JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let the maximum and minimum values of \(\left(\sqrt{8 x-x^2-12}-4\right)^2+(x-7)^2, x \in R\) be \(M\) and \(m\) respectively. Then \(\mathrm{M}^2-\mathrm{m}^2\) is equal to ...............
- A \(4600\)
- B \(4100\)
- C \(3200\)
- D \(1600\)
Answer & Solution
Correct Answer
(D) \(1600\)
Step-by-step Solution
Detailed explanation
\( (x-7)^2+(y-4)^2 \) \( y=\sqrt{8 x-x^2-12} \) \( y^2=-(x-4)^2+16-12 \) \( (x-4)^2+y^2=4\) \(Image\) \( \mathrm{m}=9 \) \( \mathrm{M}=41 \) \( \mathrm{M}^2-\mathrm{m}^2=41^2-9^2=1600\)
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