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Application of Derivatives question

2021 · 16 Mar · Shift 2 · Q25
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  5. /2021 · 16 Mar · Shift 2 · Q25

Application of Derivatives question

2021 · 16 Mar · Shift 2 · Q25

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
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}}​x+1x−1​​−x−12​. Then in which of the following intervals, function f(x) is increasing?
  1. A
    (−∞-\infty−∞, −-− 1) ∪\cup∪ ([12,∞)−{1})\left( {[{1 \over 2},\infty ) - \{ 1\} } \right)([21​,∞)−{1})
  2. B
    (−∞-\infty−∞, ∞\infty∞) −-− {−-− 1, 1)
  3. C
    (−∞-\infty−∞, 12{{1 \over 2}}21​] −-− {−-− 1}
  4. D
    (−-− 1, 12{{1 \over 2}}21​]
View written solutionFree

Correct answer: A

  1. Given function

We have

f(x)=3\ln\left|\frac{x-1}{x+1}\right|- rac{2}{x-1},

defined on

R∖{−1,1}.\mathbb R\setminus\{-1,1\}.R∖{−1,1}.

We need to find where f(x)f(x)f(x) is increasing, i.e. where

f′(x)>0.f'(x)>0.f′(x)>0.
  1. Differentiate the logarithmic term

Let

g(x)=x−1x+1.g(x)=\frac{x-1}{x+1}.g(x)=x+1x−1​.

Then

ddxln⁡∣g(x)∣=g′(x)g(x).\frac{d}{dx}\ln|g(x)|=\frac{g'(x)}{g(x)}.dxd​ln∣g(x)∣=g(x)g′(x)​.

Now,

g′(x)=(x+1)−(x−1)(x+1)2=2(x+1)2.g'(x)=\frac{(x+1)-(x-1)}{(x+1)^2}=\frac{2}{(x+1)^2}.g′(x)=(x+1)2(x+1)−(x−1)​=(x+1)22​.

So,

\frac{g'(x)}{g(x)}=\frac{2}{(x+1)^2}\cdot \frac{x+1}{x-1}= rac{2}{(x+1)(x-1)}= rac{2}{x^2-1}.

Hence,

\frac{d}{dx}\left[3\ln\left|\frac{x-1}{x+1}\right|\right]=3\cdot \frac{2}{x^2-1}= rac{6}{x^2-1}.
  1. Differentiate the second term
−2x−1=−2(x−1)−1-\frac{2}{x-1}=-2(x-1)^{-1}−x−12​=−2(x−1)−1

Therefore,

ddx(−2x−1)=2(x−1)−2=2(x−1)2.\frac{d}{dx}\left(-\frac{2}{x-1}\right)=2(x-1)^{-2}=\frac{2}{(x-1)^2}.dxd​(−x−12​)=2(x−1)−2=(x−1)22​.

So,

f′(x)=6x2−1+2(x−1)2.f'(x)=\frac{6}{x^2-1}+\frac{2}{(x-1)^2}.f′(x)=x2−16​+(x−1)22​.
  1. Simplify f′(x)f'(x)f′(x)

Since

x2−1=(x−1)(x+1),x^2-1=(x-1)(x+1),x2−1=(x−1)(x+1),

we get

f′(x)=6(x−1)(x+1)+2(x−1)2.f'(x)=\frac{6}{(x-1)(x+1)}+\frac{2}{(x-1)^2}.f′(x)=(x−1)(x+1)6​+(x−1)22​.

Taking LCM (x−1)2(x+1)(x-1)^2(x+1)(x−1)2(x+1),

f′(x)=6(x−1)+2(x+1)(x−1)2(x+1).f'(x)=\frac{6(x-1)+2(x+1)}{(x-1)^2(x+1)}.f′(x)=(x−1)2(x+1)6(x−1)+2(x+1)​.

Simplify numerator:

6(x−1)+2(x+1)=6x−6+2x+2=8x−4=4(2x−1).6(x-1)+2(x+1)=6x-6+2x+2=8x-4=4(2x-1).6(x−1)+2(x+1)=6x−6+2x+2=8x−4=4(2x−1).

Thus,

f′(x)=4(2x−1)(x−1)2(x+1).f'(x)=\frac{4(2x-1)}{(x-1)^2(x+1)}.f′(x)=(x−1)2(x+1)4(2x−1)​.
  1. Find where f′(x)>0f'(x)>0f′(x)>0

We solve

4(2x−1)(x−1)2(x+1)>0.\frac{4(2x-1)}{(x-1)^2(x+1)}>0.(x−1)2(x+1)4(2x−1)​>0.

Now:

  • 4>04>04>0
  • (x−1)2>0(x-1)^2>0(x−1)2>0 for all x≠1x\ne 1x=1

So the sign depends only on

2x−1x+1>0.\frac{2x-1}{x+1}>0.x+12x−1​>0.

Critical points are:

  • x=−1x=-1x=−1 (undefined)
  • x=12x=\frac12x=21​ (numerator zero)
  • x=1x=1x=1 also excluded from domain

Check intervals:

  • For x<−1x<-1x<−1: both 2x−1<02x-1<02x−1<0 and x+1<0x+1<0x+1<0, so ratio >0>0>0
  • For −1<x<12-1<x<\frac12−1<x<21​: 2x−1<02x-1<02x−1<0, x+1>0x+1>0x+1>0, so ratio <0<0<0
  • For x>12x>\frac12x>21​: both 2x−1>02x-1>02x−1>0 and x+1>0x+1>0x+1>0, so ratio >0>0>0

But x=1x=1x=1 is not in domain, so split there.

Hence fff is increasing on

(−∞,−1)∪(12,1)∪(1,∞).(-\infty,-1)\cup\left(\frac12,1\right)\cup(1,\infty).(−∞,−1)∪(21​,1)∪(1,∞).

This is equivalently written as

(−∞,−1)∪([12,∞)∖{1})(-\infty,-1)\cup\left([\tfrac12,\infty)\setminus\{1\}\right)(−∞,−1)∪([21​,∞)∖{1})

if one allows inclusion of the stationary point x=12x=\frac12x=21​ in the interval description. In strict terms for increasing, the interval is open at 12\tfrac1221​, but among the given options, this corresponds to Option A.

  1. Check options
  • A: (−∞,−1)∪([12,∞)∖{1})(-\infty,-1)\cup\left([\tfrac12,\infty)\setminus\{1\}\right)(−∞,−1)∪([21​,∞)∖{1}) ✅ matches the required region from the options.
  • B: includes (−1,12)(-1,\tfrac12)(−1,21​) where f′(x)<0f'(x)<0f′(x)<0 ❌
  • C: includes (−1,12)(-1,\tfrac12)(−1,21​) and misses (12,∞)(\tfrac12,\infty)(21​,∞) ❌
  • D: only (−1,12](-1,\tfrac12](−1,21​], where function is actually decreasing ❌

Therefore, the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

This matches our derived answer.

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