A) 30
B) 13
C) 10
D) 0
E) 6
Correct Answer: A
Solution :
\[f(0)=10\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x-0}=10\] \[\Rightarrow \] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=10\] \[\Rightarrow \] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0)f(h)-f(0)}{h}=10\] \[\Rightarrow \] \[f(0)\left( \underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-1}{h} \right)=10\] Now, \[f(0)=f(0)f(0)\] \[\Rightarrow \] \[f(0)[1-f(0)]=0\] \[\Rightarrow \] \[f(0)=1\] \[\therefore \] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-1}{h}=10\] Now, \[f(6)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(6+h)-f(6)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{f(h)-1}{h} \right)f(6)=10\times 3\] \[=30\]
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