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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 : 85 of 120

Marks: +1, -0

What is g[f(x) - 3x] equal to ?

Solution:

Concept:

Use the identity

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

Explanation:

Given

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

and

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

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

Apply the identity:

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

Since

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

, we have

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

Let

y = x - \frac{1}{x}

, then

f(y) = y^3 + 3y

. Hence

f(x) = x^3 + 3x

Now

f(x) - 3x = (x^3 + 3x) - 3x = x^3

Finally,

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

Thus the answer is option A.

Answer:

Option A:

x^3 - \frac{1}{x^3}

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