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Mathematics · Methods of Differentiation

JEE Main 2024 — 29 January, Shift 2 — Question 17

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

  1. Option A:

    732

  2. Option B:

    746

  3. Option C:

    742

  4. Option D:

    736

    Correct

Answer: D

Step-by-step solution

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

dydx=y′=−4x1−x4\frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^{4}}

Again, d2ydx2=y′′=−4(1+3x4)(1−x4)2\frac{d^{2} y}{d x^{2}}=y^{\prime \prime}=\frac{-4\left(1+3 x^{4}\right)}{\left(1-x^{4}\right)^{2}}

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

at x=12\mathrm{x}=\frac{1}{2},

y′−y′′=736225y^{\prime}-y^{\prime \prime}=\frac{736}{225}

Thus 225(y′−y′′)=225×736225=736225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Mathematics
Chapter
Methods of Differentiation
Topic
Methods of Differentiation