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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