Summing Extremena

The principles of this question are simple but one has to work to get the solution to this question.

First note that $f(0)=-{ a }^{ 2 }-3/4<0,f(1)=1/4-2a-{ a }^{ 2 }$

We will be dealing with 3 cases.

$Case-1 \quad a \epsilon [0,1]$

Diagram1

Global minimum of $f(x)=-2{ a }^{ 2 }-3/4$ that occurs at $x=a$

Since $\left| f(x) \right| \le 1\quad \Rightarrow -1\le f(x)\le 1$

So we have $-2{ a }^{ 2 }-3/4\ge -1 \frac { -1 }{ \sqrt { 8 } } \le a\le \frac { 1 }{ \sqrt { 8 } }$

Also $1/4-2a-{ a }^{ 2 }\le 1\quad \Rightarrow a\epsilon (-\infty ,-3/2]\cup [-1/2,\infty )$

Finally for this case $a\epsilon [0 ,\frac { 1 }{ \sqrt { 8 } } ]$

$Case-2 \quad a\epsilon (1,\infty )$

Diagram2

By diagram note that $f(1)<f(0)$ hence $f(1)$ is the minimum value.

So $f(1)\ge -1$ hence $1/4-2a-{ a }^{ 2 }\ge -1$

Solving $a\epsilon [-5/2,-1/2]$

Finally for this case $a\epsilon \phi$

$Case-3 \quad a\epsilon (-\infty ,0]$

Diagram3

Now $f(1)>f(0), f(1)\le 1,f(0)\ge -1$

$\Rightarrow -{ a }^{ 2 }-3/4\ge -1$ Solving $a\epsilon [\frac { -1 }{ 2 } ,\frac { 1 }{ 2 } ]$

Also $1/4-2a-{ a }^{ 2 }\le 1$

hence $a\epsilon (-\infty ,-3/2]\cup [-1/2,\infty )$

Finally for this case $a\epsilon [-1/2,0]$

So the final range is the intersection of all the values of $a$ in the $3$ cases $a\epsilon [-1/2,\frac { 1 }{ \sqrt { 8 } } ]$

So ${ a }_{ min }=-1/2$ and ${ a }_{ max }=\frac { 1 }{ \sqrt { 8 } }$

Hence ${ a }_{ min }+{ a }_{ max }=\frac { \sqrt { 2 } -2 }{ 4 }$

Hence $n=m=2$ and $n=4$