question_answer

The area of the region enclosed by the curve \[y=\left| x-\frac{1}{x} \right|\,\,(x>0)\] and the line \[y=2\] is

A) \[1\]                       

B)        \[2\]                    

C) \[3\]

D)        \[4\]

Correct Answer: B

Solution :

   [b]  Required area, A \[=\int\limits_{\sqrt{2}-1}^{1}{\left( 2-\left( \frac{1}{x}-x \right) \right)}dx+\int\limits_{1}^{\sqrt{2}+1}{\left( 2-\left( x-\frac{1}{x} \right) \right)}dx\] \[\underbrace{\int\limits_{\sqrt{2}-1}^{1}{\left( 2+x-\frac{1}{x} \right)dx}}_{{{A}_{1}}}+\underbrace{\int\limits_{1}^{\sqrt{2}+1}{\left( 2+\frac{1}{x}-x \right)dx}}_{{{A}_{2}}}\] \[{{A}_{1}}=2x+\left. \frac{{{x}^{2}}}{2}-In\,\,x \right]_{\sqrt{2}-1}^{1}\] \[=\frac{5}{2}-\left( 2\sqrt{2}-2+\frac{3-2\sqrt{2}}{2}-In\,\,(\sqrt{2}-1) \right)\] \[=\frac{5}{2}-2\sqrt{2}+2-\frac{3-2\sqrt{2}}{2}+In\,\,(\sqrt{2}-1)\] \[{{A}_{2}}=\left. 2x-\frac{{{x}^{2}}}{2}+In\,\,x \right]_{1}^{\sqrt{2}+1}\] \[=2\sqrt{2}+2-\frac{3+2\sqrt{2}}{2}+In\,\,(\sqrt{2}+1)-\frac{3}{2}\] \[{{A}_{1}}+{{A}_{2}}=5-3+In\,(\sqrt{2}+1)+In\,\,(\sqrt{2}-1)\] \[\therefore \,\,\,A=2\]