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JEE Main 20127 MayOnline
Exam
JEE Main
Year
2012
Date
7 May
Mode
Online
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2012 · 7 May · Online · Mathematics · Continuity and Differentiability

If x+|y|=2 y , then y as a function of x , at x=0 is

If x+∣y∣=2yx+|y|=2 y, then yy as a function of xx, at x=0x=0 is

Options

  1. Adifferentiable but not continuous
  2. Bcontinuous but not differentiable
  3. Ccontinuous as well as differentiable
  4. Dneither continuous nor differentiable

Correct answer

B. continuous but not differentiable

Step-by-step solution

Given x+∣y∣=2yx+|y|=2 y⇒x+y=2y or x−y=2y⇒x=y or x=3y\begin{aligned} & \Rightarrow \quad x+y=2 y \text{ or } x-y=2 y \\ & \Rightarrow \quad x=y \text{ or } x=3 y \end{aligned}Thisrepresentastraightlinewhichpassesthroughorigin.Hence,This represent a straight line which passes through origin. Hence,x+|y|=2 yiscontinuousatis continuous atx=0.Now,wecheckdifferentiabilityat. Now, we check differentiability atx=0 \begin{gathered} x+|y|=2 y \Rightarrow x+y=2 y, y \geq 0 \\ x-y=2 y, y < 0 \end{gathered} Thus, f(x)=\left\{x,y<0x/3,y≥0\begin{array}{ll}x, & y < 0 \\ x / 3, & y \geq 0\end{array}\right\} Now, L.H.D. =lim⁡h→0−f(x+h−)fx−h()=lim⁡h→0−x+h−x−h=−1 R.H.D =lim⁡h→0+f(x+h−)fxh()=lim⁡h→0+x+h3−x3h=lim⁡h→0+13=13\begin{aligned} & \text{ Now, L.H.D. }=\lim _{h \rightarrow 0^{-}} \frac{f(x+h-) f x}{-h}(\quad) \\ & =\lim _{h \rightarrow 0^{-}} \frac{x+h-x}{-h}=-1 \\ & \text{ R.H.D }=\lim _{h \rightarrow 0^{+}} \frac{f(x+h-) f x}{h}(\quad) \\ & =\lim _{h \rightarrow 0^{+}} \frac{\frac{x+h}{3}-\frac{x}{3}}{h}=\lim _{h \rightarrow 0^{+}} \frac{1}{3}=\frac{1}{3} \\ & \end{aligned}Since,L.H.DSince, L.H.D\neqR.H.D.atR.H.D. atx=0 \thereforegivenfunctionisnotdifferentiableatgiven function is not differentiable atx=0
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