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 \[{{A}_{2}}=\int_{\pi /4}^{\pi /2}{\cos x\,dx}\] \[=[\sin x]_{\pi /4}^{\pi /2}=\left[ 1-\frac{1}{\sqrt{2}} \right]=\frac{\sqrt{2}-1}{\sqrt{2}}\] \[\therefore \] \[{{A}_{1}}:{{A}_{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}:\frac{\sqrt{2}-1}{\sqrt{2}}=1:1\]You need to login to perform this action.
You will be redirected in
3 sec