Verified Solution • 2025–26 Session moderate

QJEEM24Mabf6a: JEE Main 2024 Mathematics - Class 12 JEEM24Mabf6a Solution: Let the maximum and minimum values of be and , respectively.…

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 _________.

Step-by-Step Verified Solution

amp; Let y=8x−x2−12⇒(x−4)2+y2=22amp;⇒d=(y−4)2+(x−7)2\begin{aligned}& \text { Let } y=\sqrt{8 x-x^2-12} \Rightarrow(x-4)^2+y^2=2^2 \\& \Rightarrow d=(y-4)^2+(x-7)^2\end{aligned}

JEE Main 2024 (Online) 5th April Evening Shift Mathematics - Application of Derivatives Question 8 English Explanation

⇒amp;M=PA2=16+25=41amp;m=PQ2=(16+9−2)2=9⇒amp;M2−m2=1681−81=1600\begin{aligned}\Rightarrow & M=P A^2=16+25=41 \\& m=P Q^2=(\sqrt{16+9}-2)^2=9 \\\Rightarrow & M^2-m^2=1681-81=1600\end{aligned}

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