Home
Class 12
MATHS
A function f : R rarr R satisfies th...

A function `f : R rarr R` satisfies the equation `f(x+y) = f(x). f(y)` for all `x y in R, f(x) ne 0`. Suppose that the function is differentiable at `x = 0` and `f'(0) = 2`, then prove that `f' = 2f(x)`.

Text Solution

Verified by Experts

The correct Answer is:
N/a
Let `f : R rarr R` satisfies the equation `f(x+y) = f(x) . f(y), AA x, y in R, f(x) ne 0`.
Let `f(x)` is differentiable at `x = 0` and `f'(0) = 2`
`rArr f'(0) = underset(xrarr0)(lim) (f(x) - f(0))/(x-0)`
`rArr 2 = underset(xrarr0)(lim) (f(x) - f(0))/(x)`
` rArr 2 = underset(hrarr0)(lim)(f(0+h)-f(0))/(h)`
`rArr 2 = underset(hrarr0)lim(f'(0).f(h)-f(0))/(h)`
`rArr 2 = underset(hrarr0)(lim) (f(0) [f(h) - 1])/(h)` , `[:' f(0) = f (h) "....."(i)]`
Also, `f'(x) = underset(hrarr0)(lim)(f(x+h)-f(x))/(h)`
` =underset(hrarr0)(lim)(f(x).f(h)-f(x))/(h), [:' f(x+y)=f(x).f(y)]`
`=underset(hrarr0)(lim)(f(x)[f(h)-1])/(h) = 2f(x)` , [using Eq. (i)]
`:. f'(x) = 2f(x)`
Promotional Banner

Topper's Solved these Questions

  • CONTINUITY AND DIFFERENTIABILITY

    NCERT EXEMPLAR ENGLISH|Exercise Objective type|28 Videos

  • CONTINUITY AND DIFFERENTIABILITY

    NCERT EXEMPLAR ENGLISH|Exercise Fillers|10 Videos

  • APPLICATION OF INTEGRALS

    NCERT EXEMPLAR ENGLISH|Exercise Objective Type Questions|22 Videos

  • DETERMINANTS

    NCERT EXEMPLAR ENGLISH|Exercise TRUE/FALSE|11 Videos

Similar Questions

Explore conceptually related problems

A function f : R rarr R satisfies the equation f(x + y) = f(x) . f(y) for all, f(x) ne 0 . Suppose that the function is differentiable at x = 0 and f'(0) = 2. Then,

A function f: R->R satisfies that equation f(x+y)=f(x)f(y) for all x ,\ y in R , f(x)!=0 . Suppose that the function f(x) is differentiable at x=0 and f^(prime)(0)=2 . Prove that f^(prime)(x)=2\ f(x) .

A function f: R->R satisfies that equation f(x+y)=f(x)f(y) for all x ,\ y in R , f(x)!=0 . Suppose that the function f(x) is differentiable at x=0 and f^(prime)(0)=2 . Prove that f^(prime)(x)=2\ f(x) .

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

If a real valued function f(x) satisfies the equation f(x +y)=f(x)+f (y) for all x,y in R then f(x) is

A function f: R->R satisfies the equation f(x+y)=f(x)f(y) for all x , y in R and f(x)!=0 for all x in Rdot If f(x) is differentiable at x=0a n df^(prime)(0)=2, then prove that f^(prime)(x)=2f(x)dot

A function f : R→R satisfies the equation f(x)f(y) - f(xy) = x + y ∀ x, y ∈ R and f (1)>0 , then

A function f(x) satisfies the relation f(x+y) = f(x) + f(y) + xy(x+y), AA x, y in R . If f'(0) = - 1, then

A function f:R->R satisfies the relation f((x+y)/3)=1/3|f(x)+f(y)+f(0)| for all x,y in R. If f'(0) exists, prove that f'(x) exists for all x, in R.

Let f:R to R such that f(x+y)+f(x-y)=2f(x)f(y) for all x,y in R . Then,