Verified Solution • 2025–26 Session moderate

QJEEAdv17Ma0365: JEE Advanced 2017 Mathematics - Class 12 JEEAdv17Ma0365 Solution: Let for x 1. Then

Let f(x)=1−x(1+∣1−x∣)∣1−x∣cos⁡(11−x)f(x) = {{1 - x(1 + |1 - x|)} \over {|1 - x|}}\cos \left( {{1 \over {1 - x}}} \right)

for x ≠\ne 1. Then

Multiple Choice Options

  • A lim⁡x→1+f(x)\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 0
  • B lim⁡x→1−f(x)\mathop {\lim }\limits_{x \to {1^ - }} f(x) does not exist
  • C lim⁡x→1−f(x)\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 0
  • D lim⁡x→1+f(x)\mathop {\lim }\limits_{x \to {1^ + }} f(x) does not exist

Step-by-Step Verified Solution

f(x)=1−x(1+∣1−x∣)∣1−x∣cos⁡(11−x)f(x) = {{1 - x(1 + |1 - x|)} \over {|1 - x|}}\cos \left( {{1 \over {1 - x}}} \right)

Now, lim⁡x→1−f(x)\mathop {\lim }\limits_{x \to {1^ - }} f(x)

=lim⁡x→1−1−x(1+1−x)1−xcos⁡(11−x)= \mathop {\lim }\limits_{x \to {1^ - }} {{1 - x(1 + 1 - x)} \over {1 - x}}\cos \left( {{1 \over {1 - x}}} \right)

=lim⁡x→1−(1−x)cos⁡(11−x)=0= \mathop {\lim }\limits_{x \to {1^ - }} (1 - x)\cos \left( {{1 \over {1 - x}}} \right) = 0

and lim⁡x→1+f(x)=lim⁡x→1+\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }}

1−x(1+1−x)x−1cos⁡(11−x){{1 - x(1 + 1 - x)} \over {x - 1}}\cos \left( {{1 \over {1 - x}}} \right)

=lim⁡x→1+−(x+1).cos(1x+1)= \mathop {\lim }\limits_{x \to {1^ + }} - (x + 1).cos\left( {{1 \over {x + 1}}} \right) , which does not exist.

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