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JEE Advanced 2017Paper 2
Exam
JEE Advanced
Year
2017
Paper
Paper 2
Subject
Mathematics
Chapter
Limits

JEE Advanced 2017 · Paper 2 · Mathematics · Limits

Let f(x) = (1 - x ( 1 + |1 - x| ))/(| 1 - x |) ((1)/(1 - x))forx ≠...

Let f(x)=1−x(1+∣1−x∣)∣1−x∣cos⁡(11−x)forx≠1,f\left(x\right) = \dfrac{1 - x ( 1 + \left|1 - x\right| )}{| 1 - x |}\cos \left(\dfrac{1}{1 - x}\right)forx \ne 1, then

Options

  1. Alim⁡x→1+f(x)\lim_{x \to {1}^{+}} f ( x ) does not exist
  2. Blim⁡x→1−f(x)\lim_{x \to {1}^{-}} f ( x ) does not exist
  3. Clim⁡x→1−f(x)=0\lim_{x \to {1}^{-}} f\left(x\right) = 0
  4. Dlim⁡x→1+f(x)=0\lim_{x \to {1}^{+}} f\left(x\right) = 0

Correct answer

A. lim⁡x→1+f(x)\lim_{x \to {1}^{+}} f ( x ) does not exist

Step-by-step solution

f\left(x\right) = \left\{{(1−x)cos11−x,x<1−(1+x)cos11−x,x>1\begin{cases}\left(1 - x\right)cos \dfrac{1}{1 - x}, & x < 1 \\ - \left(1 + x\right)cos \dfrac{1}{1 - x}, & x > 1\end{cases}\right)

lim⁡x→1+f(x)=doesnotexist\lim_{x \to {1}^{+}} f\left(x\right) = does not exist; (as cost\text{cos}\text{t} can take any value between -1 and 1,sooscillating.)</p><p></p><p>lim⁡x→1−f(x)=01, so oscillating.)</p><p> </p><p>\lim_{x \to {1}^{-}} f\left(x\right) = 0: (As cost\text{cos}\text{t} term varies between -1 and 1 for any value of t and is multiplied with zero)

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