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UPSC NDA I 2024 Mathematics Paper

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Question Numbers: 99-100

Consider the following for the items that follow:

Let

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

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Question : 99 of 120

Marks: +1, -0

What is f(x) equal to ?

$\frac{1}{8 x}-\frac{x}{8}+\frac{1}{4}$
$\frac{3}{8 x}-\frac{x}{8}+\frac{3}{4}$
$\frac{3}{8 x}+\frac{x}{8}+\frac{1}{4}$
$\frac{3}{8 x}-\frac{x}{8}+\frac{1}{4}$

Solution:

Concept:

We solve a functional equation by substituting

x \to \frac{1}{x}

and solving the resulting linear system for

f(x)

.

Explanation:

Start with

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

.

Replace

x

by

\frac{1}{x}

to get

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

.

Let

a = f(x)

and

b = f\!\left(\frac{1}{x}\right)

. Then the system is:

3a + b = \frac{1}{x} + 1

and

3b + a = x + 1

.

Multiply the first equation by

3

:

9a + 3b = \frac{3}{x} + 3

.

Subtract the second equation from this result:

(9a+3b) - (3b + a) = \left(\frac{3}{x}+3\right) - (x+1)

.

Simplify:

8a = \frac{3}{x} + 3 - x - 1 = \frac{3}{x} - x + 2

.

Hence

a = \frac{1}{8}\!\left(\frac{3}{x} - x + 2\right) = \frac{3}{8x} - \frac{x}{8} + \frac{1}{4}

.

Answer:

f(x) = \frac{3}{8x} - \frac{x}{8} + \frac{1}{4}

(Option D).

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