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UPSC NDA I 2024 Mathematics Paper

Question Numbers: 85-86

Consider the following for the items that follow :

Let f(x) and g(x) be two functions such that

g(x)=x-\frac{1}{x}

and

f og(x)=x^3-\frac{1}{x^3}

Question : 86 of 120

Marks: +1, -0

What is f"(x) equal to ?

Solution:

Concept:

Use the algebraic identity

x^{3} - \frac{1}{x^{3}} = (x - \frac{1}{x})^{3} + 3(x - \frac{1}{x})

to express

f(g(x))

in terms of

g(x)

Explanation:

We are given

g(x) = x - \frac{1}{x}

and

f(g(x)) = x^{3} - \frac{1}{x^{3}}

Apply the identity:

x^{3} - \frac{1}{x^{3}} = (x - \frac{1}{x})^{3} + 3(x - \frac{1}{x})

Thus

f(g(x)) = (x - \frac{1}{x})^{3} + 3(x - \frac{1}{x})

Since

g(x)=x - \frac{1}{x}

, we have

f(g(x)) = [g(x)]^{3} + 3[g(x)]

Let

y = g(x)

, then

f(y) = y^{3} + 3y

Therefore

f(x) = x^{3} + 3x

Differentiate

f(x)

with respect to

x

f'(x) = 3x^{2} + 3

Differentiate again:

f''(x) = 6x

Answer:

f''(x) = 6x

, which corresponds to option D.

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