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)

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