Q233 / 270 · NDA 2015 (Phase 1) · Mathematics

Given a function $f(x)=\left\{\begin{array}{ccc} -1 & \text { If } & x \leq 0 \\ a x+b & \text { If } & 0 \lt x \lt 1 \\ 1 & \text { If } & x \geq 1 \end{array}\right.$ where $\mathrm{a}, \mathrm{b}$ are constants. The function is continuous everywhere. What is the value of b?

  • Correct answer
    A
    –1
  • B
    1
  • C
    0
  • D
    2

Solution

$\begin{aligned} & \mathrm{F}(\mathrm{x})=\left\{\begin{array}{ccc} -1 & , & \mathrm{x} \leq 0 \\ \mathrm{ax}+\mathrm{b} & , & 0 \lt \mathrm{x} \lt 1 \\ 1, & \mathrm{x} \geq 1 \end{array}\right. \\ & \text { At } x=0 \text {, } \\ & \text { L.H.L. = R.HL. } \\ & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ & -1=\lim _{x \rightarrow 0}(a x+b) \\ & -1=b \end{aligned}$ Since the given function is also continuous at x = 1 L.H.L. = R.H.L. $\begin{aligned} & \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\ & a+b=1 \\ & a-1=1 \\ & a=1+1=2\end{aligned}$ Function is continuous at x = 0 $\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ & -1=a(0)+b \Rightarrow b=-1 \end{aligned}$ Now, $\mathrm{f}(\mathrm{x})$ is also continuous at $\mathrm{x}=1$ $\begin{aligned} & \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\ & a(1)+b=1 \\ & a=1-b=1+1 \\ & a=2 . \end{aligned}$ b = –1.

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

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