Q217 / 270 · NDA 2022 (Phase 1) · Mathematics

If $4 f(x)-f\left(\frac{1}{x}\right)=\left(2 x+\frac{1}{x}\right)\left(2 x-\frac{1}{x}\right)$, then what is $f(2)$ equal to?

  • A
    0
  • B
    1
  • C
    2
  • Correct answer
    D
    4

Solution

$4 f(x)-f\left(\frac{1}{x}\right)=\left(2 x+\frac{1}{x}\right)\left(2 x-\frac{1}{x}\right)$ $4 \mathrm{f}(\mathrm{x})-\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=4 \mathrm{x}^2-\frac{1}{\mathrm{x}^2}$...(i) Replace $x$ by $\frac{1}{x}$ we get $4 f\left(\frac{1}{x}\right)-f(x)=\frac{4}{x^2}-x$...(ii) from $4 \times($ i) + (ii) $16 f(x)-4 f\left(\frac{1}{x}\right)=16 x^2-\frac{4}{x^2}$ $\begin{aligned} & 16 f(x)-4 f\left(\frac{1}{x}\right)=16 x^2-\frac{4}{x^2} \\ & 4 f\left(\frac{1}{x}\right)-f(x)=\frac{4}{x^2}-x\end{aligned}$ $\begin{aligned} & 15 \mathrm{f}(\mathrm{x})=15 \mathrm{x}^2 \\ & \therefore \mathrm{f}(\mathrm{x})=\mathrm{x}^2\end{aligned}$ Now, $f(2)=2^2=4$.

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

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