Q120 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

If \(f(x), f^{\prime}(x) f^{\prime \prime}(x)\), are positive functions and \(f(0)=1, f^{\prime}(0)=2\), then the solution of the differential equation \(\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0\) is

  • Correct answer
    A
    \(e^{2 x}\)
  • B
    \(2 \sin x+1\)
  • C
    \(\sin ^2 x+2 x+1\)
  • D
    \(e^{4 x}\)

Solution

\(\begin{aligned} & \text {Since, }\left|\begin{array}{ll} f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x) \end{array}\right|=0 \\ & \Rightarrow \quad f(x) f^{\prime \prime}(x)-\left(f^{\prime}(x)\right)^2=0 \\ & \Rightarrow \quad f(x) f^{\prime \prime}(x)=\left(f^{\prime}(x)\right)^2 \\ & \Rightarrow \quad \frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)} \\ \end{aligned}\) On, integrating both sides, we get \(\begin{aligned} & \int \frac{f^{\prime \prime}(x)}{f^{\prime}(x)} d x=\int \frac{f^{\prime}(x)}{f(x)} d x \\ & \Rightarrow \quad \log _e\left(f^{\prime}(x)\right)=\log _e(f(x))+c \\ & \because \quad f(0)=1 \text { and } f^{\prime}(0)=2 \\ & \therefore \quad c=\log _e 2 \\ & \log _e\left(f^{\prime}(x)\right)=\log _e(f(x))+\log _e 2 \\ & \Rightarrow \quad f^{\prime}(x)=2 f(x) \\ & \Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=2 \\ & \Rightarrow \quad \int \frac{f^{\prime}(x)}{f(x)} d x=\int 2 d x \Rightarrow \log _e f(x)=2 x+c \\ & \because \quad f(0)=1 \text {, so } c=0 \\ & \because \quad \log _e f(x)=2 x \Rightarrow f(x)=e^{2 x} \\ \end{aligned}\) Hence, option (1) is correct.

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

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