JEE Main 2024 9 April Evening 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
∫ 0 x 1 − ( y ′ ( t ) ) 2 d t = ∫ 0 x y ( t ) d t , 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 ∫ 0 x 1 − ( y ′ ( t ) ) 2 d t = ∫ 0 x y ( t ) d t , 0 ≤ x ≤ 3 , y ≥ 0 , y ( 0 ) = 0 . Then at
x = 2 , y ′ ′ + y + 1 x=2, y^{\prime \prime}+y+1 x = 2 , y ′′ + y + 1 is equal to
Options A 1 B 2 C 2 \sqrt{2} 2 D 1 / 2 1 / 2 1/2
Step-by-step solution
1 − ( y ′ ( x ) ) 2 = y ( x ) 1 − ( d y d x ) 2 = y 2 ( d y d x ) 2 = 1 − y 2 \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} 1 − ( y ′ ( x ) ) 2 = y ( x ) 1 − ( d x d y ) 2 = y 2 ( d x d y ) 2 = 1 − y 2 d y 1 − y 2 = d x OR d y 1 − y 2 = − d x ⇒ sin − 1 y = x + c , sin − 1 y = − x + c x = 0 , y = 0 ⇒ c = 0 sin − 1 y = x , as y ≥ 0 sin x = y ⇒ d y d x = cos x d 2 y d x 2 = − 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} 1 − y 2 d y = d x OR 1 − y 2 d y = − d x ⇒ sin − 1 y = x + c , sin − 1 y = − x + c x = 0 , y = 0 ⇒ c = 0 sin − 1 y = x , as y ≥ 0 sin x = y ⇒ d x d y = cos x d x 2 d 2 y = − sin x ⇒ − sin x + sin x + 1 = 1 $
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