Advertisements
Advertisements
प्रश्न
A function f: R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f′(0) = 2. Prove that f′(x) = 2f(x).
उत्तर
Given that, f: R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0.
Let us take any point x = 0 at which the function f(x) is differentiable.
∴ f'(0) = `lim_("h" -> 0) ("f"(0 + "h") - "f"(0))/"h"`
2 = `lim_("h" -> 0) ("f"(0) * "f"("h") - "f"(0))/"h"` ......[∵ f(0) = f(h)] ....(i)
⇒ 2 = `lim_("h" -> 0) ("f"(0)["f"("h") - 1])/"h"`
Now f'(x) = `lim_("h" -> 0) ("f"(x + "h") - "f"(x))/"h"`
= `lim_("h" -> 0) ("f"(x) * "f"("h") - "f"(x))/"h"` .....[∵ f(x + y) = f(x) . f(y)]
= `lim_("h" -> 0) ("f"(x)["f"("h") - 1])/"h"`
= 2f(x)
From equation (i)
Hence, f'(x) = 2f(x).
APPEARS IN
संबंधित प्रश्न
Examine the following function for continuity:
`f (x)1/(x - 5), x != 5`
Discuss the continuity of the function f, where f is defined by `f(x) = {(2x , ","if x < 0),(0, "," if 0 <= x <= 1),(4x, "," if x > 1):}`
A function f(x) is defined as
Show that f(x) is continuous at x = 3
If \[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, when & x \neq 0 \\ 1 , when & x = 0\end{cases}\]
Find whether f(x) is continuous at x = 0.
If \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if } & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\] is continuous at x = 4, find a, b.
If \[f\left( x \right) = \begin{cases}\frac{2^{x + 2} - 16}{4^x - 16}, \text{ if } & x \neq 2 \\ k , \text{ if } & x = 2\end{cases}\] is continuous at x = 2, find k.
Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}\left| x - 3 \right|, & \text{ if } x \geq 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & \text{ if } x < 1\end{cases}\]
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & \text{ if } - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & \text{ if } 0 \leq x \leq 1\end{cases}\]
If f (x) = | x − a | ϕ (x), where ϕ (x) is continuous function, then
If \[f\left( x \right) = \begin{cases}mx + 1 , & x \leq \frac{\pi}{2} \\ \sin x + n, & x > \frac{\pi}{2}\end{cases}\] is continuous at \[x = \frac{\pi}{2}\] , then
The values of the constants a, b and c for which the function \[f\left( x \right) = \begin{cases}\left( 1 + ax \right)^{1/x} , & x < 0 \\ b , & x = 0 \\ \frac{\left( x + c \right)^{1/3} - 1}{\left( x + 1 \right)^{1/2} - 1}, & x > 0\end{cases}\] may be continuous at x = 0, are
Show that the function
(i) differentiable at x = 0, if m > 1
(ii) continuous but not differentiable at x = 0, if 0 < m < 1
(iii) neither continuous nor differentiable, if m ≤ 0
If f is defined by f (x) = x2, find f'(2).
Discuss the continuity and differentiability of f (x) = |log |x||.
Give an example of a function which is continuos but not differentiable at at a point.
If \[f\left( x \right) = \left| \log_e x \right|, \text { then}\]
Evaluate :`int Sinx/(sqrt(cos^2 x-2 cos x-3)) dx`
Discuss the continuity of f at x = 1 ,
Where f(x) = `(3 - sqrt(2x + 7))/(x - 1)` for x = ≠ 1
= `(-1)/3` for x = 1
The probability distribution function of continuous random variable X is given by
f( x ) = `x/4`, 0 < x < 2
= 0, Otherwise
Find P( x ≤ 1)
The function given by f (x) = tanx is discontinuous on the set ______.
For continuity, at x = a, each of `lim_(x -> "a"^+) "f"(x)` and `lim_(x -> "a"^-) "f"(x)` is equal to f(a).
f(x) = `{{:(3x + 5",", "if" x ≥ 2),(x^2",", "if" x < 2):}` at x = 2
f(x) = `{{:(3x - 8",", "if" x ≤ 5),(2"k"",", "if" x > 5):}` at x = 5
Examine the differentiability of f, where f is defined by
f(x) = `{{:(1 + x",", "if" x ≤ 2),(5 - x",", "if" x > 2):}` at x = 2
The set of points where the function f given by f(x) = |2x − 1| sinx is differentiable is ______.
