CLASS 11 MATH CHAPTER 13 NCERT SOLUTION

CLASS 11 MATH CHAPTER 13 NCERT SOLUTION

CH13 – LIMITS AND DERIVATIVES

Exercise 13.1 page no: 301

1. Evaluate the given limit:

Solution:

Given,

Substituting x = 3, we get

= 3 + 3

= 6

2. Evaluate the given limit:

Solution:

Given limit,

Substituting x = π, we get

  = (π – 22 / 7)

3. Evaluate the given limit:

Solution:

Given limit,

Substituting r = 1, we get

 = π(1)2

= π

4. Evaluate the given limit:

Solution:

Given limit,

Substituting x = 4, we get

 = [4(4) + 3] / (4 – 2)

= (16 + 3) / 2

= 19 / 2

5. Evaluate the given limit:

Solution:

Given limit,

Substituting x = -1, we get

= [(-1)10 + (-1)5 + 1] / (-1 – 1)

= (1 – 1 + 1) / – 2

= – 1 / 2

6. Evaluate the given limit:

Solution:

Given limit,

= [(0 + 1)5 – 1] / 0

=0

Since this limit is undefined,

Substitute x + 1 = y, then x = y – 1

7. Evaluate the given limit:

Solution:

8. Evaluate the given limit:

Solution:

9. Evaluate the given limit:

Solution:

= [a (0) + b] / c (0) + 1

= b / 1

= b

10. Evaluate the given limit:

Solution:

11. Evaluate the given limit:

Solution:

Given limit,

Substituting x = 1,

= [a (1)2 + b (1) + c] / [c (1)2 + b (1) + a]

= (a + b + c) / (a + b + c)

Given,

= 1

12. Evaluate the given limit:

Solution:

By substituting x = – 2, we get

13. Evaluate the given limit:

Solution:

Given

14. Evaluate the given limit:

Solution:

15. Evaluate the given limit:

Solution:

16. Evaluate the given limit:

Solution:

17. Evaluate the given limit:

Solution:

18. Evaluate the given limit:

Solution:

19. Evaluate the given limit:

Solution:

20. Evaluate the given limit:

Solution:

21. Evaluate the given limit:

Solution:

22. Evaluate the given limit:

Solution:

23.

Solution:

24. Find

 , where

Solution:

25. Evaluate

 , where f(x) =

Solution:

26. Find

 , where f (x) =

Solution:

27. Find

 , where

Solution:

28. Suppose

 and if

 what are the possible values of a and b?

Solution:

29. Let a1, a2,………an be fixed real numbers and define a function

f (x) = (x – a1) (x – a2) ……. (x – an).

What is

 For some a ≠ a1, a2, ……. an, compute

 .

Solution:

30. If For what value (s) of a does exist?

Solution:

31. If the function f(x) satisfies , evaluate

Solution:

32. If For what integers m and n does both and exist?

Solution:

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Exercise 13.2 page no: 312

1. Find the derivative of x2– 2 at x = 10.

Solution:

Let f (x) = x2 – 2

2. Find the derivative of x at x = 1.

Solution:

Let f (x) = x

Then,

3. Find the derivative of 99x at x = l00.

Solution:

Let f (x) = 99x,

From the first principle,

= 99

4. Find the derivative of the following functions from the first principle.

(i) x3 – 27

(ii) (x – 1) (x – 2)

(iii) 1 / x2

(iv) x + 1 / x – 1

Solution:

(i) Let f (x) = x3 – 27

From the first principle,

(ii) Let f (x) = (x – 1) (x – 2)

From the first principle,

(iii) Let f (x) = 1 / x2

From the first principle, we get

(iv) Let f (x) = x + 1 / x – 1

From the first principle, we get

5. For the function , prove that f’ (1) =100 f’ (0).

Solution:

6. Find the derivative of for some fixed real number a.

Solution:

7. For some constants a and b, find the derivative of

(i) (x − a) (x − b)

(ii) (ax2 + b)2

(iii) x – a / x – b

Solution:

(i) (x – a) (x – b)

