1. Given function

$f(x)=\frac{2x^2-3x+8}{2x^2+3x+8}$

We need the maximum and minimum values of $f(x)$ over all real $x$, then add them.

2. First check domain

Denominator:

$2x^2+3x+8$

Its discriminant is

$\Delta=3^2-4\cdot 2\cdot 8=9-64=-55<0$

Since the leading coefficient is positive, we have

$2x^2+3x+8>0 \quad \text{for all real } x$

So $f(x)$ is defined for all real $x$.

3. Let

$y=f(x)=\frac{2x^2-3x+8}{2x^2+3x+8}$

To find the range of $y$, cross-multiply:

$y(2x^2+3x+8)=2x^2-3x+8$

Rearrange:

$2yx^2+3yx+8y-2x^2+3x-8=0$

$2(y-1)x^2+3(y+1)x+8(y-1)=0$

This is a quadratic in $x$. For real $x$, its discriminant must be non-negative.

4. Apply discriminant condition

For

$2(y-1)x^2+3(y+1)x+8(y-1)=0$

we need

$[3(y+1)]^2-4\cdot 2(y-1)\cdot 8(y-1)\ge 0$

$9(y+1)^2-64(y-1)^2\ge 0$

Expand:

$9(y^2+2y+1)-64(y^2-2y+1)\ge 0$

$9y^2+18y+9-64y^2+128y-64\ge 0$

$-55y^2+146y-55\ge 0$

Multiply by $-1$:

$55y^2-146y+55\le 0$

So $y$ lies between the roots of

$55y^2-146y+55=0$

5. Find the roots

$y=\frac{146\pm\sqrt{146^2-4\cdot 55\cdot 55}}{2\cdot 55}$

$=\frac{146\pm\sqrt{21316-12100}}{110}$

$=\frac{146\pm\sqrt{9216}}{110}$

$=\frac{146\pm 96}{110}$

Thus the two roots are

$y_{\min}=\frac{146-96}{110}=\frac{50}{110}=\frac{5}{11}$

and

$y_{\max}=\frac{146+96}{110}=\frac{242}{110}=\frac{11}{5}$

Hence,

$\min f(x)=\frac{5}{11}, \qquad \max f(x)=\frac{11}{5}$

6. Sum of maximum and minimum values

$\frac{11}{5}+\frac{5}{11}=\frac{121+25}{55}=\frac{146}{55}$

So,

$m=146,\quad n=55$

and

$m+n=146+55=201$

7. Option check

$201$ corresponds to Option C.

8. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.