Question

Find the values of $a$ and $b$ such that the function defined by

$f\left(x\right)=$ $\{\begin{array}{c}5,ifx\le 2\\ ax+b,if2<x<10\\ 21,ifx\ge 10\end{array}$ is a continuous function.

Answer 1

Given definition of $f$ is

$f\left(x\right)=\{\begin{array}{c}5,ifx\le 2\\ ax+b,if2<x<10\\ 21,ifx\ge 10\end{array}$ .....(1)

Also, given $f\left(x\right)$ is a continuous function.

So, $f\left(x\right)$ is continuous at all points .

So, $f\left(x\right)$ is continuous at $x=2$

$\Rightarrow \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=f\left(2\right)$ ....(2)

Now, $RHL=\underset{x\to 2^{+}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(2+h)$

$=\underset{h\to 0}{lim}a(2+h)+b$

$\Rightarrow RHL=2a+b$

Also, $f\left(2\right)=5$

Substituting these values in (2), we get

$2a+b=5$ .......(3)

Also, $f\left(x\right)$ is continuous at $x=10$

$\therefore \underset{x\to {10}^{-}}{lim}f\left(x\right)=\underset{x\to {10}^{+}}{lim}f\left(x\right)=f\left(10\right)$ ......(4)

Now, $LHL=\underset{x\to {10}^{-}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(10-h)$

$=\underset{h\to 0}{lim}a(10-h)+b$

$=10a+b$

Also, by (1), $f\left(10\right)=21$

Substituting these values in eq (4), we get

$10a+b=21$ .....(5)

Solving eq $\left(3\right)$ from eqn $\left(5\right)$, we get

$8a=16$

$\Rightarrow a=2$

Put this value in $\left(3\right)$, we get

$2\left(2\right)+b=5$

$\Rightarrow b=1$