Let $f:[0,\infty )\to \mathrm{ℝ}$be differentiable function such that $\mathrm{f}(\mathrm{x})=1-2\mathrm{x}+{\int }_{0}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{x}-\mathrm{t}}\mathrm{f}(\mathrm{t})\mathrm{dt}$ 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 [JEE Main 2025, 4 Apr (Shift 1)]

Question

YESPYQ · Accepted Answer

Correct answer: b) $\frac{1}{2}$. $y=1−2x+{e}^{x}{\int }_{0}^{x}{e}^{−t}f(t)dt$ $\frac{dy}{dx}=−2+{e}^{−x}⋅{e}^{x}f(x)+{e}^{x}{\int }_{0}^{x}{e}^{−t}f(t)dt$ $\frac{dy}{dx}=−2+y+y+2x−1$ $\frac{dy}{dx}−2y=(2x−3)$ $I.F.={e}^{\int -2dx}={e}^{-2x}$ solution is $y{e}^{−2x}=\int (2x−3)dx⋅{e}^{−2x}$ $y{e}^{−2x}=\frac{−(2x−3)}{2}{e}^{−2x}+\int {e}^{−2x}dx$ $y{e}^{−2x}=\frac{−(2x−3)}{2}{e}^{−2x}−\frac{1}{2}{e}^{−2x}+c$ $\mathrm{f}(0)=1\Rightarrow \mathrm{c}=1−\frac{3}{2}+\frac{1}{2}=0$ $y=−\frac{(2x−3)}{2}−\frac{1}{2}$ $y=-x+1\ \Rightarrow x+y=1$ Hence area bounded by $y=f(x)$ and coordinate axes is area $=\frac{1}{2}(1)(1)=\frac{1}{2}$

Let \(f:[0,\infty )\to \mathrm{ℝ}\)be differentiable function such that \(\mathrm{f}(\mathrm{x})=1-2\mathrm{x}+{\int }_{…

Practice more JEE Maths PYQs