Question

lf the primitive of $\frac{1}{f\left(x\right)}$ is equal to $\log \left\{f\right(x){\}}^2+c$, then $f\left(x\right)$ is:

• A. $x+d$
• B. $\frac{x}{2}+d$
• C. $\frac{x^2}{2}+d$
• D. $x^2+d$

Answer 1

Answer: (B)

$\frac{x}{2}+d$

Given $\int \frac{1}{f\left(x\right)}dx={\log \left\{f\right(x\left)\right\}}^2+C$
$\Rightarrow \int \frac{1}{f\left(x\right)}dx=2\log \left\{f\right(x\left)\right\}+C$
$\Rightarrow \frac{1}{f\left(x\right)}=2\times \frac{1}{f\left(x\right)}\times f^{\prime }\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2}$
$\Rightarrow f\left(x\right)=\frac{1}{2}\int dx$
$\Rightarrow f\left(x\right)=\frac{x}{2}+d$