Q55 / 155 · WBJEE 2024 · Mathematics

Let $f$ be a differential function with $\lim _{x \rightarrow \infty} f(x)=0$. If $y^{\prime}+y f^{\prime}(x)-f(x) f^{\prime}(x)=0, \lim _{x \rightarrow \infty} y(x)=0$ then

  • A
    $y+1=e^{f(x)}+f(x)$
  • Correct answer
    B
    $y+1=e^{-f(x)}+f(x)$
  • C
    $y+2=e^{-f(x)}+f(x)$
  • D
    $y-1=e^{-f(x)}+f(x)$

Solution

\(\begin{aligned} & \frac{d y}{d x}+y f^{\prime}(x)=f(x) f^{\prime}(x) \\ & \text {I.F }=e^{\int f^{\prime}(x) d x}=e^{f(x)} \\ & \text {y. } e^{f(x)}=\int e^{f(x)} f(x) f^{\prime}(x) d x \end{aligned}\) \(\begin{aligned} & \text {Let } f(x)=t \Rightarrow f^{\prime}(x) d x=d t \\ & \Rightarrow y e^{f(x)}=\int e^t t d t \\ & \Rightarrow y e^{f(x)}=e^t(t-1)+c \\ & \Rightarrow y e^{f(x)}=e^{f(x)}(f(x)-1)+c \end{aligned}\) Given, \(\lim _{x \rightarrow \infty} f(x)=0, \lim _{x \rightarrow \infty} y(x)=0\) \(\begin{aligned} & \Rightarrow 0 . e^0=e^0(0-1)+c \Rightarrow c=1 \\ & \therefore y e^{f(x)}=e^{f(x)}(f(x)-1)+1 \\ & \Rightarrow y=f(x)-1+e^{-f(x)} \\ & \Rightarrow y+1=f(x)+e^{-f(x)} \end{aligned}\)

Hence, the correct answer is option: 'B'.

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