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 $\frac{ k \sin k }{2}+\int\limits_{ k }^\pi \sin xdx =1 \left(\right.$ using $\left.\int\limits_0^\pi \sin xdx =2\right)$
$\Rightarrow \frac{ k \sin k }{2}-\left.\cos x \right|_{ k } ^\pi=1 $
$\Rightarrow k \sin k +2[1+\cos k ]=2$
$\text { So, } k \sin k +2 \cos k =0$