JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Differential Equations

1. (A)The curve $y=f(x)$ passes through the point $(1,2)$
2. (B)The curve $y=f(x)$ passes through the point $(2,-1)$✓ Correct
3. (C)The area of the region $\left\{(x,y)\in [0,1]\times \mathbb{R}:f(x)\le y\le \sqrt{1-x^2}\right\}$ is $\frac{\pi -2}{4}$✓ Correct
4. (D)The area of the region $\left\{(x,y)\in [0,1]\times \mathbb{R}:f(x)\le y\le \sqrt{1-x^2}\right\}$ is $\frac{\pi -1}{4}$

Answer: (B), (C)

Solution

From $f(x)=1-2x+e^x{\int }_0^x{e}^{-t}f(t)dt$, we have $f(0)=1$.

Differentiating: $f^{\prime }(x)=-2+e^x{\int }_0^x{e}^{-t}f(t)dt+f(x)=-2+(f(x)+2x-1)+f(x)=2f(x)+2x-3$.

This is a linear ODE: $f^{\prime }(x)-2f(x)=2x-3$. Integrating factor $e^{-2x}$.

Solution: $f(x)e^{-2x}=\int e^{-2x}(2x-3)dx=e^{-2x}\left(\frac{3-2x}{2}-\frac{1}{2}\right)+c$.

Simplifying with $f(0)=1$: $c=0$, so $f(x)=1-x$.

$f(1)=0$, $f(2)=-1$. (B) is TRUE, (A) is FALSE.

Area $={\int }_0^1(\sqrt{1-x^2}-(1-x))dx=\frac{\pi }{4}-\frac{1}{2}=\frac{\pi -2}{4}$. (C) is TRUE.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · One or more correct options