Assertion $\left(A\right):$ The maximum value of $-x^2+3x+1$ is $\frac{11}{4}.$

Reason $\left(R\right):$ If $a<0,$ the maximum value of $a{x}^2+bx+c$ exist at $x=\frac{-b}{2a}.$

The correct option among the following is

Assertion $\left(A\right):$ Let $y=-x^2+3x+1.$

To find the maximum or minimum value of $y$ differentiate both sides w.r.to $x,$ to get

$\frac{dy}{dx}=-2x+3$

And to find the maximum or minimum value, put $\frac{dy}{dx}=0$

$\Rightarrow -2x+3=0$

$\Rightarrow x=\frac{3}{2}$ and $\frac{d^{2}y}{d{x}^2}=-2<0.$

And, we know that at the point of maxima, the second derivative of the function is negative.

Hence, $x=\frac{3}{2}$ is the point of maxima and the maximum value of the function is $-{\left(\frac{3}{2}\right)}^2+3\left(\frac{3}{2}\right)+1$

$=-\frac{9}{4}+\frac{9}{2}+1=\frac{13}{4}.$

Thus, the assertion is wrong.

Reason $\left(R\right):$ Let $y=a{x}^2+bx+c, a<0$

$\Rightarrow \frac{dy}{dx}=2ax+b=0$

$\Rightarrow x=-\frac{b}{2a}.$

And, $\frac{d^{2}y}{d{x}^2}=2a<0,$ hence the function $y=a{x}^2+bx+c,$ has a point of maxima at $x=-\frac{b}{2a}.$