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

Let \(f(x)=3 \sqrt{x-2}+\sqrt{4-x}\) be a real valued function. If \(\alpha\) and \(\beta\) are respectively the minimum and the maximum values of \(\mathrm{f}\), then \(\alpha^2+2 \beta^2\) is equal to

  1. A \(44\)
  2. B \(42\)
  3. C \(24\)
  4. D \(38\)
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Answer & Solution

Correct Answer

(B) \(42\)

Step-by-step Solution

Detailed explanation

\( \mathrm{f}(\mathrm{x})=3 \sqrt{\mathrm{x}-2}+\sqrt{4-\mathrm{x}} \) \( \mathrm{x}-2 \geq 0 \& 4-\mathrm{x} \geq 0 \) \( \therefore \mathrm{x} \in[2,4] \) \( \text { Let } \mathrm{x}=2 \sin ^2 \theta+4 \cos ^2 \theta \)…
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