Question:
If the absolute maximum value of the function
f(x)=(x2–2x+7)e(4x3−12x2−180x+31) in the interval [–3, 0] is f(α), then
If the absolute maximum value of the function
f(x)=(x2–2x+7)e(4x3−12x2−180x+31) in the interval [–3, 0] is f(α), then
f(x)=(x2–2x+7)e(4x3−12x2−180x+31) in the interval [–3, 0] is f(α), then
Updated On: Mar 25, 2026
- α = 0
- α = –3
- α∈ (–1, 0)
- α∈ (–3, –1]
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The Correct Option is B
Solution and Explanation
Given,
f(x)=(x2−2x+7)f1(x) e(4x3−12x2−180x+31)f2(x)
f1(x) = x2 – 2x + 7
f1′(z)=2z−2,
so f(x) is decreasing in [–3, 0]
and positive also
f2(x)=e4x3−12x2−180x+31
f2‘(x)=e4x3−12x2−180x+31.12x2–24x–180
=12(x−5)(x+3)x4x3−12x2−180x+31
So, f2(x) is also decreasing and positive in {–3, 0}
∴ absolute maximum value of f(x) occurs at x = –3
∴ α = -3
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