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If α\alpha and β\beta are absolute maximum and absolute minimum values of a function f(x)=x2+2xf(x)=\dfrac{x}{2}+\dfrac{2}{x} on the interval [1,2][1,2], then the value of α+β\alpha+\beta is:

Solution

✅ Correct Option: 3

f′(x)=12−2x2=0f'(x) = \dfrac{1}{2} - \dfrac{2}{x^2} = 0 gives x=2x = 2, an endpoint of the interval.

Evaluate at the endpoints: f(1)=12+2=52f(1) = \dfrac{1}{2} + 2 = \dfrac{5}{2} and f(2)=1+1=2f(2) = 1 + 1 = 2.

So α=52\alpha = \dfrac{5}{2}, β=2\beta = 2, and α+β=92\alpha + \beta = \dfrac{9}{2}.

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