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NCERT Class 11 Mathematics Chapter 12 Limits and Derivatives
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Limits and Derivatives
Chapter – 12
| Exercise 12.1 |
Evaluate the following limits in Exercises 1 to 22.
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Ans: At x = 2 the value of the given rational function takes the form 0/0.
Ans: At x = 2 the value of the given rational function takes the form 0/0 .
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At x = – 2 , the value of the given function takes the form 0/0
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At x = 0 , the value of the given function takes the form 0/0.
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At x = 0 , the value of the given function takes the form 0/0.
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At x = 0 , the value of the given function takes the form 0/0.
Now,
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At x = 0 , the value of the given function takes the form 0/0.
Now,
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Ans: At x = 0, the value of the given function takes the form 0/0.
Now,
Ans: At x = 0, the value of the given function takes the form∞-∞.
Now,
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Ans: The given function is
Ans: The given function is
Ans: The given function is
Ans: The given function is
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Ans: The given function is
⇒ a + b = 4 and b – a = 4
On solving these two equations, we obtain a = 0 and b = 4
Thus, the respective possible values of a and b are 0 and 4.
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Ans: The given function is
when a < 0,
when a > 0
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Ans: The given function is
| Exercise 12.1 |
1. Find the derivative of x² – 2 at x = 10
Ans: Let f(x) = x² – 2. Accordingly,
Thus, the derivative of x² – 2 at x = 10 is 20.
2. Find the derivative of x at x = 1
Ans: Letf(x) = x. Accordingly,
Thus, the derivative of x at x = 1 is 1.
3. Find the derivative of 99x at x = 100
Ans: Let f(x) = 99x . Accordingly,
Thus, the derivative of 99x at x = 100 is 99.
4. Find the derivative of the following functions from first principle.
(i) x³ – 27
Ans: Let f(x) = x3 – 27 According to the first principle,
(ii) (x – 1)(x – 2)
Ans: Let f(x) = (x – 1)(x – 2) Thus according to first principle,
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5. For the function
Prove that f'(1) = 100f'(0) .
Ans: The given function is
= x⁹⁹ + x⁹⁸ +…+x+1
∴ f’ (x)= x⁹⁹ + x⁹⁸ +…+x+1
At x = 0
f’ (0) = 1,
At x = 1,
f'(1) = 1⁹⁹+1⁹⁸ +…+1+1=[1+1+…+1+1]₁₀₀ ₜₑᵣₘₛ
=1×100=100
Thus, f’ (1) = 100 × f’ (0)
6. Find the derivative of xⁿ + axⁿ⁻¹ + a²xⁿ⁻² +…+aⁿ⁻¹x+aⁿ for some fixed real number a.
Ans: Let f(x)= xⁿ + axⁿ⁻¹ + a² xⁿ⁻² +…+aⁿ⁻¹ x + aⁿ
f’ (x)= nxⁿ⁻¹ + a(n – 1) xⁿ⁻² + a² (n – 2) xⁿ⁻³ +…+aⁿ⁻¹ +aⁿ (0)
= nxⁿ⁻¹ + a(n – 1) xⁿ⁻² + a² (n – 2) xⁿ⁻³ +…+aⁿ⁻¹
7. For some constants a and b, find the derivative of
(i) (x – a)(x – b)
Ans: Let f(x) = (x – a)(x – b)
⇒ f(x) = x² – (a + b)x + ab
f’ (x) = 2x – (a + b) + 0 = 2x – a – b
(ii) (ax² + b)²
Ans: Let f(x) = (ax²+b)²
⇒ f(x) = a²x⁴ + 2abx² + b²
f’ (x) = a² (4x³) + 2ab(2x) + b² (0)
= 4a²x³ + 4abx
= 4ax(ax² + b)
