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JEE Main 201323 AprilOnline
Exam
JEE Main
Year
2013
Date
23 April
Mode
Online
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2013 · 23 April · Online · Mathematics · Continuity and Differentiability

Let f be a composite function of x defined by f(u)=(1)/(u^2+u-2),...

Let ff be a composite function of xx defined by f(u)=1u2+u−2,u(x)=1x−1f(u)=\frac{1}{u^2+u-2}, u(x)=\frac{1}{x-1}. Then the number of points xx where ff is discontinuous is :

Options

  1. A4
  2. B3
  3. C2
  4. D1

Correct answer

B. 3

Step-by-step solution

μ(x)=1x−1\mu(x)=\frac{1}{x-1}, which is discontinous at x=1x=1 f(u)=1u2+u−2=1(u+2)(u−1),\frac{1}{u^2+u-2}=\frac{1}{(u+2)(u-1)},which is discontinous at u=−2,1u=-2,1 when u=−2u=-2, then 1x−1=−2⇒x=12\frac{1}{x-1}=-2 \Rightarrow x=\frac{1}{2} when u=1u=1, then 1x−1=1⇒x=2\frac{1}{x-1}=1 \Rightarrow x=2 Hence given composite function is discontinous at three points, x=1,12x=1, \frac{1}{2} and 2 .
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