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UPSC NDA II 2024 Mathematics Paper

Question Numbers: 95-96

Direction : Consider the following for the items that follow :

Let

f : (-1, 1) \to R

be a differentiable function with

f(0) = -1

and

f'(0) = 1

Let

h(x) = f(2f(x) + 2)

and

g(x) = (h(x))^{2}

Question : 95 of 120

Marks: +1, -0

What is h’(0) equal to?

Solution:

Concept:

Apply the chain rule to differentiate the composite function and evaluate at

x = 0

using the given values.

Explanation:

Given

f(0) = -1

and

f'(0) = 1

Let

h(x) = f(2f(x) + 2)

Differentiate:

h'(x) = f'(2f(x) + 2) \cdot 2 f'(x)

Substitute

x = 0

h'(0) = f'(2f(0) + 2) \cdot 2 f'(0)

Now

2f(0) + 2 = 2(-1) + 2 = 0

, so

f'(2f(0)+2) = f'(0) = 1

Also

2 f'(0) = 2 \cdot 1 = 2

Thus

h'(0) = 1 \cdot 2 = 2

Answer:

2

(Option D)

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