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

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

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

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

Correct Option: , 4

Solution:

The given function

$f(x)=\left(3 x^{2}+a x-2-a\right) e^{x}$

$f^{\prime}(x)=(6 x+a) e^{x}+\left(3 x^{2}+a x-2-a\right) e^{x}$

$f^{\prime}(x)=\left[3 x^{2}+(a+6) x-2\right] e^{x}$

$\because x=1$ is critical point:

$\therefore f^{\prime}(1)=0$

$\Rightarrow(3+a+6-2) \cdot e=0$

$\Rightarrow a=-7$ $(\because e>0)$

$\therefore f^{\prime}(x)=\left(3 x^{2}-x-2\right) e^{x}$

$=(3 x+2)(x-1) e^{x}$

\begin{tabular}{c}

$+_{1}-_{1}+$ \\

\hline$-2 / 3 \quad 1$

\end{tabular}

$\therefore x=-\frac{2}{3}$ is point of local maxima.

and $x=1$ is point of local minima.