Question

A figure is bounded by the curves $y=\left|\sqrt{2}\sin \frac{\pi x}{4}\right|$ $y=0,x=2$ & $x=4$. At what angles to the positive $x$-axis straight lines must be drawn through $(4,0)$, so that these lines divide the figure into three parts of the same size

• A. $\pi -{\tan }^{-1}\frac{2\sqrt{2}}{3\pi },\pi +{\tan }^{-1}\frac{4\sqrt{2}}{3\pi }$
• B. $\pi -{\tan }^{-1}\frac{2\sqrt{2}}{2\pi },\pi -{\tan }^{-1}\frac{4\sqrt{2}}{2\pi }$
• C. $\pi -{\tan }^{-1}\frac{2\sqrt{2}}{3\pi },\pi -{\tan }^{-1}\frac{4\sqrt{2}}{3\pi }$
• D. $\pi +{\tan }^{-1}\frac{2\sqrt{2}}{2\pi },\pi -{\tan }^{-1}\frac{4\sqrt{2}}{2\pi }$

Answer 1

Answer: (C)

$\pi -{\tan }^{-1}\frac{2\sqrt{2}}{3\pi },\pi -{\tan }^{-1}\frac{4\sqrt{2}}{3\pi }$

Let equation of line is $y=mx-4m$
$A={\int }_2^4\sqrt{2}\sin \frac{\pi }{4}xdx={[-\sqrt{2}\frac{4}{\pi }\cos \frac{\pi x}{4}]}_2^4=\frac{4\sqrt{2}}{\pi }....\left(i\right)$

Also area of $\Delta ABC=\frac{1}{2}.2.(-2m_1)=-2m_1...\left(ii\right)$ from (i) and (ii)
$-2m_1=\frac{4\sqrt{2}}{3\pi }\Rightarrow m_1=\frac{-2\sqrt{2}}{3\pi }$

$\Rightarrow \tan (\pi -{\theta }_1)=\frac{-2\sqrt{2}}{3\pi }\Rightarrow \pi -{\theta }_1={\tan }^{-1}\frac{2\sqrt{2}}{3\pi }$

$\Rightarrow {\theta }_1=\pi -{\tan }^{-1}\frac{2\sqrt{2}}{3\pi }$ or $\frac{1}{2}.\left(2\right)(-2m_2)=\frac{8\sqrt{2}}{3\pi }$

$\Rightarrow m_2=\frac{-4\sqrt{2}}{3\pi }\Rightarrow \tan (\pi -{\theta }_2)=\frac{-4\sqrt{2}}{3\pi }$

$\Rightarrow {\theta }_2=\pi -{\tan }^{-1}\frac{4\sqrt{2}}{3\pi }$