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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Application of Derivatives

Let the maximum and minimum values of ( 8 x-x^2-12 -4)^2+(x-7)^2, x...

Let the maximum and minimum values of (8x−x2−12−4)2+(x−7)2,x∈R\left(\sqrt{8 x-x^2-12}-4\right)^2+(x-7)^2, x \in \mathbf{R} be M\mathrm{M} and m\mathrm{m}, respectively. Then M2−m2\mathrm{M}^2-\mathrm{m}^2 is equal to _________Let the maximum and minimum values of ( 8 x-x^2-12 -4)^2+(x-7)^2, x ∈ R be M and m , respectively. Then M^2-m^2 is

Correct answer

0

Step-by-step solution

(x−7)2+(y−4)2y=8x−x2−12y2=−(x−4)2+16−12(x−4)2+y2=4\begin{aligned} & (x-7)^2+(y-4)^2 \\ & y=\sqrt{8 x-x^2-12} \\ & y^2=-(x-4)^2+16-12 \\ & (x-4)^2+y^2=4\end{aligned} Let the maximum and minimum values of ( 8 x-x^2-12 -4)^2+(x-7)^2, x ∈ R be M and m , respectively. Then M^2-m^2 is
m=9M=41M2−m2=412−92=1600\begin{aligned} & m=9 \\ & M=41 \\ & M^2-m^2=41^2-9^2=1600\end{aligned}
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