Q249 / 270 · NDA 2024 (Phase 1) · Mathematics

Let $3 f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}+1$ What is $f(x)$ equal to?

  • A
    $\frac{1}{8 x}-\frac{x}{8}+\frac{1}{4}$
  • B
    $\frac{3}{8 x}-\frac{x}{8}+\frac{3}{4}$
  • C
    $\frac{3}{8 x}+\frac{x}{8}+\frac{1}{4}$
  • Correct answer
    D
    $\frac{3}{8 x}-\frac{x}{8}+\frac{1}{4}$

Solution

Given: $\quad 3 f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}+1$ $\quad ....(i)$ Replacing ( $x$ ) by $\left(\frac{1}{x}\right)$, we get $3 f\left(\frac{1}{x}\right)+f(x)=x+1$ $\quad ....(ii)$ Now, (i) $\times 3-(i i)$, we get $8 f(x)=\frac{3}{x}+3-x-1$ $\Rightarrow \quad f(x)=\frac{3}{8 x}-\frac{x}{8}+\frac{1}{4}$ $\quad ....(iii)$

Hence, the correct answer is option: 'D'.

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