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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Definite Integration

Let ∫_0^x 1-(y^ (t))^2 d t=∫_0^x y(t) d t, 0 ≤ x ≤ 3, y ≥ 0, y(0)=0...

Let ∫0x1−(y′(t))2dt=∫0xy(t)dt,0≤x≤3,y≥0,y(0)=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. Then at x=2,y′′+y+1x=2, y^{\prime \prime}+y+1 is equal to

Options

  1. A1
  2. B2
  3. C2\sqrt{2}
  4. D1/21 / 2

Correct answer

A. 1

Step-by-step solution

1−(y′(x))2=y(x)1−(dydx)2=y2(dydx)2=1−y2\begin{aligned} & \sqrt{1-\left(y^{\prime}(x)\right)^2}=y(x) \\ & 1-\left(\frac{d y}{d x}\right)^2=y^2 \\ & \left(\frac{d y}{d x}\right)^2=1-y^2\end{aligned}dy1−y2=dx OR  dy1−y2=−dx⇒sin⁡−1y=x+c,sin⁡−1y=−x+cx=0,y=0⇒c=0sin⁡−1y=x, as y≥0sin⁡x=y⇒dydx=cos⁡xd2ydx2=−sin⁡x⇒−sin⁡x+sin⁡x+1=1\begin{aligned} & \frac{d y}{\sqrt{1-y^2}}=d x \text{\text{ OR } } \frac{d y}{\sqrt{1-y^2}}=-d x \\ & \Rightarrow \sin ^{-1} y=x+c, \sin ^{-1} y=-x+c \\ & x=0, y=0 \Rightarrow c=0 \\ & \sin ^{-1} y=x, \text{ as } y \geq 0 \\ & \sin x=y \\ & \Rightarrow \frac{d y}{d x}=\cos x \\ & \frac{d^2 y}{d x^2}=-\sin x \\ & \Rightarrow-\sin x+\sin x+1=1\end{aligned}$
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