Question

Two lines drawn through the point $A(4,0)$ divide the area bounded by the curve $y=\sqrt{2}\sin \left(\pi x/4\right)$ and $x$ - axis between the lines $x=2$ and $x=4$ into three equal parts. Sum of the slopes of the drawn lines is:

• A. $-4\sqrt{2}/\pi$
• B. $-\sqrt{2}/\pi$
• C. $-2\sqrt{2}/\pi$
• D. None

Answer 1

The correct option is C $-2\sqrt{2}/\pi$

Area bounded by $y=\sqrt{2}\sin \left(\frac{\pi x}{4}\right)$ and $x-axis$ between the lines $x=2$ and $x=4,$

$\Delta =\sqrt{2}{\int }_2^4\sin \frac{\pi x}{4}dx$

$={[-\frac{4\sqrt{2}}{\pi }.\cos \frac{\pi x}{4}]}_2^4$

$=\frac{4\sqrt{2}}{\pi }uni{t}^2$

Let the drawn lines are $L_1:y-m_1(x-4)=0$ and $L_2:y-m_2(x-4)=0,$ meeting the line $x=2$ at the points $A$ and $B,$ respectively.

Clearly, $A=(2,-2m_1);B=(2,-2m_2)$ [ Fig ]

Now,

${\Delta }_{ACD}=\frac{\Delta }{3}\Rightarrow \frac{4\sqrt{2}}{3\pi }=\frac{1}{2}.2.(-2m_1)$

$\Rightarrow$ $m_1=-\frac{2\sqrt{2}}{3\pi }$

Also, ${\Delta }_{BCD}=\frac{2\Delta }{3}$

$\Rightarrow$ $\frac{8\sqrt{2}}{3\pi }=\frac{1}{2}\times 2-2m_2$

$\Rightarrow$ $m_2=\frac{-4\sqrt{2}}{3\pi }$

$\Rightarrow$ Required sum $=-\frac{2\sqrt{2}}{3\pi }+\left(\frac{-4\sqrt{2}}{3\pi }\right)$

$=\frac{-6\sqrt{2}}{3\pi }$

$=\frac{-2\sqrt{2}}{\pi }$