Question

The area bounded by $y=\frac{1}{x}$ and $y=\frac{1}{2x-1}$ 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 $\frac{1}{x}>\frac{1}{2x-1}\forall x\in \left(1,2\right)$
$\therefore$ the required area is $A={\int }_1^2\left(\frac{1}{x}-\frac{1}{2x-1}\right)dx$
$={\left(\left[lnx\right]\right)}_1^2-\frac{1}{2}{\left(\left[ln\left(2x-1\right)\right]\right)}_1^2$
$=ln2-\frac{1}{2}ln3=ln2-ln\sqrt{3}$
$=ln\left(\frac{2}{\sqrt{3}}\right)$ sq. units
$\therefore a=\frac{2}{\sqrt{3}}\Rightarrow 3a^2=4$