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JEE Mathematics 2024 Question with Solution

Let y=ln⁡(1−x21+x2)y = \ln\left(\frac{1 - x^2}{1 + x^2}\right), −1<x<1-1 < x < 1. Then at x=12x = \frac{1}{2}, the value of 225(y′−y′′)225(y' - y'') is equal to:

  • A

    732732

  • B

    746746

  • C

    742742

  • D

    736736

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: y=log⁡e(1−x21+x2)y = \log_e\left(\frac{1 - x^2}{1 + x^2}\right) and x=12x = \frac{1}{2}

Find: 225(y′−y′′)225(y' - y'')

Rewrite using logarithm properties:

y=log⁡e(1−x2)−log⁡e(1+x2)y = \log_e(1-x^2) - \log_e(1+x^2)

Differentiate to get the first derivative:

y′=ddx[log⁡e(1−x2)−log⁡e(1+x2)]y' = \frac{d}{dx}[\log_e(1-x^2) - \log_e(1+x^2)] y′=−2x1−x2−2x1+x2y' = \frac{-2x}{1-x^2} - \frac{2x}{1+x^2} y′=−2x(11−x2+11+x2)y' = -2x\left(\frac{1}{1-x^2} + \frac{1}{1+x^2}\right) y′=−2x(1+x2+1−x2(1−x2)(1+x2))y' = -2x\left(\frac{1+x^2 + 1-x^2}{(1-x^2)(1+x^2)}\right) y′=−2x(21−x4)=−4x1−x4y' = -2x\left(\frac{2}{1-x^4}\right) = \frac{-4x}{1-x^4}

Now differentiate again:

y′′=ddx(−4x1−x4)y'' = \frac{d}{dx}\left(\frac{-4x}{1-x^4}\right)

Using the quotient rule with u=−4xu = -4x and v=1−x4v = 1-x^4,

y′′=v⋅dudx−u⋅dvdxv2y'' = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}

where dudx=−4\frac{du}{dx} = -4 and dvdx=−4x3\frac{dv}{dx} = -4x^3.

Substitute and simplify:

y′′=(1−x4)(−4)−(−4x)(−4x3)(1−x4)2y'' = \frac{(1-x^4)(-4) - (-4x)(-4x^3)}{(1-x^4)^2} y′′=−4(1−x4)−16x4(1−x4)2y'' = \frac{-4(1-x^4) - 16x^4}{(1-x^4)^2} y′′=−4+4x4−16x4(1−x4)2y'' = \frac{-4 + 4x^4 - 16x^4}{(1-x^4)^2} y′′=−4−12x4(1−x4)2y'' = \frac{-4 - 12x^4}{(1-x^4)^2}

Evaluate at x=12x = \frac{1}{2}:

y′(12)=−4(12)1−(12)4=−21−116y'\left(\frac{1}{2}\right) = \frac{-4\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^4} = \frac{-2}{1-\frac{1}{16}} =−21516=−3215= \frac{-2}{\frac{15}{16}} = \frac{-32}{15}

Also,

y′′(12)=−4−12(12)4(1−(12)4)2y''\left(\frac{1}{2}\right) = \frac{-4 - 12\left(\frac{1}{2}\right)^4}{\left(1-\left(\frac{1}{2}\right)^4\right)^2} =−4−1216(1516)2= \frac{-4 - \frac{12}{16}}{\left(\frac{15}{16}\right)^2} =−4−34225256= \frac{-4 - \frac{3}{4}}{\frac{225}{256}} =−194225256=−19⋅2564⋅225= \frac{-\frac{19}{4}}{\frac{225}{256}} = \frac{-19 \cdot 256}{4 \cdot 225} =−19⋅64225= \frac{-19 \cdot 64}{225}

Now compute:

225(−3215−−19⋅64225)225 \left(\frac{-32}{15} - \frac{-19 \cdot 64}{225}\right) =225(−32×15−(−19×64)225)= 225 \left(\frac{-32 \times 15 - (-19 \times 64)}{225}\right) =225(−480+1216225)= 225 \left(\frac{-480 + 1216}{225}\right) =225(736225)= 225 \left(\frac{736}{225}\right) =736= 736

Therefore, the value of 225(y′−y′′)225(y' - y'') is 736736. The correct option is D.

Alternative Extracted Approach

Given: y=log⁡e(1−x21+x2)y = \log_e\left(\frac{1 - x^2}{1 + x^2}\right)

Find: 225(y′−y′′)225(y' - y'') at x=12x = \frac{1}{2}

From the extracted approach:

dydx=y′=−4x1−x4\frac{dy}{dx} = y' = \frac{-4x}{1 - x^4} y′′=−4(1+3x4)(1−x4)2y'' = \frac{-4(1 + 3x^4)}{(1 - x^4)^2}

Then

y′−y′′=−4x1−x4+4(1+3x4)(1−x4)2y' - y'' = \frac{-4x}{1 - x^4} + \frac{4(1 + 3x^4)}{(1 - x^4)^2}

Substituting x=12x = \frac{1}{2} and simplifying gives:

225(y′−y′′)=736225(y' - y'') = 736

Hence, the correct answer is D.

Common mistakes

  • A common mistake is differentiating ln⁡(1−x21+x2)\ln\left(\frac{1-x^2}{1+x^2}\right) directly without first using logarithm properties. This often causes sign errors in the derivative. Rewrite it as log⁡e(1−x2)−log⁡e(1+x2)\log_e(1-x^2) - \log_e(1+x^2) before differentiating.

  • Students often make an error while combining −2x1−x2−2x1+x2\frac{-2x}{1-x^2} - \frac{2x}{1+x^2} by using an incorrect denominator. The correct combined denominator is (1−x2)(1+x2)=1−x4(1-x^2)(1+x^2) = 1-x^4. Use algebra carefully before simplifying.

  • Another mistake is applying the quotient rule incorrectly for y′=−4x1−x4y' = \frac{-4x}{1-x^4}, especially the signs in v⋅u′−u⋅v′v \cdot u' - u \cdot v'. Keep track of the negative signs in both u=−4xu = -4x and v′=−4x3v' = -4x^3.

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