Question

Function $f\left(x\right)$ is defined as follows
$f\left(x\right)=\{\begin{array}{c}ax-b,x\le 1\\ 3x,1<x<2\\ b{x}^2-a,x\ge 2\end{array}$
If $f\left(x\right)$ is continuous at $x=1$, but discontinuous at $x=2$, then the locus of the point $(a,b)$ is a straight line excluding the point where it cuts the line

• A. $y=3$
• B. $y=2$
• C. $y=0$
• D. $y=1$

Answer 1

The correct option is A $y=3$

Given: $f\left(x\right)$ is continuous at $x=1$
$\therefore f\left(1\right)=$ RHL
$\Rightarrow f\left(1\right)=\underset{x\to 1}{lim}f\left(x\right)\Rightarrow f\left(x\right)=\underset{h\to 0}{lim}f(1+h)$
$\Rightarrow a-b=\underset{h\to 0}{lim}3(1+h)\Rightarrow a-b=3$ ...(i)
Again, given $f\left(x\right)$ is discontinuous at $x=2$.
$\therefore$ LHL $\ne f\left(x\right)$
$\Rightarrow \underset{x\to 2^{-}}{lim}f\left(x\right)\ne f\left(2\right)\Rightarrow \underset{h\to 0}{lim}f(2-h)\ne f\left(2\right)$
$\Rightarrow \underset{h\to 0}{lim}3(2-h)\ne 4b-a\Rightarrow 6\ne 4b-a$ ...(ii)
Assume, $6=4b-a$
then from (i) and (ii), we get $b=3$.
Thus locus is $y=3$.
Which is impossible .....$(\text{Since}6\ne 4b-a)$
Hence, locus of $(a,b)$ is $x-y=3$ excluding the point when it cuts the line $y=3$.