Question:
If $x=1$ is a critical point of the function $f(x)=\left(3 x^{2}+a x-2-a\right) e^{x}$, then :
Correct Option: 1
Solution:
$f(x)=\left(3 x^{2}+a x-2-a\right) e^{x}$
$f^{\prime}(x)=\left(3 x^{2}+a x-2-a\right) e^{x}+e^{x}(6 x+a)$
$=e^{x}\left(3 x^{2}+x(6+a)-2\right)$
$f^{\prime}(x)=0$ at $x=1$
$\Rightarrow 3+(6+a)-2=0$
$a=-7$
$f^{\prime}(x)=e^{x}\left(3 x^{2}-x-2\right)$
$=e^{x}(x-1)(3 x+2)$
$x=1$ is point of local minima
$x=\frac{-2}{3}$ is point of local maxima
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