JEE Main 2012 — Application Of Derivatives Question with Solution

Equation of a line passing through $$\left(x_1,y_1\right)$$ having

slope $$m$$ is given by $$y-y_1=m\left(x-x_1\right)$$

Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its

equation is

$$\left(y-2\right)=m\left(x-1\right)$$

where $$m$$ is the slope of the line $$PQ$$.

Now, point $$P\left(x,0\right)$$ will also satisfy the equation of $$PQ$$

$$\therefore$$ $$y-2=m\left(x-1\right)$$

$$\Rightarrow 0-2=m\left(x-1\right)$$

$$\Rightarrow -2=m\left(x-1\right)$$

$$\Rightarrow x-1=\frac{-2}{m}$$

$$\Rightarrow x=\frac{-2}{m}+1$$

Also, $$OP=\sqrt{\left(x-0\right){}^2+{\left(0-0\right)}^2}=x=\frac{-2}{m}+1$$

Similarly, point $$Q\left(0,y\right)$$ will satisfy equation of $$PQ$$

$$\therefore$$ $$y-2=m\left(x-1\right)$$

$$\Rightarrow y-2=m\left(-1\right)\Rightarrow y=2-m$$b

and $$OQ=y=2-m$$

Area of $$\Delta POQ=\frac{1}{2}\left(OP\right)\left(OQ\right)=\frac{1}{2}\left(1-\frac{2}{m}\right)\left(2-m\right)$$

( As Area of $$\Delta =\frac{1}{2}\times$$ base $$\times$$ height )

$$=\frac{1}{2}\left[2-m-\frac{4}{m}+2\right]=\frac{1}{2}\left[4-\left(m+\frac{4}{m}\right)\right]$$

$$=2-\frac{m}{2}-\frac{2}{m}$$



Let Area $$=f\left(m\right)=2-\frac{m}{2}-\frac{2}{m}$$

Now, $$f^{\prime }\left(m\right)=\frac{-1}{2}+\frac{2}{m^2}$$

Put $$f^{\prime }\left(m\right)=0$$

$$\Rightarrow m^2=4\Rightarrow m=\pm 2$$

Now, $$f^{\prime \prime }\left(m\right)=\frac{-4}{m^3}$$

$${f^{\prime \prime }\left(m\right)|}_{m=2}=-\frac{1}{2}<0$$

$${f^{\prime \prime }\left(m\right)|}_{m=-2}=\frac{1}{2}>0$$

Area will be least at $$m=-2$$

Hence, slope of $$PQ$$ is $$-2.$$