Q65 / 125 · BITSAT 2012 · Mathematics

If the function $f(x)=\begin{cases}A x-B, & x \leq 1 \\ 3 x & , 1<x<2 \\ B x^{2}-A, & x \geq 2\end{cases}$ is continuous at $x=2$ and $4,$ then the values of a and $b$ are

  • A
    3,5
  • Correct answer
    B
    3,-5
  • C
    0,3
  • D
    0,5

Solution

Since $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=2$ $\therefore \mathrm{f}(2)=\lim _{\mathrm{x} \rightarrow 2^{+}} \mathrm{f}(\mathrm{x}) \Rightarrow 1=\lim _{\mathrm{x} \rightarrow 2^{+}}(\mathrm{ax}+\mathrm{b})$ $\therefore 1=2 \mathrm{a}+\mathrm{b}$ Again $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=4$ $\therefore \mathrm{f}(4)=\lim _{x \rightarrow 4^{-}} \mathrm{f}(\mathrm{x}) \Rightarrow 7=\lim _{\mathrm{x} \rightarrow 4^{-}}(\mathrm{ax}+\mathrm{b})$ $\therefore 7=4 \mathrm{a}+\mathrm{b}$ Solving (1) and $(2),$ we get $a=3, b=-5$

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

Relevant Questions