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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2015 · Mathematics · Continuity and Differentiability

Given a function f(x)=\ array ccc -1 & If & x ≤ 0 a x+b & If & 0 <...

Given a function f(x)=\left\{−1 If x≤0ax+b If 0<x<11 If x≥1\begin{array}{ccc} -1 & \text{ If } & x \leq 0 \\ a x+b & \text{ If } & 0 < x < 1 \\ 1 & \text{ If } & x \geq 1 \end{array}\right. where a,bareconstants.Thefunctioniscontinuouseverywhere.Whatisthevalueofb?\mathrm{a}, \mathrm{b} are constants. The function is continuous everywhere. What is the value of b?

Options

  1. A–1
  2. B1
  3. C0
  4. D2

Correct answer

A. –1

Step-by-step solution

F(x)={−1,x≤0ax+b,0<x<11,x≥1 At x=0,  L.H.L. = R.HL. lim⁡x→0−f(x)=lim⁡x→0+f(x)−1=lim⁡x→0(ax+b)−1=b\begin{aligned} & \mathrm{F}(\mathrm{x})=\left\{\begin{array}{ccc} -1 & , & \mathrm{x} \leq 0 \\ \mathrm{ax}+\mathrm{b} & , & 0 < \mathrm{x} < 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. lim⁡x→1−f(x)=lim⁡x→1+f(x)a+b=1a−1=1a=1+1=2\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 lim⁡x→0−f(x)=lim⁡x→0+f(x)−1=a(0)+b⇒b=−1\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ & -1=a(0)+b \Rightarrow b=-1 \end{aligned} Now, f(x)\mathrm{f}(\mathrm{x}) is also continuous at x=1\mathrm{x}=1lim⁡x→1−f(x)=lim⁡x→1+f(x)a(1)+b=1a=1−b=1+1a=2.\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.
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