Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2
(I) Differentiate the following w.r.t. x
Question 1
Maharashtra Board Solution
y =
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Question 2
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y = √x + tan x – x3
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Question 3
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y = log x – cosec x +
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Question 4
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y =
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Question 5
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y = 7x + x7 – x√x – log x + 77
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Question 6
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y = 3 cot x – 5ex + 3 log x –
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(II) Diffrentiate the following w.r.t. x
Question 1
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y = x5 tan x
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Question 2
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y = x3 log x
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Question 3
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y = (x2 + 2)2 sin x
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Question 4
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y = ex log x
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Question 5
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y =
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Question 6
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y =
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(III) Diffrentiate the following w.r.t. x
Question 1
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y = x2√x + x4 log x
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Question 2
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y = ex sec x – log x
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Question 3
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y = x4 + x√x cos x – x2 ex
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Question 4
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y = (x3 – 2) tan x – x cos x + 7x. x7
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Question 5
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y = sin x log x + ex cos x – ex √x
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Question 6
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y = ex tan x + cos x log x – √x 5x
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(IV) Diffrentiate the following w.r.t.x.
Question 1
Maharashtra Board Solution
y =
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Question 2
Maharashtra Board Solution
y =
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Question 3
Maharashtra Board Solution
y =
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Question 4
Maharashtra Board Solution
y =
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y =
Question 5
Maharashtra Board Solution
y =
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Question 6
Maharashtra Board Solution
y =
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(V).
Question 1
Maharashtra Board Solution
If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12, then find f(x).
Solution & Step-by-Step Answer:
Let f(x) = ax2 + bx + c …..(i) ∴ f(0) = a(0)2 + b(0) + c ∴ f(0) = c But, f(0) = 3 …..(given) ∴ c = 3 …..(ii) Differentiating (i) w.r.t. x, we get f'(x) = 2ax + b ∴ f'(2) = 2a(2) + b ∴ f'(2) = 4a + b But, f'(2) = 2 …..(given) ∴ 4a + b = 2 …..(iii) Also, f'(3) = 2a(3) + b ∴ f'(3) = 6a + b But, f'(3) = 12 …..(given) ∴ 6a + b = 12 …..(iv) equation (iv) – equation (iii), we get 2a = 10 ∴ a = 5 Substituting a = 5 in (iii), we get 4(5) + b = 2 ∴ b = -18 ∴ a = 5, b = -18, c = 3 ∴ f(x) = 5x2 – 18x + 3
Check:
If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x2– 18x + 3 and f'(x) = 10x – 18
f(0) = 5(0)2– 18(0) + 3 = 3
f'(2) = 10(2) – 18 = 2
f'(3) = 10(3) – 18 = 12
Thus, our answer is correct.
Question 2
Maharashtra Board Solution
If f(x) = a sin x – b cos x, f'() = √2 and f'() = 2, then find f(x).
Solution & Step-by-Step Answer:
f(x) = a sin x – b cos x Differentiating w.r.t. x, we get f'(x) = a cos x – b (- sin x) ∴ f'(x) = a cos x + b sin x Now, f(x) = a sin x – b cos x ∴ f(x) = (√3 + 1) sin x + (√3 – 1) cos x
VI. Fill in the blanks. (Activity Problems)
Question 1
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y = ex. tan x Diff. w.r.t. x
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Question 2
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y = diff. w.r.t. x
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Question 3
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y = (3x2 + 5) cos x Diff. w.r.t. x
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Question 4
Maharashtra Board Solution
Differentiate tan x and sec x w.r.t. x using the formulae for differentiation of and respectively.
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