If $x=1$ is a critical point of the function $f\left(x\right)=\left(3x^2+ax-2-a\right)e^x$, then:

The correct option is A

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

Explanation for the correct option:

Step 1: Find the value of $a$:

Given,

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

On differentiating

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

Since, $x=1$ is a critical point, then

$\begin{array}{rcl}f\text{'}\left(1\right)& =& 0\\ & \Rightarrow & \left(6+a\right)e^1+\left(3+a-2-a\right)e^1=0\\ & \Rightarrow & \left(6+a+3-2\right)e=0\\ & \Rightarrow & a=-7\end{array}$

Substitute the value of $a$ in the given function.

$f\left(x\right)=\left(3x^2-7x+5\right)e^x...\left(1\right)$

Step 2: Find the value of $x$:

On differentiating equation $\left(1\right)$, we get

$f\text{'}\left(x\right)=\left(6x-7\right)e^x+\left(3x^2-7x+5\right)e^x$

Now to find the local maxima and local minima, make $f\text{'}\left(x\right)=0$

Then,

$\begin{array}{rcl}\left(6x-7\right)e^x+\left(3x^2-7x+5\right)e^x& =& 0\\ & \Rightarrow & e^x\left(3x^2-x-2\right)=0\\ & \Rightarrow & 3x^2-x-2=0\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{e}}^{\mathbf{x}}\mathbf{\ne }\mathbf{0}\mathbf{]}\\ & \Rightarrow & 3x^2-3x+2x-2=0\\ & \Rightarrow & 3x\left(x-1\right)+2\left(x-1\right)=0\\ & \Rightarrow & x=1,\frac{-2}{3}\end{array}$

Step 3: Find the local maxima and local minima:

To check for local maxima and local minima find the double derivative of $x$, if $f\text{'}\text{'}\left(x\right)$ is positive then at that point it has local minima, if $f\text{'}\text{'}\left(x\right)$ is negative then at that point it has local maxima.

So,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=e^x\left(3x^2-x-2\right)+e^x\left(6x-1\right) \\ f\text{'}\text{'}\left(1\right)=e\left(3-1-2\right)+e\left(6-1\right) \\ =5e>0 \\ f\text{'}\text{'}\left(\frac{-2}{3}\right)=e^{\frac{-2}{3}}\left(\frac{4}{3}+\frac{2}{3}-2\right)+e^{\frac{-2}{3}}\left(6\left(\frac{-2}{3}\right)-1\right) \\ =-5e^{\frac{-2}{3}}<0 \end{gathered}$$

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

Hence, the correct option is A.