Let f - R - R be a differentiable function satisfying f y f x - y - f x - x - y - R and f 10- p - f 15- q then f 5 is- p2B- q-qD- q-p

The correct option is C y=0⇒ f(0)=1 and x=0⇒ f( y)=1/f(y) Hence f(x+y)=f(x)f(y) f^1(x)=h → 0Limf(x+h) f(x)/h=f(x)h → 0Ltf(x) 1/h=f(x).f^1(0)=p f(x) put x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Domain

Let f : R → R...

Question

Let $f : R \to R$ be a differentiable function satisfying $f (y) f (x - y) = f (x) \forall x, y ϵ R$ and $f^{1} (0) = p, f^{1} (5) = q t h e n f (5)$ is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$y = 0 \Rightarrow f (0) = 1 a n d x = 0 \Rightarrow f (- y) = \frac{1}{f (y)} H e n c e f (x + y) = f (x) f (y) f^{1} (x) = L i m h \to 0 \frac{f (x + h) - f (x)}{h} = f (x) L t h \to 0 \frac{f (x) - 1}{h} = f (x) . f^{1} (0) = p f (x) p u t x = 5 f (5) = \frac{q}{p}$

Suggest Corrections

Similar questions

Let

f : R \to R

be a differentiable function satisfying

f (\frac{x + y}{3}) = \frac{2 + f (x) + f (y)}{3} \forall x, y \in R

and

f^{'} (2) = 2

, then answer the following questions:

The function

h (x) = | f (| x |) - 4 |

Q. A function

f : R \to R

satisfies the equation

f (x + y) = f (x), f (y)

for all

x, y ϵ R, f (x) \neq 0.

Suppose that the function is differentiable at

x = 0

and

f^{'} (0) = 2,

then

f^{'} (x) =

Q. If

f : R \to R

be a differentiable function, such that

f (x + 2 y) = f (x) + f (2 y) + 4 x y \forall x, y ϵ R,

then

Q. Let f(x) be a continuous and differentiable function and f(y)f(x + y) = f(x) for all real values of x and y. If f(5) = 3 and f'(3) = 7 then the value of f'(8) is

Q. A function

f : R ⟶ R

satisfies the equation

f (x + y) = f (x), f (y)

for all

x, y ϵ R, f (x) \neq 0

Suppose that the function is differentiable at x=0 and

f^{^{'}} (0) = 2

prove that

f^{^{'}} (x) = 2 f (x)

Join BYJU'S Learning Program

Let f:R→R be a differentiable function satisfying f(y)f(x−y)=f(x)∀x,yϵR and f1(0)=p, f1(5)=q then f(5) is

The correct option is C y=0⇒f(0)=1 and x=0⇒f(−y)=1f(y)Hence f(x+y)=f(x)f(y) f1(x)=Limh→0f(x+h)−f(x)h=f(x)Lth→0f(x)−1h=f(x).f1(0)=pf(x) put x=5 f(5)=qp

Let $f : R \to R$ be a differentiable function satisfying $f (y) f (x - y) = f (x) \forall x, y ϵ R$ and $f^{1} (0) = p, f^{1} (5) = q t h e n f (5)$ is

The correct option is C
$y = 0 \Rightarrow f (0) = 1 a n d x = 0 \Rightarrow f (- y) = \frac{1}{f (y)} H e n c e f (x + y) = f (x) f (y) f^{1} (x) = L i m h \to 0 \frac{f (x + h) - f (x)}{h} = f (x) L t h \to 0 \frac{f (x) - 1}{h} = f (x) . f^{1} (0) = p f (x) p u t x = 5 f (5) = \frac{q}{p}$