Question 74

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 $$f$$, then $$\alpha^2 + 2\beta^2$$ is equal to

f(x)=3√(x-2)+√(4-x). Domain [2,4].

f'(x)=3/(2√(x-2))-1/(2√(4-x))=0. 3√(4-x)=√(x-2). 9(4-x)=x-2. 36-9x=x-2. 10x=38. x=3.8.

f(3.8)=3√1.8+√0.2=3·1.342+0.447=4.472=2√5. β=2√5.

f(2)=0+√2=√2. f(4)=3√2+0=3√2. Check: f(2)=√2≈1.41, f(4)=3√2≈4.24, f(3.8)≈4.47.

α=√2 (minimum), β=2√5 (maximum).

α²+2β²=2+40=42.

The answer is Option (1): 42.

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