Q100 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

If $f(x)=\left\{\begin{array}{ll}3 a x-2 b, & x\gt1 \\ a x+b+1, & x \lt 1\end{array}\right.$ and $\lim _{x \rightarrow 1} f(x)$ exists, then the relation between $a$ and $b$ is

  • A
    $3 a-2 b=1$
  • Correct answer
    B
    $2 a-3 b=1$
  • C
    $2 a+3 b=1$
  • D
    $2 a+3 b=-1$

Solution

Given, $f(x)= \begin{cases}3 a x-2 b, & x\gt1 \\ a x+b+1, & x \lt 1\end{cases}$ $\begin{aligned} & \text { Since, } \lim _{x \rightarrow 1} f(x) \text { exist } \Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\ & \Rightarrow \lim _{x \rightarrow 1^{-}}(a x+b+1)=\lim _{x \rightarrow 1^{+}}(3 a x-2 b) \\ & \Rightarrow a+b+1=3 a-2 b \Rightarrow 2 a-3 b=1\end{aligned}$

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

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