The sum of values of a and b such that the function f(x) defined by

$f\left(x\right)=\{\begin{array}{cc}3,& x\le 1\\ ax+b,& 1<x<5\\ 10,& x\ge 5\end{array}\text{is a continuous function is}$

The correct answer/option is 3.

To find the values of ^a and ^b that make the function continuous, we must ensure that the function is continuous at the transition points, which are $x=1$ and $x=5$.

For a function $f\left(x\right)$ to be continuous at a point $x=c$, the left-hand limit, right-hand limit, and the value of the function at that point must all be equal:
$\underset{x\to c^{-}}{lim}f\left(x\right)=\underset{x\to c^{+}}{lim}f\left(x\right)=f\left(c\right)$

Step 1: Check continuity at $x=1$
The left-hand limit and value of the function at $x=1$ is:
$\underset{x\to 1^{-}}{lim}f\left(x\right)=f\left(1\right)=3$
The right-hand limit at $x=1$ is:
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(ax+b)=a\left(1\right)+b=a+b$
For continuity at $x=1$, we equate these two values:
$a+b=3$ (Equation 1)

Step 2: Check continuity at $x=5$
The left-hand limit at $x=5$ is:
$\underset{x\to 5^{-}}{lim}f\left(x\right)=\underset{x\to 5^{-}}{lim}(ax+b)=a\left(5\right)+b=5a+b$
The right-hand limit and value of the function at $x=5$ is:
$\underset{x\to 5^{+}}{lim}f\left(x\right)=f\left(5\right)=10$
For continuity at $x=5$, we equate these two values:
$5a+b=10$ (Equation 2)

Step 3: Solve for ^a and ^b
We have a system of two linear equations:
1) $a+b=3$
2) $5a+b=10$
Subtracting Equation 1 from Equation 2:
$(5a+b)-(a+b)=10-3$
$4a=7$
$a=\frac{7}{4}$
Substituting $a=\frac{7}{4}$ back into Equation 1:
$\frac{7}{4}+b=3$
$b=3-\frac{7}{4}=\frac{12}{4}-\frac{7}{4}=\frac{5}{4}$

Step 4: Find the sum of the values of ^a and ^b
We need to find $a+b$:
$a+b=\frac{7}{4}+\frac{5}{4}=\frac{12}{4}=3$

Thus, the sum of the values of ^a and ^b is indeed 3.