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Mathematics · Definite Integration

JEE Main 2024 — 9 April, Shift 2 — Question 3

Let ∫0x1−(y′(t))2dt=∫0xy(t)dt,0≤x≤3,y≥0\int_{0}^{x} \sqrt{1-\left(y^{\prime}(t)\right)^{2}} d t=\int_{0}^{x} y(t) d t, 0 \leq x \leq 3, y \geq 0 y(0)=0\mathrm{y}(0)=0. Then at x=2,y′′+y+1\mathrm{x}=2, \mathrm{y}^{\prime \prime}+\mathrm{y}+1 is equal

  1. Option A:

    11

    Correct
  2. Option B:

    22

  3. Option C:

    2\sqrt{2}

  4. Option D:

    1/21 / 2

Answer: A

Step-by-step solution

1−(y′(x))2=y(x)\sqrt{1-\left(y^{\prime}(x)\right)^{2}}=y(x)

1−(dydx)2=y21-\left(\frac{d y}{d x}\right)^{2}=y^{2}

(dydx)2=1−y2\left(\frac{d y}{d x}\right)^{2}=1-y^{2}

dy1−y2=dx\frac{d y}{\sqrt{1-y^{2}}}=d x OR dy1−y2=−dx\frac{d y}{\sqrt{1-y^{2}}}=-d x

⇒sin⁡−1y=x+c,sin⁡−1y=−x+c\Rightarrow \sin ^{-1} y=x+c, \sin ^{-1} y=-x+c

x=0,y=0⇒c=0x=0, y=0 \Rightarrow c=0

sin⁡−1y=x\sin ^{-1} \mathrm{y}=\mathrm{x}, as y≥0\mathrm{y} \geq 0 sin⁡x=y\sin x=y

⇒dydx=cos⁡x\Rightarrow \frac{d y}{d x}=\cos x

d2ydx2=−sin⁡x\frac{d^{2} y}{d x^{2}}=-\sin x

⇒−sin⁡x+sin⁡x+1=1\Rightarrow-\sin x+\sin x+1=1

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Mathematics
Chapter
Definite Integration
Topic
Leibnitz rule & its application in limits

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