MHT CET · Maths · Differentiation
If \(2 \mathrm{f}(\mathrm{x})=\mathrm{f}^{\prime}(x)\) and \(\mathrm{f}(0)=3\), then the value of \(\mathrm{f}(2)\) is
- A \(3 e^{2}\)
- B \(2 e^{3}\)
- C \(4 e^{3}\)
- D \(3 e^{4}\)
Answer & Solution
Correct Answer
(D) \(3 e^{4}\)
Step-by-step Solution
Detailed explanation
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 \(x=2\),
\(
\begin{aligned}
& \log |\mathrm{f}(2)|=2(2)+\log 3=4+\log 3 \\
\therefore & \mathrm{f}(2)=\mathrm{e}^{4+\log 3}=\mathrm{e}^{4} \cdot \mathrm{e}^{\log 3}=3 \mathrm{e}^{4}
\end{aligned}
\)
\(
\begin{array}{l}
\therefore \int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{d} \mathrm{x}=\int 2 \mathrm{~d} \mathrm{x} \\
\therefore \log |\mathrm{f}(\mathrm{x})|=2 \mathrm{x}+\mathrm{c}
\end{array}
\)
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 \(x=2\),
\(
\begin{aligned}
& \log |\mathrm{f}(2)|=2(2)+\log 3=4+\log 3 \\
\therefore & \mathrm{f}(2)=\mathrm{e}^{4+\log 3}=\mathrm{e}^{4} \cdot \mathrm{e}^{\log 3}=3 \mathrm{e}^{4}
\end{aligned}
\)
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