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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 : 43 of 120

Marks: +1, -0

What is f(0.5) equal to ?

Solution:

Concept:

Solving a functional equation by substituting

x = 1/x

and eliminating

f(1/x)

Explanation:

We are given:

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

. (Equation 1)

Replace

x

1/x

in Equation 1:

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

. (Equation 2)

Multiply Equation 1 by 2:

4f(x) + 2f\!\left(\frac{1}{x}\right) = 2x + 6

Subtract Equation 2 from this:

[4f(x) + 2f(1/x)] - [2f(1/x) + f(x)] = (2x+6) - \left(\frac{1}{x} + 3\right)

This simplifies to

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

Now set

x = 0.5

3f(0.5) = 2(0.5) - \frac{1}{0.5} + 3 = 1 - 2 + 3 = 2

Thus

f(0.5) = \frac{2}{3}

Answer:

\frac{2}{3}

(Option B)

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