Question

A fighter plane of enemy is flying along the curve given by $y=x^2+7.$ A soldier placed at (3, 7) wants to shoot down the plane when it is nearest to him, then the nearest distance is

• A. $2$
• B. $\sqrt{3}$
• C. $\sqrt{5}$
• D. $\sqrt{6}$

Answer 1

Answer: (C)

$\sqrt{5}$

Let the position of the plane at any time be $(x,x^2+7)$.
The position of the soldier is (3,7).
So, the distance between the plane and the soldier $=\sqrt{(x-3{)}^2+x^4}$
Let $f\left(x\right)=(x-3{)}^2+x^4$.
$f^{\prime }\left(x\right)=2(x-3)+4x^3$
$f^{\prime }\left(x\right)=2(x-1)(2x^2+2x+3)$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=1,2x^2+2x+3=0$
First we will check if $x=1$ is the point of minima.
$f^{\prime \prime }\left(x\right)=12x^2+2$
$f^{\prime \prime }\left(x\right)>0$ at $x=1$
So, the minimun distance $=\sqrt{(1-3{)}^2+1^4}=\sqrt{5}$.