An enemy Apache helicopter is flying along th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For each value of $x$,the helicopter's position is at point $(x, x^{2} + 7)$.Therefore,the distance $d$ between the helicopter and the soldier at

Vedclass · Accepted Answer

$\sqrt{5}$

Vedclass · Answer

(A) For each value of $x$,the helicopter's position is at point $(x, x^{2} + 7)$.Therefore,the distance $d$ between the helicopter and the soldier at $(3, 7)$ is given by $d = \sqrt{(x - 3)^{2} + (x^{2} + 7 - 7)^{2}} = \sqrt{(x - 3)^{2} + x^{4}}$.Let $f(x) = (x - 3)^{2} + x^{4}$. To minimize the distance,we minimize $f(x)$.Taking the derivative,$f'(x) = 2(x - 3) + 4x^{3} = 2(x - 1)(2x^{2} + 2x + 3)$.Setting $f'(x) = 0$,we get $x = 1$ (since $2x^{2} + 2x + 3 = 0$ has no real roots as its discriminant $D = 4 - 24 = -20 The value of $f(x)$ at $x = 1$ is $f(1) = (1 - 3)^{2} + (1)^{4} = 4 + 1 = 5$.Thus,the minimum distance is $\sqrt{f(1)} = \sqrt{5}$.

An enemy Apache helicopter is flying along the curve given by $y = x^{2} + 7$. $A$ soldier,placed at $(3, 7)$,wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.

Similar Questions

The condition for the function $f(x) = x^3 + px^2 + qx + r$ $(x \in R)$ to have no extreme value is:

The sum of all local minimum values of the function $f(x) = \begin{cases} 1-2x, & x < -1 \\ \frac{1}{3}(7+2|x|), & -1 \leq x \leq 2 \\ \frac{11}{18}(x-4)(x-5), & x > 2 \end{cases}$ is:

The greatest and least values of $f(x)=\tan ^{-1} x-\frac{1}{2} \ln x$ on $\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$ are

The maximum value of the function $f(x)=3x^3-18x^2+27x-40$ on the set $S=\{x \in R : x^2+30 \leq 11x\}$ is

The function $f(x) = x^{25}(1 - x)^{75}$ for $x \in [0, 1]$ attains its maximum at $x = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module