Let f ( x ) = 3 x - 2 + 4 - x be a real valued function. If and are respectively the minimum and the maximum values of f, then is equal to [2024]

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Solution

Q.
Let $f (x) = 3 \sqrt{x - 2} + \sqrt{4 - x}$ be a real valued function. If $α$ and $β$ are respectively the minimum and the maximum values of $f,$ then $α^{2} + 2 β^{2}$ is equal to [2024]

1 42

2 44

3 24

4 38

Ans.
(1)

$f (x) = 3 \sqrt{x - 2} + \sqrt{4 - x}$

$f' (x) = \frac{3}{2 \sqrt{x - 2}} - \frac{1}{2 \sqrt{4 - x}}$

$∴$ $f' (x) = 0 \Rightarrow 9 (4 - x) = (x - 2) \Rightarrow x = \frac{19}{5} \in [2, 4]$

Now, $f (2) = \sqrt{2}, f (\frac{19}{5}) = 2 \sqrt{5}, f (4) = 3 \sqrt{2}$

$∴$ $α = \sqrt{2}, β = 2 \sqrt{5} \Rightarrow α^{2} + 2 β^{2} = 2 + 40 = 42$

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