Question

If $3f\left(x\right)-2f\left(\frac{1}{x}\right)=x$, then $f^{{}^{\prime }}\left(x\right)$ is equal to

• A. $\frac{2}{7}$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{7}{2}$

Answer 1

Answer: (B)

$\frac{1}{2}$

$3f\left(x\right)-2f\left(\frac{1}{x}\right)=x$ ....(i)
Let $\frac{1}{x}=y$, then $3f\left(\frac{1}{y}\right)-2f\left(y\right)=\frac{1}{y}$
$\Rightarrow -2f\left(y\right)+3f\left(\frac{1}{y}\right)=\frac{1}{y}$
$\Rightarrow -2f\left(x\right)+3f\left(\frac{1}{x}\right)=\frac{1}{x}$ ...(ii)
On multiplying equation (i) by $3$ and equation (ii) by $2$ and then adding them, we get
$9f\left(x\right)-6f\left(\frac{1}{x}\right)-4f\left(x\right)+6f\left(\frac{1}{x}\right)=3x+\frac{2}{x}$
$\Rightarrow 5f\left(x\right)=3x+\frac{2}{x}$
$\Rightarrow f\left(x\right)=\frac{1}{5}[3x+\frac{2}{x}]$
$\therefore f^{{}^{\prime }}\left(x\right)=\frac{1}{5}[3-\frac{2}{x^2}]$
Then, $f^{{}^{\prime }}\left(2\right)=\frac{1}{5}[3-\frac{2}{4}]=\frac{1}{2}$