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JEE Main 20199 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2019 · 9 January · Morning Shift · Mathematics · Continuity and Differentiability

Let f: R → R be a function defined as f(x) = \ array ll 5, ifx ≤ 1...

Let f:R→Rbeafunctiondefinedas<br/><br/>f: R \to R be a function defined as<br /><br />f\left(x\right) = \left\{5,ifx≤1a+bx,if1<x<3b+5x,if3≤x<530,ifx≥5\begin{array}{ll}5, ifx \le 1 \\ a + bx, if1 < x < 3 \\ b + 5x, if3 \le x < 5 \\ 30, ifx \ge 5\end{array}\right)

Then ff is:

Options

  1. Acontinuous if a=−5a = - 5 and b=10b = 10
  2. Bcontinuous if a=0a = 0 and b=5b = 5
  3. Cnot continuous for any values of aa and bb
  4. Dcontinuous if a=5a = 5 and b=5b = 5

Correct answer

C. not continuous for any values of aa and bb

Step-by-step solution

f\left(x\right) = \left\{5:x≤1a+bx:1<x<3b+5x:3≤x<530:x≥5\begin{array}{ll}5: x \le 1 \\ a + bx: 1 < x < 3 \\ b + 5x: 3 \le x < 5 \\ 30: x \ge 5\end{array}\right)

For f(x)f ( x ) to be continuous at x=5x = 5

lim⁡x→5−f(x)=lim⁡x→5+f(x)=f(5)\lim_{x \to {5}^{-}}f\left(x\right) = \lim_{x \to {5}^{+}}f\left(x\right) = f\left(5\right)

⇒lim⁡x→5−(b+5x)=30⇒ \lim_{x \to {5}^{-}}\left(b + 5x\right) = 30

⇒b+25=30⇒b=5⇒ b + 25 = 30 ⇒ b = 5

For f(x)f ( x ) to be continuous at x=1x = 1

lim⁡x→1+f(x)=lim⁡x→1−f(x)=f(1)\lim_{x \to {1}^{+}}f\left(x\right) = \lim_{x \to {1}^{-}}f\left(x\right) = f\left(1\right)

⇒lim⁡x→1+(a+5x)=5⇒ \lim_{x \to {1}^{+}}\left(a + 5x\right) = 5

⇒a=0⇒ a = 0

Now at x=3,x = 3,

LHL=lim⁡x→3−(0+5x)=15LHL = \lim_{x \to {3}^{-}}\left(0 + 5x\right) = 15

RHL=lim⁡x→3+(5+5x)=20RHL = \lim_{x \to {3}^{+}}\left(5 + 5x\right) = 20

Hence function f(x)f ( x ) is discontinuous at x=3x = 3 for all a,b∈Ra, b \in R .
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