on x-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 13.
Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Solution:
Since, ends of major axis (±3, 0) lie on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 14.
Ends of major axis (0, ), ends of minor axis (±1, 0)
Solution:
Since, ends of major axis (0, ) lie on i-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 15.
Length of major axis 26, foci (±5, 0)
Solution:
Since the foci (±5, 0) lie on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 16.
Length of major axis 16, foci (0, ±6)
Solution:
Since the foci (0, ±6) lie on y-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 17.
Foci (±3, 0) a = 4
Solution:
since the foci (±3, 0) on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 18.
b = 3, c = 4, centre at the origin; foci on the x axis.
Solution:
Since the foci lie on x-axis.
∴ The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 19.
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6)
Solution:
Since the major axis is along y-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 20.
Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Solution:
Since the major axis is along the x-axis.
∴ The equation of ellipse in standard form is

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.3, drop a comment below and we will get back to you at the earliest.

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, eccentricity and the length of the latus rectum of the hyperbolas.

Ex 11.4 Class 11 Maths Question 1.

Solution:
Given equation of hyperbola is

Ex 11.4 Class 11 Maths Question 2.

Solution:
Given equation of hyperbola is

Ex 11.4 Class 11 Maths Question 3.
9y² – 4x² = 36
Solution:
Given equation of hyperbola is9y² – 4x² = 36

Ex 11.4 Class 11 Maths Question 4.
16x² – 9y² = 576
Solution:
Given equation of hyperbola is16x² – 9y² = 576

Ex 11.4 Class 11 Maths Question 5.
5y² – 9x² = 36
Solution:
Given equation of hyperbola is
5y² – 9x² = 36

Ex 11.4 Class 11 Maths Question 6.
49y² – 16x² = 784
Solution:
Given equation of hyperbola is49y² – 16x² = 784

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

Ex 11.4 Class 11 Maths Question 7.
Vertices (±2,0), foci (±3,0)
Solution:
Vertices are (±2, 0) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 8.
Vertices (0, ±5), foci (0, ±8)
Solution:
Vertices are (0, ±5) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 9.
Vertices (0, ±3), foci (0, ±5)
Solution:
Vertices are (0, ±3) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 10.
Foci (±5, 0), the transverse axis is of length 8.
Solution:
Here foci are (±5, 0) which lie on x-axis. So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 11.
Foci (0, ±13), the conjugate axis is of length 24.
Solution:
Here foci are (0, ±13) which lie on y-axis.
So the equation of hyperbola in standard

Ex 11.4 Class 11 Maths Question 12.
Foci (,0) , the latus rectum is of length 8.
Solution:
Here foci are (, 0) which lie on x-axis.
So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 13.
Foci (±4, 0), the latus rectum is of length 12.
Solution:
Here foci are (±4, 0) which lie on x-axis.
So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 14.
Vertices (+7, 0), e = 
Solution:
Here vertices are (±7, 0) which lie on x-axis.
So, the equation of hyperbola in standard

Ex 11.4 Class 11 Maths Question 15.
Foci (0, ), passing through (2, 3).
Solution:
Here foci are (0, ) which lie on y-axis.
So the equation of hyperbola in standard form

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.4, drop a comment below and we will get back to you at the earliest.