JEE Advanced 2013 · Paper 1 · Mathematics Q45

MathematicsSingle correct+2Question 45 of 60

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0,\frac\pi2\right]$ is

4\left(\sqrt2-1\right)

2\sqrt2\left(\sqrt2-1\right)

2\left(\sqrt2+1\right)

2\sqrt2\left(\sqrt2+1\right)

Answer and worked solution

Official answer: B

Split at π/4, where $\cos x=\sin x$ .

$[0,\frac\pi4]$ : $(\sin x+\cos x)-(\cos x-\sin x)=2\sin x$ , area $2\left(1-\frac1{\sqrt2}\right)$ .
$[\frac\pi4,\frac\pi2]$ : $(\sin x+\cos x)-(\sin x-\cos x)=2\cos x$ , area $2\left(1-\frac1{\sqrt2}\right)$ .

Total $4-2\sqrt2=2\sqrt2\left(\sqrt2-1\right)$ .

(B)

Topics:Area between curves

JEE Advanced 2013 · Paper 1, Mathematics question 45