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Let $f(x) = x - 1$ and $g(x) = e^x$ for $x \in \mathbb{R}$. If $\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f(f(x))\right) - \frac{y}{\sqrt{x}} \right)$, $y(0) = 0$, then $y(1)$ is

JEE · Math · previous-year question

  1. A.$\frac{1 - e^3}{e^4}$
  2. B.$\frac{e-1}{e^4}$correct
  3. C.$\frac{1 - e^2}{e^4}$
  4. D.$\frac{2e - 1}{e^3}$

Answer

B. $\frac{e-1}{e^4}$

Explanation

$$\begin{aligned} & f(x)=x-1 \\ & f(f(x))=f(x)-1=x-1-1=x-2 \\ & g(f(f(x)))=e^{x-2} \\ & \therefore \frac{d y}{d x}=e^{-2 \sqrt{x}} \times e^{x-2}-\frac{1}{\sqrt{x}} y \\ & \frac{d y}{d x}+\frac{1}{\sqrt{x}} y=e^{x-2 \sqrt{x}-2} \text { which is L.D.E } \\ & \text { I.F. }=e^{\int \frac{d y}{\sqrt{x}}}=e^{2 \sqrt{x}} \end{aligned}$$ Its solution is $$\begin{aligned} & y \times e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \times e^{x-2 \sqrt{x}-2} d x+c \\ & y \times e^{2 \sqrt{x}}=\int e^{x-2} d x+c \\ & y \times e^{2 \sqrt{x}}=e^{x-2}+c \end{aligned}$$ Given $\mathrm{x}=0, \mathrm{y}=0 \Rightarrow 0=\mathrm{e}^{-2}+\mathrm{c} \quad ; \mathrm{c}=-\mathrm{e}^{-2}$ $$\therefore \mathrm{y} \times \mathrm{e}^{2 \sqrt{x}}=\mathrm{e}^{\mathrm{x}-2}-\mathrm{e}^{-2}$$ when $\mathrm{x}=1, \mathrm{y} \times \mathrm{e}^2=\mathrm{e}^{-1}-\mathrm{e}^{-2}$ $$y=\frac{e^{-1}-e^{-2}}{e^2}=\frac{\frac{1}{e}-\frac{1}{e^2}}{e^2}=\frac{e^2-e}{e^5}=\frac{e-1}{e^4}$$ Option (1) is correct

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