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

A function `f : R rarr R` satisfies the equation f(x+y)=f(x)f(y) for all `x,y in R" and "f(x)ne0" for all "x in R.` If f(x) is differenitable at x=0 and f'(0)=2, then prove that `f'(x)=2f(x).`

Text Solution

Verified by Experts

We have f(x+y) =f(x)f(y) for all `x, y in R`
`therefore" "f(0)=f(0)f(0) or f(0) {f(0)-1}=0`
`or" "f(0)=1" "[becausef(0)ne0]`
Now, f'(0)=2
`or" "underset(hrarr0)lim(f(0+h)-f(0))/(h)=2`
`or" "underset(hrarr0)lim(f(h)-1)/(h)=2" "(becausef(0)=1)" (1)"`
`"Now, "f'(x)=underset(hrarr0)lim(f(x+h)-f(x))/(h)`
`=underset(hrarr0)lim(f(x)f(h)-f(x))/(h)" "[becausef(x+y)=f(x)f(y)]`
`=f(x)(underset(hrarr0)lim(f(h)-1)/(h))=2f(x)" [Using (1)]"`
Promotional Banner

Topper's Solved these Questions

  • DIFFERENTIATION

    CENGAGE ENGLISH|Exercise Solved Examples|28 Videos

  • DIFFERENTIATION

    CENGAGE ENGLISH|Exercise Concept Application 3.1|1 Videos

  • DIFFERENTIAL EQUATIONS

    CENGAGE ENGLISH|Exercise Matrix Match Type|5 Videos

  • DOT PRODUCT

    CENGAGE ENGLISH|Exercise DPP 2.1|15 Videos

Similar Questions

Explore conceptually related problems

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 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) .

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

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)f(y) - f(xy) = x + y ∀ x, y ∈ R and f (1)>0 , 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

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