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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 43-44

Consider the following for the next items that follow:

Let a function f be defined on ℝ - {0} and

2f(x)+f(\frac{1}{x})=x+3

Question : 44 of 120

Marks: +1, -0

If f is differentiable, then what is f'(0.5) equal to?

Solution:

Concept:

Use a functional equation to solve for

f(x)

and then differentiate.

Explanation:

Given:

2f(x) + f(1/x) = x + 3

... (1)

Replace

x

with

1/x

2f(1/x) + f(x) = 1/x + 3

... (2)

Multiply (1) by 2:

4f(x) + 2f(1/x) = 2x + 6

Subtract (2) from this:

(4f(x) + 2f(1/x)) - (2f(1/x) + f(x)) = (2x + 6) - (1/x + 3)

Simplify:

3f(x) = 2x - 1/x + 3

Thus,

f(x) = \frac{2}{3}x - \frac{1}{3x} + 1

Differentiate with respect to

x

f'(x) = \frac{2}{3} + \frac{1}{3x^2}

(since derivative of

-\frac{1}{3x}

+\frac{1}{3x^2}

)

Put

x = 0.5

f'(0.5) = \frac{2}{3} + \frac{1}{3(0.5)^2} = \frac{2}{3} + \frac{1}{3 \cdot 0.25} = \frac{2}{3} + \frac{1}{0.75} = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2

Answer:

The value of

f'(0.5)

2

. (Option C)

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