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JEE Main 201910 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2019 · 10 January · Morning Shift · Mathematics · Limits

For each t ∈ R, let [t] be the greatest integer less than or equal...

For each t∈R,t \in R, let \left[t\right]bethegreatestintegerlessthanorequaltobe the greatest integer less than or equal tot.Then,. Then,\lim_{x \to {1}^{+}}\dfrac{\left(1 - \left|x\right| + \sin \left|1 - x\right|\right)\sin \left(\left[1 - x\right]\dfrac{\pi}{2}\right)}{\left|1 - x\right|\left[1 - x\right]}

Options

  1. Aequals 00
  2. Bequals −1- 1
  3. Cdoes not exist
  4. Dequal 11

Correct answer

A. equals 00

Step-by-step solution

lim⁡x→1+(1−∣x∣+sin⁡∣1−x∣)sin⁡([1−x]π2)∣1−x∣[1−x]\lim_{x \to {1}^{+}}\dfrac{\left(1 - \left|x\right| + \sin \left|1 - x\right|\right)\sin \left(\left[1 - x\right]\dfrac{\pi}{2}\right)}{\left|1 - x\right|\left[1 - x\right]}

=lim⁡h→0(1−∣(1+h)∣+sin⁡∣1−(1+h)∣)sin⁡(π2[1−(1+h)])∣1−(1+h)∣[1−(1+h)]= \lim_{h \to 0}\dfrac{\left(1 - \left|\left(1 + h\right)\right| + \sin \left|1 - \left(1 + h\right)\right|\right)\sin \left(\dfrac{\pi}{2}\left[1 - \left(1 + h\right)\right]\right)}{\left|1 - \left(1 + h\right)\right|\left[1 - \left(1 + h\right)\right]}

=lim⁡h→0(1−h−1+sin⁡∣(1−h−1)∣)sin⁡(π2[1−1−h])∣(1−1−h)∣[1−1−h]= \lim_{h \to 0}\dfrac{\left(1 - h - 1 + \sin \left|\left(1 - h - 1\right)\right|\right)\sin \left(\dfrac{\pi}{2}\left[1 - 1 - h\right]\right)}{\left|\left(1 - 1 - h\right)\right|\left[1 - 1 - h\right]}

=lim⁡h→0(−h+sinh)sin⁡(π2[−h])∣−h∣[−h]= \lim_{h \to 0}\dfrac{\left(- h + sinh\right)\sin \left(\dfrac{\pi}{2}\left[- h\right]\right)}{\left|- h\right|\left[- h\right]}

=lim⁡h→0(−h+sinh)sin⁡(−π2)h(−1)=lim⁡h→0(−h+sinh)(−1)−h= \lim_{h \to 0}\dfrac{\left(- h + sinh\right)\sin \left(- \dfrac{\pi}{2}\right)}{h\left(- 1\right)} = \lim_{h \to 0}\dfrac{\left(- h + sinh\right)\left(- 1\right)}{- h}

=1−1=0= 1 - 1 = 0.

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