question_answer

The line \[x=\frac{\pi }{4}\]divides the area of the region bounded by \[y=\text{sin }x,\] \[y=\text{cos }x\] and x-axis \[\left( 0\le x\le \frac{\pi }{2} \right)\] into two regions of areas \[{{A}_{1}}\] and \[{{A}_{2}}\]. Then\[{{A}_{1}}:{{A}_{2}}\]equals

A)  4 : 1            

B)  3 : 1

C)  2 : 1            

D)  1 : 1

Correct Answer: D

Solution :

Area, \[{{A}_{1}}=\int_{0}^{\pi /4}{\sin x\,dx}\] \[=-[\cos x]_{0}^{\pi /4}=1-\frac{1}{\sqrt{2}}\] \[=\frac{\sqrt{2}-1}{\sqrt{2}}\] and area \[{{A}_{2}}=\int_{\pi /4}^{\pi /2}{\cos x\,dx}\] \[=[\sin x]_{\pi /4}^{\pi /2}=\left[ 1-\frac{1}{\sqrt{2}} \right]=\frac{\sqrt{2}-1}{\sqrt{2}}\] \[\therefore \] \[{{A}_{1}}:{{A}_{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}:\frac{\sqrt{2}-1}{\sqrt{2}}=1:1\]