Question:
If \(x=1\) is a critical point of the function \( f(x)=(3x^2+ax-2-a)e^x \), then :
If \(x=1\) is a critical point of the function \( f(x)=(3x^2+ax-2-a)e^x \), then :
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Find \(a\) from \(f'(1)=0\), factorise \(f'(x)\) and study the sign change (or use \(f''\)).
Updated On: Nov 5, 2026
- \(x=1\) is a local maxima and \(x=-\frac23\) is a local minima of f.
- \(x=1\) and \(x=-\frac23\) are local maxima of f.
- \(x=1\) and \(x=-\frac23\) are local minima of f.
- \(x=1\) is a local minima and \(x=-\frac23\) is a local maxima of f.
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The Correct Option is D
Solution and Explanation
\(f'(x)=e^x\left[(3x^2+ax-2-a)+(6x+a)\right]=e^x\left(3x^2+(a+6)x-2\right)\).
\(f'(1)=0\Rightarrow 3+a+6-2=0\Rightarrow a=-7\).
\(f'(x)=e^x(3x^2-x-2)=e^x(3x+2)(x-1)\), critical points \(x=1\) and \(x=-\frac23\).
Sign of \(f'\): positive for \(x<-\frac23\), negative for \(-\frac23<x<1\), positive for \(x>1\).
So \(f'\) changes + to - at \(x=-\frac23\) (local maxima) and - to + at \(x=1\) (local minima).
Answer: (4) \(x=1\) is a local minima and \(x=-\frac23\) is a local maxima of f.
\(f'(1)=0\Rightarrow 3+a+6-2=0\Rightarrow a=-7\).
\(f'(x)=e^x(3x^2-x-2)=e^x(3x+2)(x-1)\), critical points \(x=1\) and \(x=-\frac23\).
Sign of \(f'\): positive for \(x<-\frac23\), negative for \(-\frac23<x<1\), positive for \(x>1\).
So \(f'\) changes + to - at \(x=-\frac23\) (local maxima) and - to + at \(x=1\) (local minima).
Answer: (4) \(x=1\) is a local minima and \(x=-\frac23\) is a local maxima of f.
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