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By quotient rule,
8. Find the derivative of for some constant a.
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9. Find the derivative of
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= 2 – 0
= 2
(ii) (5x³ + 3x – 1)(x – 1)
Ans: Let f(x) = (5x³ + 3x – 1)(x – 1)
By Leibnitz product rule,
= (5x³+3x-1)(1)+(x-1)(5.3x² +3-0)
= (5x³ + 3x – 1) + (x – 1)(15x² + 3)
= 5x³ + 3x – 1 + 15x³ + 3x – 15x² – 3
= 20x³ – 15x² + 6x – 4
(iii) x⁻³ (5 + 3x)
Ans: Let f(x) = x⁻³ (5 + 3x)
By Leibnitz product rule,
= x⁻³ (0 + 3) + (5 + 3x)(- 3x⁻³⁻¹)
= x⁻³ (3) + (5 + 3x)(- 3x⁻⁴)
= 3x⁻³ – 15x⁻⁴ – 9x⁻³
= – 6x⁻³ – 15x⁻⁴
(iv) x⁵ (3 – 6x⁻⁹)
Ans: Let f(x) = x⁵ (3 – 6x⁻⁹).
By Leibnitz product rule,
= x⁵ {0 – 6(- 9) x⁻⁹⁻¹} + (3 – 6x⁻⁹)(5x⁴)
= x⁵ (54x⁻¹⁰) + 15x⁴ – 30x⁻⁵
= 54x⁻⁵ + 15x⁴ – 30x⁻⁵
= 24x⁻⁵ + 15x⁴
(v) x⁻⁴ (3 – 4x⁻⁵)
Ans: Let f(x) = x⁻⁴ (3 – 4x⁻⁵)
By Leibnitz product rule
= x⁻⁴ {0 – 4(- 5) x⁻⁵⁻¹} + (3 – 4x⁻⁵)(- 4) x⁻⁴⁻¹
= x⁻⁴ (20x⁻⁶) + (3 – 4x⁻⁵)(- 4x⁻⁵)
= 20x⁻¹⁰ – 12x⁻⁵ + 16x⁻¹⁰
= 36x⁻¹⁰ – 12x⁻⁵
Ans: Let f(x) =
10. Find the derivative of cos x from first principle.
Ans: Let f(x) = cos x Accordingly, from the first principle,
11. Find the derivative of the following functions:
(i) sin x cos x
Ans: Letf (x) = sin x cos x. Accordingly, from the first principle,
(ii) sec x
Ans: Letf (x) = sec x. Accordingly, from the first principle,
(iii) 5sec(x) + 4cos x
Ans: Letf(x) = 5sec(x) + 4cos x Accordingly, from the first principle,
(iv) cosec x
Ans: Let f (x) = cosec x. Accordingly, from the first principle,
(v) 3 cot x + 5 cosec x
Ans: Let f(x) = 3 cot x + 5 cosec x Accordingly, from the first principle,
= – cosec x cot x …..(3)
From (1), (2), and (3), we obtain
f’ (x) = – 3cosec² x – 5 cosec x cot x.
(vi) 5sin x – 6cos x + 7
Ans: Let f(x) = 5sin x – 6cos x + 7.Accordingly, from the first principle,
= 5 cos x.1-6[(-cosx).(0)-sinx.1]
= 5 cosx+6 sin x
(vii) 2 tan x – 7 sec x
Ans: Let f(x) = 2 tan x – 7 sec x.Accordingly, from the first principle,
| Miscellaneous Exercise on Chapter 12 |
1. Find the derivative of the following functions from first principle:
(i) -x
Ans: Let f(x) = -x.Accordingly, f(x+h) = -(x+h)
By first principle
(ii) (- x)⁻¹
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(iii) sin(x + 1)
Ans: Let f(x) = sin(x + 1) . Accordingly, f(x + h) = sin(x + h + l)
By first principle,
Ans: let
By first principal,
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)
Ans: Let f(x) = x + a. Accordingly, f(x+h)=x+h+a
By first principle,
3.
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4. (a x + b) (cx + d)²
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Let
By quotient rule,
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11. 4√x – 2.
Ans: Let f (x) = 4√x – 2.
12. (a x + b)ⁿ.
