Let f(x) be a positive function such that the area bounded by y=f(x), y=0 from x=0 to x=a>0 is e^ -a +4 a^2+a-1 . Then the differential equation, whose general solution is y=c_1 f(x)+c_2 , where c_1 and c_2 are

Correct answer: D. (8 e^x+1) (d^2 y)/(d x^2)+(d y)/(d x)=0 aligned & ∫_0^a f(x) d x=e^ -a +4 a^2+a-1 & f(a)=-e^ -a +8 a+1 & f(x)=-e^ -x +8 x+1 aligned Now y=C_1 f(x)+C_2 dy dx =C_1 f^ (x)=C_1 (e^ -x +8) ...(1) (d^2 y)/(d x^2)=-C_1 e^ -x -e^x (d^2 y)/(d x^2) Put in equation (1) aligned & (d y)/(d x)=-e^x (d^2 y)/(d x^2)(e^ -x +8) & (8 e^x+1) (d^2 y)/(d x^2)+(d y)/(d x)=0 aligned

JEE Main 20248 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Area Under Curves

Let f(x) be a positive function such that the area bounded by...

Let

f(x)

be a positive function such that the area bounded by

y=f(x), y=0

from

x=0

x=a>0

e^{-a}+4 a^2+a-1

. Then the differential equation, whose general solution is

y=c_1 f(x)+c_2

, where

c_1

and

c_2

are arbitrary constants, is

Options

A $\left(8 e^x-1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$
B $\left(8 e^x-1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$
C $\left(8 e^x+1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$
D $\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$

Correct answer

D. $\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$

Step-by-step solution

\begin{aligned} & \int_0^a f(x) d x=e^{-a}+4 a^2+a-1 \\ & f(a)=-e^{-a}+8 a+1 \\ & f(x)=-e^{-x}+8 x+1\end{aligned}

Now

y=C_1 \mathrm{f}(\mathrm{x})+\mathrm{C}_2 \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{C}_1 \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{C}_1

\left(\mathrm{e}^{-\mathrm{x}}+8\right)

...(1)

\frac{d^2 y}{d x^2}=-C_1 e^{-x} \Rightarrow-e^x \frac{d^2 y}{d x^2}

Put in equation (1)

\begin{aligned} & \frac{d y}{d x}=-e^x \frac{d^2 y}{d x^2}\left(e^{-x}+8\right) \\ & \left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0\end{aligned}

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