(ii) (ax2 + b)2

= 4ax (ax2 + b)

(iii) x – a / x – b

8. Find the derivative of for some constant a.

Solution:

9. Find the derivative of

(i) 2x – 3 / 4

(ii) (5x3 + 3x – 1) (x – 1)

(iii) x-3 (5 + 3x)

(iv) x5 (3 – 6x-9)

(v) x-4 (3 – 4x-5)

(vi) (2 / x + 1) – x2 / 3x – 1

Solution:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

10. Find the derivative of cos x from the first principle.

Solution:

11. Find the derivative of the following functions.

(i) sin x cos x

(ii) sec x

(iii) 5 sec x + 4 cos x

(iv) cosec x

(v) 3 cot x + 5 cosec x

(vi) 5 sin x – 6 cos x + 7

(vii) 2 tan x – 7 sec x

Solution:

(i) sin x cos x

(ii) sec x

(iii) 5 sec x + 4 cos x

(iv) cosec x

(v) 3 cot x + 5 cosec x

(vi)5 sin x – 6 cos x + 7

(vii) 2 tan x – 7 sec x

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Miscellaneous exercise page no: 317

1. Find the derivative of the following functions from the first principle.

(i) –x

(ii) (–x)–1

(iii) sin (x + 1)

(iv)

Solution:

(ii) (-x)-1

= 1 / x2

(iii) sin (x + 1)

(iv)

We get,

Find the derivative of the following functions. (It is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants, and m and n are integers.)

2. (x + a)

Solution:

3. (px + q) (r / x + s)

Solution:

4. (ax + b) (cx + d)2

Solution:

5. (ax + b) / (cx + d)

Solution:

6. (1 + 1 / x) / (1 – 1 / x)

Solution:

7. 1 / (ax2 + bx + c)

Solution:

8. (ax + b) / px2 + qx + r

Solution:

9. (px2 + qx + r) / ax + b

Solution:

10. (a / x4) – (b / x2) + cos x

Solution:

11.

Solution:

12. (ax + b)n

Solution:

13. (ax + b)n (cx + d)m

Solution:

14. sin (x + a)

Solution:

15. cosec x cot x

Solution:

So, we get

16.

Solution:

17.

Solution:

18.

Solution:

19. sinn x

Solution:

20.

Solution:

21.

Solution:

22. x4 (5 sin x – 3 cos x)

Solution:

23. (x2 + 1) cos x

Solution:

24. (ax2 + sin x) (p + q cos x)

Solution:

25.

Solution:

26.

Solution:

27.

Solution:

28.

Solution:

29. (x + sec x) (x – tan x)

Solution:

30.

Solution:

Now, let us substitute the values, and we get

= 6

∴ The coordinates of the required point are (4, -2, 6).

6. If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.

Solution:

Given:

Points A (3, 4, 5) and B (-1, 3, -7)

x1 = 3, y1 = 4, z1 = 5;

x2 = -1, y2 = 3, z2 = -7;

PA2 + PB2 = k2 ……….(1)

Let the point be P (x, y, z).

Now, by using the distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

So,

And

Now, substituting these values in (1), we have

[(3 – x)2 + (4 – y)2 + (5 – z)2] + [(-1 – x)2 + (3 – y)2 + (-7 – z)2] = k2

[(9 + x2 – 6x) + (16 + y2 – 8y) + (25 + z2 – 10z)] + [(1 + x2 + 2x) + (9 + y2 – 6y) + (49 + z2 + 14z)] = k2

9 + x2 – 6x + 16 + y2 – 8y + 25 + z2 – 10z + 1 + x2 + 2x + 9 + y2 – 6y + 49 + z2 + 14z = k2

2x2 + 2y2 + 2z2 – 4x – 14y + 4z + 109 = k2

2x2 + 2y2 + 2z2 – 4x – 14y + 4z = k2 – 109

2 (x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109

(x2 + y2 + z2 – 2x – 7y + 2z) = (k2 – 109)/2

Hence, the required equation is (x2 + y2 + z2 – 2x – 7y + 2z) = (k2 – 109)/2.


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