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JEE Mains · Maths · STD 12 - 6. Application of derivatives

ધારો કે \(\left(\sqrt{8 x-x^2-12}-4\right)^2+(x-7)^2, x \in {R}\) ની મહત્તમ તથા ન્યૂનતમ કિંમતો અનુક્રમે \(\mathrm{M}\) અને \(\mathrm{m}\) છે. તો \(\mathrm{M}^2-\mathrm{m}^2\) = ............

  1. A \(4600\)
  2. B \(4100\)
  3. C \(3200\)
  4. D \(1600\)
Verified Solution

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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