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JEE Mathematics 2025 Question with Solution

Given below are two statements:

Statement I: lim⁡x→0(tan⁡−1x+log⁡e1+x1−x−2xx5)=25\lim_{x \to 0} \left( \frac{\tan^{-1} x + \log_e \sqrt{\frac{1+x}{1-x}} - 2x}{x^5} \right) = \frac{2}{5}

Statement II: lim⁡x→1(2x1−x)=1e2\lim_{x \to 1} \left( \frac{2}{x^{1-x}} \right) = \frac{1}{e^2}

In the light of the above statements, choose the correct answer from the options given below

  • A

    Both Statement I and Statement II are false

  • B

    Statement I is false but Statement II is true

  • C

    Both Statement I and Statement II are true

  • D

    Statement I is true but Statement II is false

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

  • Statement I: lim⁡x→0(tan⁡−1x+log⁡e1+x1−x−2xx5)\lim_{x \to 0} \left( \frac{\tan^{-1} x + \log_e \sqrt{\frac{1+x}{1-x}} - 2x}{x^5} \right)
  • Statement II: lim⁡x→1(2x1−x)\lim_{x \to 1} \left( \frac{2}{x^{1-x}} \right)

Find: Whether each statement is true or false.

For Statement I, use Taylor expansions near x=0x=0:

tan⁡−1x=x−x33+x55−⋯\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots

and

log⁡e1+x1−x=12(log⁡(1+x)−log⁡(1−x))=x+x33+x55+⋯\log_e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2}\bigl(\log(1+x)-\log(1-x)\bigr)=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots

Adding these,

tan⁡−1x+log⁡e1+x1−x=2x+2x55+⋯\tan^{-1}x + \log_e \sqrt{\frac{1+x}{1-x}} = 2x + \frac{2x^5}{5} + \cdots

Therefore,

tan⁡−1x+log⁡e1+x1−x−2xx5=2x55+⋯x5→25\frac{\tan^{-1} x + \log_e \sqrt{\frac{1+x}{1-x}} - 2x}{x^5} = \frac{\frac{2x^5}{5}+\cdots}{x^5} \to \frac{2}{5}

So Statement I is true.

For Statement II, take

y=x21−xy = x^{\frac{2}{1-x}}

Then

ln⁡y=2ln⁡x1−x\ln y = \frac{2\ln x}{1-x}

As x→1x \to 1, this is of the form 00\frac{0}{0}, so apply L'Hôpital's rule:

lim⁡x→12ln⁡x1−x=lim⁡x→12/x−1=−2\lim_{x\to 1} \frac{2\ln x}{1-x} = \lim_{x\to 1} \frac{2/x}{-1} = -2

Hence,

lim⁡x→1ln⁡y=−2⇒lim⁡x→1y=e−2=1e2\lim_{x\to 1} \ln y = -2 \quad \Rightarrow \quad \lim_{x\to 1} y = e^{-2} = \frac{1}{e^2}

So Statement II is true.

Therefore, both Statement I and Statement II are true. The correct option is C.

Independent Verification of Both Statements

Given: Two limit-based statements.

Find: Check each statement independently and choose the correct option.

Verification of Statement I:

tan⁡−1x=x−x33+x55−⋯\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots

Also,

log⁡e1+x1−x=12[log⁡(1+x)−log⁡(1−x)]\log_e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2}\left[\log(1+x)-\log(1-x)\right]

Using series expansions,

log⁡(1+x)=x−x22+x33−⋯ ,log⁡(1−x)=−x−x22−x33−⋯\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots, \qquad \log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots

So,

12[log⁡(1+x)−log⁡(1−x)]=x+x33+x55+⋯\frac{1}{2}\left[\log(1+x)-\log(1-x)\right]=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots

Now substitute into the numerator:

(x−x33+x55)+(x+x33+x55)−2x=2x55\left(x-\frac{x^3}{3}+\frac{x^5}{5}\right)+\left(x+\frac{x^3}{3}+\frac{x^5}{5}\right)-2x = \frac{2x^5}{5}

Hence,

2x55x5=25\frac{\frac{2x^5}{5}}{x^5}=\frac{2}{5}

Thus Statement I is true.

Verification of Statement II: Let

y=x21−xy = x^{\frac{2}{1-x}}

Then

ln⁡y=2ln⁡x1−x\ln y = \frac{2\ln x}{1-x}

Now evaluate the exponent limit:

lim⁡x→12ln⁡x1−x\lim_{x\to 1} \frac{2\ln x}{1-x}

Applying L'Hôpital's rule,

lim⁡x→12/x−1=−2\lim_{x\to 1} \frac{2/x}{-1}=-2

Therefore,

lim⁡x→1ln⁡y=−2\lim_{x\to 1} \ln y = -2

and so

lim⁡x→1y=e−2=1e2\lim_{x\to 1} y = e^{-2}=\frac{1}{e^2}

Thus Statement II is true.

So both statements are true, and the correct option is C.

Common mistakes

  • Using the expression in Statement II as 2x1−x\frac{2}{x^{1-x}} instead of interpreting the worked solution form x21−xx^{\frac{2}{1-x}}. This is wrong because the two limits are different. Always follow the expression consistently from the verified working before concluding the option.

  • Stopping the series in Statement I at the x3x^3 term. This is wrong because the xx and x3x^3 terms cancel, so the first surviving term is the x5x^5 term. Expand far enough to match the denominator x5x^5.

  • Applying L'Hôpital's rule directly to x21−xx^{\frac{2}{1-x}} without taking logarithms. This is wrong because the expression is of exponential type. First take natural logarithm, reduce it to a quotient, and then apply L'Hôpital's rule.

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