Verified Solution • 2025–26 Session moderate

QJEEM21Mae41e: JEE Main 2021 Mathematics - Class 12 JEEM21Mae41e Solution: Let f be a real valued function, defined on R { 1, 1} and given …

Let f be a real valued function, defined on R −- { −- 1, 1} and given by

f(x) = 3 loge ∣x−1x+1∣−2x−1\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}} .

Then in which of the following intervals, function f(x) is increasing?

Multiple Choice Options

  • A ( −- ∞\infty , −- 1) ∪\cup ([12,∞)−{1})\left( {[{1 \over 2},\infty ) - \{ 1\} } \right)
  • B ( −- ∞\infty , ∞\infty ) −- { −- 1, 1)
  • C ( −- ∞\infty , 12{{1 \over 2}} ] −- { −- 1}
  • D ( −- 1, 12{{1 \over 2}} ]

Step-by-Step Verified Solution

f(x) = 3 loge ∣x−1x+1∣−2x−1\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}

f'(x) = {{3(x + 1)} \over {x - 1}} \times {{(x + 1) - (x - 1)} \over {{{(x + 1)}^2}}} + {2 \over {{{(x - 1)}^2}}} > 0

= {6 \over {{x^2} - 1}} + {2 \over {{{(x - 1)}^2}}} > 0

= {{2(3(x - 1) + (x + 1))} \over {{{(x - 1)}^2}(x + 1)}} = {{4(2x - 1)} \over {{{(x - 1)}^2}(x + 1)}} > 0

JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Application of Derivatives Question 102 English Explanation

⇒x∈(−∞,−1)∪[12,∞)−{1}\Rightarrow x \in ( - \infty , - 1) \cup \left[ {{1 \over 2},\infty ) - \{ 1\} } \right.

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