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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Application of Derivatives

Let f(x)=3 x-2 + 4-x be a real valued function. If α and β are...

Let f(x)=3x−2+4−xf(x)=3 \sqrt{x-2}+\sqrt{4-x} be a real valued function. If α\alpha and β\beta are respectively the minimum and the maximum values of ff, then α2+2β2\alpha^2+2 \beta^2 is equal to

Options

  1. A42
  2. B38
  3. C24
  4. D44

Correct answer

A. 42

Step-by-step solution

f(x)=3x−2+4−xx−2≥04−x≥0∴x∈[2,4]\begin{aligned} & f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\ & x-2 \geq 0 & 4-x \geq 0 \\ & \therefore x \in[2,4] \end{aligned}
Let x=2sin⁡2θ+4cos⁡2θx=2 \sin ^2 \theta+4 \cos ^2 \theta∴f(x)=32∣cos⁡θ∣+2∣sin⁡θ∣∴2≤32∣cos⁡θ∣+2∣sin⁡θ∣≤9×2+22≤32∣cos⁡θ∣+2∣sin⁡θ∣≤20∴α=2β=20α2+2β2=2+40=42\begin{aligned} & \therefore \mathrm{f}(\mathrm{x})=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \\ & \therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2} \\ & \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20} \\ & \therefore \alpha=\sqrt{2} \quad \beta=\sqrt{20} \\ & \alpha^2+2 \beta^2=2+40=42 \end{aligned}$$
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