Question

Find the natural number a for which  where the function f satisfies the relation f (x + y) = f (xf (y) for all natural number xy and further f (1) = 2.

 

Solution

Correct option is

a = 3

Since the function satisfies the relation 

          f (y) = f (xf (y)

It must be an exponential function. 

Let the base of this exponential function be a.  

Thus f (x) = ax  

It is given that f (1) = 2. So we can make

          f (1) = a1 = 2 ⇒     a = 2

Hence, the function is f (x) =2x                    … (i)

                                                [Alternatively, we have] 

     f (x) = f (x – 1 + 1) = f (x – 1) f (1) 

            = f (x – 2 + 1) f (1)  

            = f (x – 2) [f (1)]2 = … = [f (1)]x = 2x]  

Using equation (i), the given expression reduces to:  

            

   

SIMILAR QUESTIONS

Q1

Discuss the continuity of the function .

Q2

Let f : → R, such that f’ (0) = 1 and f (x +2y) = f (x) + f (2y) + ex+2y (x + 2y) – x. ex – 2y. e2y + 4xy∀ xy Ïµ R. Find f (x).

Q3

If g(x) is continuous function in [0, ∞) satisfying g(1) = 1. If

                          .

Q4

Let f is a differentiable function such that 

          .

Q5

Let f : R+ → R satisfies the functional equation  

          .  

If f’(1) = e, determine f (x).

Q6

Let f is a differentiable function such that   

     .

Q7

Let f be a function such that  .   

.

Q8

Find and b so that the function:    

 

   

Q9

 is continuous at x = 0, find the values of Aand B. Also find f (0).

Q10

Find the derivative of y = log x wrt x from first principles.