Question

For what value of 'a' is the area of the figure bounded by $y=\frac{1}{x},y=\frac{1}{2x-1}$ $x=2$ & $x=a$ equal to $ln\frac{4}{\sqrt{5}}$?

• A. $a=4$
• B. $a=8$
• C. $a=4or\frac{2}{5}(6-\sqrt{21})$
• D. none of these

Answer 1

The correct option is B $a=8$

The required area is
$={\int }_2^a\frac{1}{x}-\frac{1}{2x-1}.dx$
$=[lnx-\frac{ln(2x-1)}{2}{]}_2^a$
$=[ln\frac{x}{\sqrt{2x-1}}{]}_2^a$
$=-ln\frac{\left(2\right)}{\sqrt{3}}+ln\frac{a}{\sqrt{2a-1}}$
$=ln\frac{4}{\sqrt{5}}$
Hence
$ln\left(\frac{a}{\sqrt{2a-1}}\right)=ln\frac{4}{\sqrt{5}}+ln\frac{\left(2\right)}{\sqrt{3}}$
$\Rightarrow ln\left(\frac{a}{\sqrt{2a-1}}\right)=ln\frac{8}{\sqrt{15}}$
Hence
$2a-1=15$
Hence
$a=8$.
Therefore a=8 is one solution.
Now
$\frac{a^2}{2a-1}=\frac{64}{15}$
$\Rightarrow 15a^2=128a-64$
$\Rightarrow 15a^2-128a+64=0$

Hence $a=8$ and $a=\frac{8}{\sqrt{15}}$