Question

In the shown figure, half a period of $\sin x$ from 0 to $\pi$ is split into two regions (light and dark shaded) of equal area by a line through the origin. If the line and the sine function intersect at a point whose $x$ co-ordinate is $k$, then $k$ satisfies the equation

• A. $k\cos k+2\sin k=0$
• B. $k\sin k+2\cos k=0$
• C. $k\sin k+2\cos k-2=0$
• D. $2\cos k+k\sin k+2=0$

Answer 1

Answer: (B)

Clearly, we have $\frac{k\sin k}{2}+\underset{k}{\overset{\pi }{\int }}\sin xdx=1($ using $\underset{0}{\overset{\pi }{\int }}\sin xdx=2)$
$\Rightarrow \frac{k\sin k}{2}-{\cos x|}_k^{\pi }=1$
$\Rightarrow k\sin k+2[1+\cos k]=2$
$\text{ So, }k\sin k+2\cos k=0$