Question

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

• A. $2/7$
• B. $1/2$
• C. $2$
• D. $7/2$

Answer 1

Answer: (B)

$1/2$

Differentiate $3f\left(x\right)-2f\left(\frac{1}{x}\right)=x$ with respect to $x$,

$3f^{\prime }\left(x\right)-2f^{\prime }\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)=1$

$3f^{\prime }\left(x\right)+\frac{2}{x^2}{f}^{\prime }\left(\frac{1}{x}\right)=1$ (1)

Put $x=2$ in equation (1),

$3f^{\prime }\left(2\right)+\frac{2}{2^2}{f}^{\prime }\left(\frac{1}{2}\right)=1$

$3f^{\prime }\left(2\right)+\frac{1}{2}{f}^{\prime }\left(\frac{1}{2}\right)=1$ (2)

Put $x=\frac{1}{2}$ in equation (1),

$3f^{\prime }\left(\frac{1}{2}\right)+\frac{2}{{\left(\frac{1}{2}\right)}^2}{f}^{\prime }\left(\frac{1}{\frac{1}{2}}\right)=1$

$3f^{\prime }\left(\frac{1}{2}\right)+\frac{2}{\frac{1}{4}}{f}^{\prime }\left(2\right)=1$

$3f^{\prime }\left(\frac{1}{2}\right)+8f^{\prime }\left(2\right)=1$

$3f^{\prime }\left(\frac{1}{2}\right)=1-8f^{\prime }\left(2\right)$

$f^{\prime }\left(\frac{1}{2}\right)=\frac{1}{3}-\frac{8}{3}{f}^{\prime }\left(2\right)$

Substitute above value in equation (2),

$3f^{\prime }\left(2\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{8}{3}{f}^{\prime }\left(2\right)\right)=1$

$3f^{\prime }\left(2\right)+\frac{1}{6}-\frac{4}{3}{f}^{\prime }\left(2\right)=1$

$3f^{\prime }\left(2\right)-\frac{4}{3}{f}^{\prime }\left(2\right)=1-\frac{1}{6}$

$\frac{9f^{\prime }\left(2\right)-4f^{\prime }\left(2\right)}{3}=\frac{6-1}{6}$

$5f^{\prime }\left(2\right)=\frac{5}{2}$

$f^{\prime }\left(2\right)=\frac{1}{2}$

Therefore, the value of $f^{\prime }\left(2\right)$ is $\frac{1}{2}$.