CAT 2019 Slot 1 — QA Question 33
Basics (Functions)Easy
Consider a function f satisfying 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(2n – 1) then a is equal to
Answer & solution
Answer: 3
Solution
Given, f(1) = 2
Now,
f(2) = f(1 + 1) = f(1) × f(1) = 2 × 2 = 4 = 22.
f(3) = f(1 + 2) = f(1) × f(2) = 2 × 22 = 23.
f(3) = f(1 + 3) = f(1) × f(3) = 2 × 23 = 24.
So, f(n) = 2n.
f(a + 1) + f(a + 2) +…+ f(a + n) = 16(2n – 1)
∴ 2(a + 1) + 2(a + 2) + ... + 2(a + n) = 24(2n – 1)
∴ 2(a + 1)[1 + 2 + ... 2(n−1)] = 24(2n – 1)
∴ 2(a + 1)(2n – 1) = 24(2n – 1)
∴ 2(a + 1) = 24
So, a + 1 = 4
∴ a = 3.
Hence, 3.