A) \[4\]
B) \[1\]
C) \[\frac{1}{2}\]
D) \[8\]
Correct Answer: D
Solution :
We have\[f(xy)=f(x),\,\,f(y)\]all\[x,\,\,y\in R\] Putting\[x=y=1\], we get \[f(1)=f(1)f(1)\] \[\Rightarrow \]\[f(1)[1-f(1)]=0\] \[\Rightarrow \]\[f(1)=1\] \[[\because \,\,f(1)\ne 0]\] Now,\[f(1)=2\] \[\Rightarrow \]\[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1+h)-f(1)}{h}=2\] \[\Rightarrow \]\[f(1)\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-1}{h}=2\] \[\Rightarrow \]\[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1)f(h)-(f)1}{h}=2\][using\[f(1)=1\]]? (i) \[\Rightarrow \]\[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-1}{h}=2\] Now\[f(4)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(4+h)-f(4)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(4)\cdot f(h)-f(4)}{h}\] \[=\left\{ \underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-1}{h} \right\}\cdot f(4)=2f(4)\] {from (i)}
You need to login to perform this action.
You will be redirected in
3 sec
