Q80 / 200 · Manipal (MET) 2013 · Mathematics

Function $f(x)$ is defined as follows $f(x)=\left\{\begin{array}{cc}a x-b, & x \leq 1 \\ 3 x, & 1 \lt x \lt 2 \\ b x^2-a, & x \geq 2\end{array}\right.$ If $f(x)$ 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

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

Solution

Given; $f(x)$ is continuous at $x=1$ $\begin{aligned} & \therefore f(1)=\text { RHL } \\ & \Rightarrow f(1)=\lim _{x \rightarrow 1}+f(x) \Rightarrow f(x)=\lim _{h \rightarrow 0} f(1+h)\end{aligned}$ $\Rightarrow a-b=\lim _{h \rightarrow 0} 3(1+h) \Rightarrow a-b=3$ ...(i) Again, given $f(x)$ is discontinuous at $x=2$. $\begin{aligned} & \therefore \quad \mathrm{LHL} \neq f(x) \\ & \Rightarrow \lim _{x \rightarrow 2^{-}} f(x) \neq f(2) \Rightarrow \lim _{h \rightarrow 0} f(2-h) \neq f(2)\end{aligned}$ $\Rightarrow \lim _{h \rightarrow 0} 3(2-h) \neq 4 b-a \Rightarrow 6 \neq 4 b-a$ ....(ii) Assume, $6=4 b-a$ then from (i) and (ii), we get $b=3$ $\therefore \quad$ locus $\quad y=3$ Which is impossile $(\because 6 \neq 4 b-a)$ Hence, locus of $(a, b)$ is $x-y=3$ excluding the point when it cuts the line $y=3$.

Hence, the correct answer is option: 'A'.

Relevant Questions