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  • <img alt="\mathrm{Let ~f(x y)=f(x) f(y) for ~all~ x, y \in R. If ~f^{\prime}(1)=2~ and ~f(4)=4, then ~f^{\prime}(4)~ equal~ to}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BLet

\mathrm{Let ~f(x y)=f(x) f(y) for ~all~ x, y \in R. If ~f^{\prime}(1)=2~ and ~f(4)=4, then ~f^{\prime}(4)~ equal~ to}

Option: 1

4


Option: 2

1


Option: 3

\frac{1}{2}


Option: 4

8


Answers (1)

best_answer

We have \mathrm{f(x y)=f(x) f(y)}  for all \mathrm{x, y \in R} . Putting x=y=1, we get
\mathrm{ f(1)=f(1) f(1) \Rightarrow f(1)(1-f(1))=0 \Rightarrow f(1)=1[\because f(1) \neq 0] }
\mathrm{ \text { Now, } f^{\prime}(1)=2 \Rightarrow \lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=2 \\ }

\mathrm{ \Rightarrow \lim _{h \rightarrow 0} \frac{f(1) f(h)-f(1)}{h}=2 }
\mathrm{ \Rightarrow f(1) \lim _{h \rightarrow 0} \frac{f h)-1}{h}=2 \\ } 
\mathrm{ \left.\Rightarrow \lim _{h \rightarrow 0} \frac{f(h)-1}{h}=2 \quad \text { [Using } f(1)=1\right] \\ }.....................(i)
\mathrm{ \text { Now, } f^{\prime}(4)=\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h}=\lim _{h \rightarrow 0} \frac{f(4) f(h)-f(4)}{h} }
\mathrm{ =\left(\lim _{h \rightarrow 0} \frac{f(h)-1}{h}\right) f(4)=2 f(4)[\operatorname{from}(j)] \\ }

\mathrm{ =2 \times 4=8 . }

Posted by

Devendra Khairwa

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