Q. The area bounded by $y = 2 - |2 - x|$ and $y = \frac{3}{|x|}$ is

A) $\frac{4 + 3\log 3}{2}\ sq\ unit$
B) $\frac{4 - 3\log 3}{2}\ sq\ unit$
C) $\frac{3}{2}\log 3\ sq\ unit$
D) $\frac{1}{2} + \log 3sq\ unit$
Explanation

(b)
Required area $= \int_{\sqrt{3}}^{3}{(2 - 12 - x)dx - \int_{\sqrt{3}}^{3}\frac{3}{|x|}}dx$
$= \int_{\sqrt{3}}^{2}{x\ dx + \int_{2}^{3}{(4 - x)dx -}}\int_{\sqrt{3}}^{3}{\frac{3}{x}dx}$
$= \left\lbrack \frac{x^{2}}{2} \right\rbrack_{\sqrt{3}}^{2} + \left\lbrack 4x - \frac{x^{2}}{2} \right\rbrack_{2}^{3} - \lbrack 3{\log{x\rbrack}}_{\sqrt{3}}^{3}$
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$= \frac{1}{2}\lbrack 4 - 3\rbrack + \left\lbrack 12 - \frac{9}{2} - (8 - 2) \right\rbrack - 3\lbrack\log{3 - \log\sqrt{3}\rbrack}$
$= \frac{1}{2} + \frac{3}{2} - 3\log{\frac{3}{\sqrt{3}} = \frac{4}{2} - \frac{3}{2}\log 3}$
$= \frac{4 - 3\log 3}{2}\ sq\ unit$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.