Question

Let mean value of $f\left(x\right)=\frac{1}{x+c}$ over interval $(0,2)$ is $\frac{1}{2}\ell n3$ then positive values of $c$ is

• A. $12$
• B. $1$
• C. $2$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $1$

$<f\left(x\right)>=\frac{{\int }_0^{2}f\left(x\right)dx}{{\int }_0^{2}dx}$

$=\frac{1}{2}{\int }_0^2\frac{1}{x+c}dx$

$=\frac{1}{2}{\left|n\right(x+c\left)\right|}_0^2$

$=\frac{1}{2}ln\left(\frac{2+c}{c}\right)=\frac{1}{2}\ell n3$

$\Rightarrow \frac{2+c}{c}=3$

$\Rightarrow 2+c=3c$

$\Rightarrow c=1$