Question

The area bounded by $y=\frac{1}{x}$ and $y=\frac{1}{2 x - 1}$ from $x=1$ to $x=2$ is $ln \left(a\right)$ sq. units, then $3a^{2}$ is equal to

• A. $\frac{1}{2}$
• B. $4$
• C. $1$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

As $\frac{1}{x}>\frac{1}{2 x - 1}\forall x\in \left(1 , 2\right)$
$\therefore $ the required area is $A=\displaystyle \int _{1}^{2} \left(\frac{1}{x} - \frac{1}{2 x - 1}\right)dx$
$=\left(\left[ln x\right]\right)_{1}^{2}-\frac{1}{2}\left(\left[ln ⁡ \left(2 x - 1\right)\right]\right)_{1}^{2}$
$=ln 2-\frac{1}{2}ln ⁡ 3=ln ⁡ 2-ln ⁡ \sqrt{3}$
$=ln \left(\frac{2}{\sqrt{3}}\right)$ sq. units
$\therefore a=\frac{2}{\sqrt{3}}\Rightarrow 3a^{2}=4$