If $2 \mathrm{f}(\mathrm{x})=\mathrm{f}^{\prime}(x)$ and $\mathrm{f}(0)=3$, then the value of…
If $2 \mathrm{f}(\mathrm{x})=\mathrm{f}^{\prime}(x)$ and $\mathrm{f}(0)=3$, then the value of $\mathrm{f}(2)$ is
- $3 e^{2}$
- $2 e^{3}$
- $4 e^{3}$
- $3 e^{4}$
Solution
We have $f^{\prime}(x)=2 f(x)$
$\begin{array}{l}
\therefore \int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{d} \mathrm{x}=\int 2 \mathrm{~d} \mathrm{x} \\
\therefore \log |\mathrm{f}(\mathrm{x})|=2 \mathrm{x}+\mathrm{c}
\end{array}$
Now $f(0)=3$
$\begin{array}{l}
\therefore|\log 3|=0+c \Rightarrow c=\log 3 \\
\therefore \log |f(x)|=2 x+\log 3
\end{array}$
When $x=2$,
$\begin{aligned}
& \log |\mathrm{f}(2)|=2(2)+\log 3=4+\log 3 \\
\therefore & \mathrm{f}(2)=\mathrm{e}^{4+\log 3}=\mathrm{e}^{4} \cdot \mathrm{e}^{\log 3}=3 \mathrm{e}^{4}
\end{aligned}$
Asked in: MHT CET 2020 (15 Oct Shift 2)
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