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At x = 1, the function f(x)=x4−48x2+αx+120f(x)=x^4-48x^2+\alpha x+120 attains its maximum value in the interval [0, 2], then the value of α\alpha is:

Solution

✅ Correct Option: 2

Since x=1x=1 is interior to [0,2][0,2], we need f′(1)=0f'(1)=0.

f′(x)=4x3−96x+αf'(x)=4x^3-96x+\alpha, so 4−96+α=04-96+\alpha=0 and α=92\alpha=92.

Check: f′′(1)=12−96<0f''(1)=12-96<0, and f(1)=165f(1)=165 exceeds f(0)=120f(0)=120 and f(2)=128f(2)=128, confirming the maximum.

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