At \( x=1 \), the function \( f(x)=x^4-48x^2+\alpha x+120 \) attains its maximum value in the interval [0, 2], then the value of \( \alpha \) is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 92
Since x = 1 is an interior point of [0, 2], a maximum there needs \( f'(1)=0 \):
\( \displaystyle f'(x)=4x^3-96x+\alpha \)
\( \displaystyle f'(1)=4-96+\alpha=0 \)
\( \displaystyle \alpha=92 \)
Check that it is a maximum: \( f''(x)=12x^2-96 \), so \( f''(1)=-84\lt 0 \). In fact \( f''(x)\lt 0 \) on the whole of [0, 2], so f′ is decreasing there and x = 1 gives the maximum on the interval.