Find the values of a and b such that the function defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{5,}&{if}&{x \le 2}\\{ax + b,}&{if}&{2 < x < 10}\\{21,}&{if}&{x \ge 10}\end{array}} \right.$ is a continuous function.}}

Since f is continuous at all x, so f is continuous at x $=$ 2, 10.

At x $=$ 2 :
$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} (5) = 5$

$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} (ax + b)$

$= \mathop {\lim }\limits_{\scriptstyle x \to 2 + h\atop\scriptstyle h \to 0} (a(2 + h) + b) = a(2 + 0) + b = 2a + b$
and $f(2) = 5$

For continuity, $\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2)$

$\Rightarrow$ $5 = 2a + b = 5 \Rightarrow 2a + b = 5$ …..(i)

At x $=$ 10 :

$\mathop {\lim }\limits_{x \to {{10}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{10}^ - }} (ax + b)$

$= \mathop {\lim }\limits_{\scriptstyle x \to 10 + h\atop\scriptstyle h \to 0} (a(10 - h) + b) = a(10 - 0) + b = 10a + b$

$\mathop {\lim }\limits_{x \to {{10}^ + }} f(x) = \mathop {\lim }\limits_{x \to {{10}^ + }} (21) = 21$

$\Rightarrow f(10) = 21$

For continuity,$\mathop {\lim }\limits_{x \to {{10}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{10}^ + }} f(x) = f(10)$

$\Rightarrow$ $10a + b = 21 \Rightarrow 10a + b = 21$ ...(ii)

Subtracting (i) from (ii), we get 8a $=$ 16 $\Rightarrow$ a$=$2

Putting a $=$ 2 in (i), we get 2(2) + b $=$ 5 $\Rightarrow$ b $=$ 5$-$ 4 $=$ 1
Hence, a $=$2, b $=$ 1.