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Suppose f: R \rightarrow R^{+}be a differentiable function such that 3 f(x+y)=f(x) f(y) \forall x, y \in R withf(1)=6. Then the value of f(2) is
 

Option: 1

6


Option: 2

9


Option: 3

12


Option: 4

15


Answers (1)

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\begin{aligned} & 3 f(x+y)=f(x) f(y) \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \quad \quad \quad \quad \quad \quad (i) \\ & =\lim _{h \rightarrow 0} \frac{\frac{f(x) \cdot f(h)}{3}-f(x)}{h} \quad \quad \quad \quad (\text{ using f rule }) \\ & =\lim _{h \rightarrow 0} \frac{f(x) \cdot f(h)-3 f(x)}{3 h} \\ \end{aligned}\begin{aligned} & \Rightarrow \quad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x)}{3} \cdot \frac{[f(h)-3]}{h} \\ & \text { Put } \quad x=0, y=1 \\ & \therefore \quad 3 f(1)=f(0) f(1) \\ & \therefore \quad f(0)=3 \\ & \therefore \quad f^{\prime}(x)=\frac{f(x)}{3} \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ \end{aligned}\begin{aligned} & \Rightarrow \quad f^{\prime}(x)=\frac{f(x)}{3} f^{\prime}(0) \text {; } \\ & \Rightarrow \quad 3 \cdot \frac{f^{\prime}(x)}{f(x)}=f^{\prime}(0)=k \text { (say) } \\ \end{aligned}


\Rightarrow  Integrating, we get 3 \log (f(x))=k x+C;
Using f(0)=3 \Rightarrow 3 \log 3=C

\begin{aligned} & \Rightarrow \quad 3 \log (f(x))=k x+3 \log 3 \\ & \Rightarrow \quad 3 \log \left(\frac{f(x)}{3}\right)=k x ; \end{aligned}
\begin{aligned} & \text { Using } f(1)=6 \Rightarrow 3 \log 2=k \\ & \Rightarrow \quad f(x)=3 \cdot 2^x \\ & \Rightarrow \quad f(2)=12 \end{aligned}
 

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HARSH KANKARIA

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