Question

$A$ function $f\left(x\right)$ is defined as
$f\left(x\right)=\{\begin{array}{cc}ax-b& x\le 1\\ 3x,& 1<x<2\\ b{x}^2-a& x\ge 2\end{array}$ is continuous at
$x=1,2$ then:

• A. $a=5,b=2$
• B. $a=6,b=3$
• C. $a=7,b=4$
• D. $a=8,b=5$

Answer 1

Answer: (B)

$a=6,b=3$

Given $f\left(x\right)$ is continuous at $x=1$

$\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$

Now $LHL=\underset{x\to 1^{-}}{lim}f\left(x\right)$
$=\underset{x\to 1^{-}}{lim}ax-b$
$LHL=a-b$

And $f\left(1\right)=3$

$\Rightarrow a-b=3$ .....(i)

Given $f\left(x\right)$ is continuous at $x=2.$

$\underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)$

Now $LHL=\underset{x\to 2^{-}}{lim}f\left(x\right)$
$=\underset{x\to 2^{-}}{lim}3x$
$LHL=6$

And $f\left(2\right)=4b-a$

$\Rightarrow 4b-a=6$ .....(ii)

Solving (i) and (ii)
$\Rightarrow b=3,a=6$