Ans: Let f (x) = (ax+b)ⁿ. Accordingly. f (x+h) = {a(x+h)+b}ⁿ = (ax+ah+b)ⁿ
By first principle,
13. (ax + b)ⁿ (cx + d)ᵐ
Ans: Let f(x) = (ax+b)ⁿ (cx+d)ᵐ
By Leibnitz product rule,
Therefore, from (1), (2), and (3), we obtain
f’ (x) = (ax + b)ⁿ {mc(cx + d)ᵐ⁻¹} + (cx + d)ᵐ {na (ax + b)ⁿ⁻¹}
= (ax + b)ⁿ⁻¹ (cx + d)ᵐ⁻¹ [mc(ax + b) + na(cx + d)]
14. sin(x + a)
Ans: Let f(x) = sin(x + a)
f(x + h) = sin(x + h + a)
By first principle,
15. cosec x cot x
Ans: Let f(x) cosec x cot x
By Leibnitz product rule,
f’ (x) = cosec x (cot x)’ + cot x(cosec x)’ …(1)
Let f₁ (x) = cot x . Accordingly. f₁ (x + h) = cot(x + h)
By first principle,
=-cosec²x
∴ (cot x)’ = – cosec² x ……(2)
Now, let f₂(x) = cosec x. Accordingly, f₂(x+h) = cosec(x+h)
By first principle,
= – cos ecx . cot x
∴ (cosec x)’ =-cos x.cot x …..(3)
From (1), (2), and (3), we obtain
f’ (x) = cosec x (- cosec² x) + cot x (- cosec x cot x)
= – coseck³ x – cot² x cosec x.
Ans:
Ans: Let
By quotient rule,
Ans:
By quotient rule,
19. sinⁿ x
Ans: Let y = sinⁿ x.
Accordingly, for n = 1 , y = sin x.
For n = 2 , y = sin² x.
= (sin x)’ sin x + sin x (sin x)’ [By Leibnitz product rule]
= cos x sin x + sin x cos x
= 2sin x cos x ……(1)
For n = 3 , y = sin³ x
= (sin x)’ sin² x + sin x (sin² x)’ [By Leibnitz product rule]
= cos x sin² x + sin x (2sin x cos x) [Using (1)]
= cos x sin² x + 2sin² x cos x
= 3sin² x cos x
We assert that
Let our assertion be true for n = k
= (sin x)’ sinᵏ x + sin x (sinᵏ x)’ [By Leibnitz product rule]
= cos x sinᵏ x + sin x (k sinᵏ⁻¹ x cos x) [Using (2)]
= cos x sinᵏ x + k sinᵏ x cos x
= (k + 1) sinᵏ x cos x
Thus, our assertion is true for n = k + 1
Hence, by mathematical induction
Ans:
Ans:
Let g(x) = sin(x+a). Accordingly, g(x+h) = sin(x+h+a)
By first principle,
= cos(x + a) ……(ii)
From (i) and (ii), we obtain
22. x⁴ (5sin x – 3cos x)
Ans: Let f(x) = x⁴ (5sin x – 3cos x)
By product rule,
= x⁴ [5cos x – 3(- sin x)] + (5sin x – 3cos x)(4x³)
= x³ [5x cos x + 3x sin x + 20sin x – 12cos x]
23. (x² + 1) cos x
Ans: Let f(x) = (x² + 1) cos x
By product rule,
= (x² + 1)(- sin x) + cos x (2x)
= – x² sin x – sin x + 2x cos x
24. (ax² + sin x)(p + q cos x).
Ans: Let f(x) = (ax² + sin x)(p + q cos x)
By product rule,
= (ax² + sin x)(- q sin x) + (p + q cos x)(2ax + cos x)
= – q sin x (ax² + sin x) + (p + q cos x)(2ax + cos x)
25. (x + cos x)(x – tan x)
Ans: Let f(x) = (x+cosx)(x-tanx)
By product rule,
Let g(x) = tan x Accordingly, g(x + h) = tan(x + h)
By first principle,
= sec² x … (ii)
Therefore, from (i) and (ii), we obtain
f’ (x) = (x + cos x)(1 – sec² x) + (x – tan x)(1 – sin x)
= (x + cos x)(- tan² x) + (x – tan x)(1 – sin x)
= – tan² x (x + cos x) + (x – tan x)(l – sin x)
Ans: Let
By quotient rule,
Ans: Let
By quotient rule,
Ans:
Let g(x) = 1 + tan x Accordingly, g(x + h) = 1 + tan(x + h)
By first principle,
From (i) and (ii), we obtain
29. (x + sec(x))(x – tan x)
Ans: Let f(x) = (x + sec x)(x – tan x)
By product rule,
From (i), (ii), and (iii), we obtain
f’ (x) = (x + sec x)(l – sec² x) + (x – tan x)(1 + sec x tan x)
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