Let f- satisfy the equation fx-y -fx-fy for all x-y- and fx0 for any x- If the function f is differentiable at x - 0 and f-0 - 3- then limh-01hfh-1 is equal to

F(x+y) =f(x)·f(y) Putting x = y = 0 f(0) = (f(0))^2 ⇒ f(0) = 1 (∵ f(x) 0 ∀ x∈ℝ) Now, nbsp; lim_h→0f(h) 1h =lim_h→0f(0+h) f(0)h nbsp;= f'(0) = 3 ...

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Byju's Answer

Standard XII

Mathematics

Variable Separable Method

Let f:ℝ→ℝ sat...

Question

Let $f : R \to R$ satisfy the equation $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in R$ and $f (x) \neq 0$ for any $x \in R .$ If the function $f$ is differentiable at $x = 0$ and $f^{'} (0) = 3,$ then $lim h \to 0 \frac{1}{h} (f (h) - 1)$ is equal to

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Solution

$f (x + y) = f (x) \cdot f (y)$
Putting $x = y = 0$
$f (0) = (f (0))^{2} \Rightarrow f (0) = 1 (∵ f (x) \neq 0 \forall x \in R)$
Now,
$lim h \to 0 \frac{f (h) - 1}{h} = lim h \to 0 \frac{f (0 + h) - f (0)}{h} = f^{'} (0) = 3$

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Let f:R→R satisfy the equation f(x+y)=f(x)⋅f(y) for all x,y∈R and f(x)≠0 for any x∈R. If the function f is differentiable at x=0 and f′(0)=3, then limh→01h(f(h)−1) is equal to

f(x+y)=f(x)⋅f(y) Putting x=y=0 f(0)=(f(0))2⇒f(0)=1 (∵f(x)≠0 ∀x∈R) Now, limh→0f(h)−1h=limh→0f(0+h)−f(0)h=f′(0)=3

Let $f : R \to R$ satisfy the equation $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in R$ and $f (x) \neq 0$ for any $x \in R .$ If the function $f$ is differentiable at $x = 0$ and $f^{'} (0) = 3,$ then $lim h \to 0 \frac{1}{h} (f (h) - 1)$ is equal to

$f (x + y) = f (x) \cdot f (y)$
Putting $x = y = 0$
$f (0) = (f (0))^{2} \Rightarrow f (0) = 1 (∵ f (x) \neq 0 \forall x \in R)$
Now,
$lim h \to 0 \frac{f (h) - 1}{h} = lim h \to 0 \frac{f (0 + h) - f (0)}{h} = f^{'} (0) = 3$