Question

Let $A_1$ be the rea of the region bounded by the curve $y=\sin x,y=\cos x$ and $y-$axis in the first quadrant. Also, let $A_2$ be the area of the region bounded by the curves $y=\sin x,y=\cos x,x-$axis and $x=\frac{\pi }{2}$ in the first quadrant. Then,

• A. $A_1=A_2\text{and}{A}_1+A_2=\sqrt{2}$
• B. $A_1:A_2=1:2\text{and}{A}_1+A_2=1$
• C. $2A_1=A_2\text{and}{A}_1+A_2=1+\sqrt{2}$
• D. $A_1:A_2=1:\sqrt{2}\text{and}{A}_1+A_2=1$

Answer 1

The correct option is D $A_1:A_2=1:\sqrt{2}\text{and}{A}_1+A_2=1$

$A_1+A_2={\int }_0^{\frac{\pi }{2}}\cos x\times dx=\sin x{|}_0^{\frac{\pi }{2}}=1$


$A_1={\int }_0^{\frac{\pi }{4}}(\cos x-\sin x)dx=(\sin x+\cos x){|}_0^{\frac{\pi }{2}}=\sqrt{2}-1$
$\therefore A_2=1-(\sqrt{2}-1)=2-\sqrt{2}$
$\therefore \frac{A_1}{A_2}=\frac{\sqrt{2}-1}{\sqrt{2}(\sqrt{2}-1)}=\frac{1}{\sqrt{2}}$