Q194 / 270 · NDA 2023 (Phase 2) · Mathematics

Let a function f be defined on $\mathbb{R}-\{0\}$ and $2 f(x)+f\left(\frac{1}{x}\right)=x+3$ If f is differentiable, then what is $\quad(0.5)$ equal to?

  • A
    $\frac{1}{4}$
  • B
    $\frac{2}{3}$
  • Correct answer
    C
    2
  • D
    4

Solution

$\begin{aligned} & \text { Since, } 2 \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{x}+3 \\ & \Rightarrow 2 \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime}\left(\frac{1}{\mathrm{x}}\right)\left(-\frac{1}{\mathrm{x}^2}\right)=1 \\ & \text { Put } \mathrm{x}=2, \frac{1}{2} \\ & \Rightarrow 2 \mathrm{f}^{\prime}(2)-\frac{1}{4} \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=1...(i) \\ & \text { and, } 2 \mathrm{f}^{\prime}\left(\frac{1}{2}\right)-4 \mathrm{f}^{\prime}(2)=1...(ii) \end{aligned}$ after solving (i) \& (ii), we get $\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=2, \mathrm{f}^{\prime}(2)=\frac{3}{4}$ So, $\mathrm{f}^{\prime}(0.5)=2$

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

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