If f\left ( x+y \right )=f\left ( x \right ).f\left ( y \right )\forall\: x,y\: and\: f\left ( 5 \right )= 2,f{}'\left ( 0 \right )= 3,

Then {f}'\left ( 5 \right ) is

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    6

  • Option 4)

    2

 

Answers (1)

As we learnt in 

Condition for differentiable -

A function  f(x) is said to be differentiable at  x=x_{\circ }  if   Rf'(x_{\circ })\:\:and\:\:Lf'(x_{\circ })   both exist and are equal otherwise non differentiable

-

 

 \\ f(x+y)=f(x)\cdot f(y) \\ \\ f(5)=2 \: \: \: \: \: \Rightarrow put :\ x=0, \: y=0 \\ \\ {f}'(0)=3 \: \: \: \: \: \therefore f(0)=1

\lim_{h \to 0} \: \frac{f(0+h)-f(0)}{h}={f}'\left ( 0 \right )

\lim_{h \to 0} \: \frac{f(0)\times f(h)-f(0)}{h}= \: f(0) \: \lim_{h \to 0} \: \frac{f(h)-1}{h}=3

\Rightarrow \lim_{h \to 0} \: \frac{f(h)-1}{h}=3

Now,

{f}'(5)=\lim_{h \to 0} \frac{f(5+h)-f(5)}{h}

          =\lim_{h \to 0} \frac{f(5)\times f(h)-f(5)}{h}

           =\lim_{h \to 0} \: \frac{f(5)\left [ f(h)-1 \right ]}{h}

           = f(5) \lim_{h \to 0} \: \frac{f(h)-1}{h}

           =2\times 3= 6


Option 1)

0

This option is incorrect

Option 2)

1

This option is incorrect

Option 3)

6

This option is correct

Option 4)

2

This option is incorrect

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