JEE Mains · Maths · STD 12 - 9. differential equations
Let \(f:[0, \infty) \rightarrow \mathbb{R}\) be differentiable function such that \(f(\mathrm{x})=1-2 \mathrm{x}+\int_0^x e^{x-t} f(t) \mathrm{dt}\) for all \(\mathrm{x} \in[0, \infty)\).
Then the area of the region bounded by \(\mathrm{y}=f(\mathrm{x})\) and the coordinate axes is
- A \(\sqrt{5}\)
- B \(\frac{1}{2}\)
- C \(\sqrt{2}\)
- D 2
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & y=1-2 x+e^x \int_0^x e^{-t} f(t) d t \\ & \frac{d y}{d x}=-2+e^{-x} \cdot e^x f(x)+e^x \int_0^x e^{-t} f(t) d t \\ & \frac{d y}{d x}=-2+y+y+2 x-1 \\ & \frac{d y}{d x}-2 y=(2 x-3) \\ & y e^{-2 x}=\int(2 x-3) d x \cdot e^{-2 x} \\ & y e^{-2 x}=\frac{-(2 x-3)}{2}…
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