JEE Main 2024 (Online) 5th April Evening ShiftQuestion 1
Let the maximum and minimum values of $\left(\sqrt{8x-x^{2}-12}-4\right)^{2}+(x-7)^{2}$, $x\in\mathbf{R}$ be $M$ and $m$, respectively. Then $M^{2}-m^{2}$ is equal to $\_\_\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$1600$
Solution
Concept. $y=\sqrt{8x-x^2-12}$ is the upper half of the circle with centre $(4,0)$ and radius $2$. The expression is the squared distance from $(7,4)$ to a point of that semicircle. Formulas used. (1) $8x-x^2-12=4-(x-4)^2$(2) $12\cos\theta+16\sin\theta=20\sin(\theta+\varphi)$ with $\tan\varphi=\frac34$Step 1 - Parametrise. By (1), points are $(4+2\cos\theta,\ 2\sin\theta)$, $\theta\in[0,\pi]$: $$(2\sin\theta-4)^2+(2\cos\theta-3)^2=29-12\cos\theta-16\sin\theta.$$Step 2 - Minimum. By (2), $12\cos\theta+16\sin\theta$ reaches $20$ inside $[0,\pi]$ (at the point nearest to $(7,4)$). So $m=29-20=9$. Step 3 - Maximum. On $[0,\pi]$ the smallest value of $12\cos\theta+16\sin\theta$ is at $\theta=\pi$, namely $-12$ (the end $(2,0)$). So $M=29+12=41$. Step 4 - Value. $M^2-m^2=1681-81=1600$. Answer: 1600