If the function $f(x)= \begin{cases}3 a x+b, & \text { for } x 1\end{cases}$ $x=1$. Then, the values of $a$…
If the function $f(x)= \begin{cases}3 a x+b, & \text { for } x < 1 \\ 11, & \text { for } x=1 \text { is continuous at } \\ 5 a x-2 b, & \text { for } x>1\end{cases}$ $x=1$. Then, the values of $a$ and $b$ are
$a=2, b=3$
$a=3, b=3$
$a=2, b=2$
$a=3, b=2$
Solution
$\begin{aligned}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 3 a x+b=3 a+b \\
& \lim _{x \rightarrow 1^{+}} f(x) \lim _{x \rightarrow 1^{+}} 5 a x-2 b=5 a-2 b
\end{aligned}$
$f(1)=11$ and function is continuous at $x=1$, so we write
$3 \mathrm{a}+\mathrm{b}=11=5 \mathrm{a}-2 \mathrm{~b} \Rightarrow 2 \mathrm{a}=3 \mathrm{~b}$
This condition is satisfied when $\mathrm{a}=3, \mathrm{~b}=2$