Question

The distance of that point on $y=x^4+3x^2+2x$ which is nearest to the line $y=2x-1$ is

• A. $\frac{3}{\sqrt{5}}$
• B. $\frac{4}{\sqrt{5}}$
• C. $\frac{2}{\sqrt{5}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

Answer: (D)

$y=x^4+3x^2+2x\dots \left(i\right)$
$y=2x-1\dots \left(ii\right)$
Let $P\left(x,y\right)$ be a point on $\left(i\right)$
$\therefore S=$ Distance of $P$ from $\left(ii\right)=\frac{|2x-1-y|}{\sqrt{5}}$
$=\frac{|2x-1-x^4-3x^2-2x|}{\sqrt{5}}$ ($\because P$ lies on $\left(i\right)$)
$=\frac{|-x^4-3x^2-1|}{\sqrt{5}}$
$=\frac{x^4+3x^2+1}{\sqrt{5}}$
$\frac{dS}{dx}=\frac{4x^3+6x}{\sqrt{5}}$,
$\frac{d^{2}S}{d{x}^2}=\frac{12x^3+6}{\sqrt{5}}$
$\therefore S$ is min. when $\frac{dS}{dx}=0$
i.e., $4x^3+6x=0$
$\Rightarrow x=0$ (Real)
$\frac{d^{2}S}{d{x}^2}{|}_{x=0}=\frac{6}{\sqrt{5}}>0$
Hence, min. value of $S$ is $\frac{1}{\sqrt{5}}$.