Functions 2 Question 2

3. Let k=110f(a+k)=16(2101), where the function f satisfies f(x+y)=f(x)f(y) for all natural numbers x,y and f(1)=2. Then, the natural number ’ a ’ is

(2019 Main, 9 April I)

(a) 2

(b) 4

(c) 3

(d) 16

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Answer:

Correct Answer: 3. (c)

Solution:

  1. Given, f(x+y)=f(x)f(y)

Let f(x)=λx

[where λ>0 ]

f(1)=2

(given)

λ=2

 So, k=110f(a+k)=k=110λa+k=λak=110λk=2a[21+22+23+..+210]=2a2(2101)21

[by using formula of sum of n-terms of a GP having first term ’ a ’ and common ratio ’ r ‘, is

Sn=a(rn1)r1, where r>1

2a+1(2101)=16(2101) (given) 2a+1=16=24a+1=4a=3



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