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QJEEM24Mad21c: JEE Main 2024 Mathematics - Class 12 JEEM24Mad21c Solution: Let be a real valued function. If and are respectively the…

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

Multiple Choice Options

  • A 42
  • B 38
  • C 24
  • D 44

Step-by-Step Verified Solution

amp;f(x)=3x−2+4−xamp; Let x=2sin⁡2θ+4cos⁡2θamp;=32sin⁡2θ+4cos⁡2θ−2+4−2sin⁡2θ−4cos⁡2θamp;=32cos⁡2θ+2sin⁡2θamp;=32∣cos⁡θ∣+2∣sin⁡θ∣\begin{aligned}& f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\& \text { Let } x=2 \sin ^2 \theta+4 \cos ^2 \theta \\& =3 \sqrt{2 \sin ^2 \theta+4 \cos ^2 \theta-2}+\sqrt{4-2 \sin ^2 \theta-4 \cos ^2 \theta} \\& =3 \sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta} \\& =3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta|\end{aligned}

amp;⇒32cos⁡θ+2sin⁡θ≤18+2amp;⇒32cos⁡θ+2sin⁡θ≤20\begin{aligned}& \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{18+2} \\& \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{20}\end{aligned}

Minimum value exists when θ=π2\theta=\frac{\pi}{2}

amp; So, minimum value =2amp;⇒α=2 and β=20amp;⇒α2+2β2=2+40amp;=42\begin{aligned}& \text { So, minimum value }=\sqrt{2} \\& \Rightarrow \alpha=\sqrt{2} \text { and } \beta=\sqrt{20} \\& \Rightarrow \alpha^2+2 \beta^2=2+40 \\& =42\end{aligned}

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