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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)

best_answer

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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prateek

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