Home
Class 12
MATHS
A function f : R to R satisfies the equ...

A function `f : R to R` satisfies the equation `f(x+y) = f (x) f(y), AA x, y ` in R and `f (x) ne 0` for any x in R . Let the function be differentiable at x = 0 and f'(0) = 2. Show that` f'(x) = 2 f(x), AA x ` in R. Hence, determine f(x)

Text Solution

Verified by Experts

The correct Answer is:
`e^(2x)`
We have, ` f(x+y) = f(x) * f(y) , AA x, y in R`.
`:. F(0) = f(0)* f(0) rArr f(0) {f(0) - 1} =0 `
`rArr" " f(0) = 1" "[:' f(0) ne 0]`
Since, `f'(0) = 2rArr underset( h to 0) lim (f(0+h)-f(0))/ h =2`
` rArr" " underset( h to 0) lim (f(h)-1)/ h = 2" "[:' f(0) = 1]` ....(i)
Also, ` f'(x) = underset( h to 0) lim (f(x+h)-f(x))/h `
`= underset( h to 0) lim (f(x)*f(h)-f(x))/h ` ,
[using, `f(x+y) = f(x)* f(y)`]
`= f(x) {underset( h to 0) lim (f(h)-1)/h}`
`:." "f'(x) = 2 f(x) " "`[from Eq. (i)]
` rArr" " (f'(x))/(f(x)) = 2`
On integrating both sides between 0 to x, we get
` " " int_(0)^(x) (f'(x))/(f(x)) dx = 2x`
` rArr" " log_(e) |f(x)-log_(e) |f(0)| = 2x`
`rArr" " log_(e) |f(x) | = 2x " " [:' f(0) = 1]`
` rArr" " log_(e) | f(0) | =0 `
`rArr" " f(x) = e^(2x) `
Promotional Banner

Similar Questions

Explore conceptually related problems

A function f: R -> R satisfy the equation f (x)f(y) - f (xy)= x+y for all x, y in R and f(y) > 0 , then

Let f(x) = ax^(2) - bx + c^(2), b ne 0 and f(x) ne 0 for all x in R . Then

A function f:RtoR is such that f(x+y)=f(x).f(y) for all x.y inR and f(x)ne0 for all x inR . If f'(0)=2 then f'(x) is equal to

If a function f: R ->R be such that f(x-f(y)) = f(f(y) )+xf(y)+f(x) -1 AA x , y in R then f(2)=

If F :R to R satisfies f(x +y ) =f(x) + f(y) for all x ,y in R and f (1) =7 , then sum_(r=1)^(n) f(R ) is

If f (x/y)= f(x)/f(y) , AA y, f (y)!=0 and f' (1) = 2 , find f(x) .

If for a function f : R to R f (x +y ) =F(x ) + f(y) for all x and y then f(0) is

If a function satisfies (x-y)f(x+y)-(x+y)f(x-y)=2(x^2 y-y^3) AA x, y in R and f(1)=2 , then

If the function / satisfies the relation f(x+y)+f(x-y)=2f(x),f(y)AAx , y in R and f(0)!=0 , then (a) f(x) is an even function (b) f(x) is an odd function (c) If f(2)=a ,then f(-2)=a (d) If f(4)=b ,then f(-4)=-b

Consider a differentiable f:R to R for which f(1)=2 and f(x+y)=2^(x)f(y)+4^(y)f(x) AA x , y in R. The minimum value of f(x) is