For all possible integers n satisfying 2.25≤ 2+2^(n+2)≤ 202 , the number of integer values of 3+3^(n+1) is

CAT 2021 Slot 2 · Quant · Question 15

CAT 2021 Slot 2 previous year question on Surds and Indices, with the answer and a step-by-step solution.

Surds and IndicesAlgebraType in the answer (TITA)Medium

For all possible integers n satisfying 2.25\le 2+{2}^{n+2}\le 202, the number of integer values of 3+{3}^{n+1} is

Answer & solution

Answer: 7

First find which integers n are allowed. Subtracting 2 throughout: 0.25\le{2}^{n+2}\le 200.

Since 0.25={2}^{-2}, the left inequality gives n+2\ge -2, that is n\ge -4.

For the right, {2}^{7}=128\le 200 but {2}^{8}=256>200, so n+2\le 7, that is n\le 5.

So n runs over -4,-3,\ldots ,5, which is 10 values.

Now the question asks how many of these make 3+{3}^{n+1} an integer. {3}^{n+1} is an integer only when n+1\ge 0, that is n\ge -1.

That leaves n=-1,0,1,2,3,4,5, giving 3+{3}^{n+1}=4,6,12,30,84,246,732

So, 7 distinct integer values.