A) \[64(log\text{ }2)\]
B) \[128\text{ }(log\text{ }2)\]
C) 256
D) \[256\text{ }(log\text{ }2)\]
E) \[1024(log\text{ }2)\]
Correct Answer: A
Solution :
Given, \[f(x+y)=2f(x)f(y)\] Now, \[f(0+0)=2f(0)f(0)\] \[\Rightarrow \] \[f(0)=\frac{1}{2}\] \[f(5)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5+h)-f(5)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,2f(5)\left[ \frac{f(h)-\frac{1}{2}}{h} \right]\] \[\Rightarrow \] \[1025\log 2=2f(5)f(0)\] ...(i) Again Now, \[f(2+3)=2f(2)f(3)\] \[\Rightarrow \] \[\frac{1024\log 2}{2f(0)}=2\times 8\times f(3)\] \[\Rightarrow \] \[f(3)=\frac{32\log 2}{f(0)}\] ?(ii) \[\therefore \] \[f(3)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(3+h)-f(3)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2f(3)f(h)-f(3)}{h}\] \[=2\underset{h\to 0}{\mathop{\lim }}\,f(3)\left[ \frac{f(h)-\frac{1}{2}}{h} \right]\] \[=2f(3)f(0)\] \[=2\times \frac{32\log 2f(0)}{f(0)}=64\log 2\] [from Eq. (ii)]
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