TITA questions can prove to be time-guzzlers. They are attractive because there is no negative marks. However you need to realize that managing time is even more important. Here is a nice and tough functions question that appeared in CAT 2019. Needless to say students can choose to attend TITA questions very carefully. The absence of answer choices also make it easier to those silly errors which may prove to be the difference between a 90th & 80th percentile.If you are looking to solve CAT level questions for free, visit **2IIM's CAT Question Bank**.

Question 33 : Consider a function f(x+y) = f(x) f(y) where x , y are positive integers, and f(1) = 2. If f (a+1) + f (a+2) + ..... + f(a+n) = 16 (2^{n} - 1) then a is equal to. [TITA]

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We know that f(x+y) = f(x) f(y)

Let x = a and y = 1,

f (a + 1) = f(a). f (1)

f(a+1) = f(a). 2

f(a+2) = f(a+1) f(1)

f(a+2) = f(a) x 2 x 2 = f(a) x 2^{2}

Similarly, f(a+n) = f(a) x 2^{n}

(a+1) + f (a+2) + ..... + f(a+n) = f(a) {2^{1} + 2^{2} + 2^{3}+...+2^{n}}

(a+1) + f (a+2) + ..... + f(a+n) = 2f(a) {1 + 2 + 2^{2} + 2^{3}+...+2^{n-1}}, which is an infinite GP series

On solving,

(a+1) + f (a+2) + ..... + f(a+n) = 2f(a) {2^{n} - 1}

We know that, (a+1) + f (a+2) + ..... + f(a+n) = 16 (2^{n} - 1)

So, 2f(a) {2^{n} - 1} = 16 (2^{n} - 1)

2 f(a) = 16

f(a) = 8

We know that, f (1) = 2 , f(2) = 4 and f(3) = 8

So, a = 3

The question is **" Consider a function f(x+y) = f(x) f(y) where x , y are positive integers, and f(1) = 2. If f (a+1) + f (a+2) + ..... + f(a+n) = 16 (2 ^{n} - 1) then a is equal to. [TITA]" **

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