If the absolute maximum value of the function f(x)=(x²-2 x+7) e^((4 x³-12 x²-180 x+31)) in the interval [-3,0] is f(α), then :

Solution

Maths sir · 4 steps

Do not hunt for critical points here. Show that f' stays negative on [-3,0], so the peak sits at x=-3.

1. Differentiate. Write f(x) = p(x) eᵍ⁽ˣ⁾ with p(x) = x²-2x+7 and g(x) = 4x³-12x²-180x+31. The product rule gives f'(x) = eᵍ⁽ˣ⁾[p'(x) + p(x) g'(x)].

   f'(x) = eᵍ⁽ˣ⁾[(2x-2) + 12 (x²-2x+7)(x-5)(x+3)]

2. Check each sign. On [-3,0], 2x-2 lies between -8 and -2, so it is negative. The factor x²-2x+7 is always positive because its discriminant 4-28 is negative. Also x-5<0 and x+3 ≥ 0, so the second term is ≤ 0. That makes the bracket negative everywhere on the interval.
3. Decreasing function. Since eᵍ⁽ˣ⁾>0, we get f'(x)<0 on all of [-3,0]. So f is strictly decreasing there and has no interior maximum. The largest value is at the left end: f(-3) = 22 e³⁵⁵, far bigger than f(0) = 7 e³¹.
4. So. The absolute maximum is f(-3), so α = -3. That is option (b).

The idea to keep. Before solving f'(x)=0 on a closed interval, check the sign of f' there. If it never changes sign, the function is monotonic and the extreme values sit at the endpoints.

Answer from the NTA final answer key. Solution by PrepWiser.