If x=1 is a critical point of the function
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 :

1. $x=1$ is a local minima and $x=-\frac{2}{3}$ is a local maxima of $f$.

2. $x=1$ is a local maxima and $x=-\frac{2}{3}$ is a local minima of $f$.

3. $x=1$ and $x=-\frac{2}{3}$ are local minima of $f$.

4. $x=1$ and $x=-\frac{2}{3}$ are local maxima of $f$.

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