JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
Let x=(8 √3+13)¹³ and y=(7 √2+9)⁹. If [t] denotes the greatest integer ≤ t, then :
- (a) [x] is odd but [y] is even
- (b) [x] and [y] are both odd
- (c) [x]+[y] is even
- (d) [x] is even but [y] is odd
Answer and solution
Answer (c) [x]+[y] is even
Solution
Maths sir · 4 steps
The conjugates here are negative, so subtract them: each number is an even integer plus a small fraction.
- Find the small partner. 8√3 ≈ 13.86 and 7√2 ≈ 9.90. So f = (8√3-13)¹³ and g = (7√2-9)⁹ are both positive and lie between 0 and 1.
(a+b)ⁿ - (a-b)ⁿ = 2Σ_(k odd) C(n, k) aⁿ⁻ᵏ bᵏ
- Even integer. Take a = 8√3, b = 13, n = 13. Every surviving term has k odd, so n-k is even and (8√3)ⁿ⁻ᵏ is an integer. Hence x - f is 2 times an integer, an even integer. The same argument with a = 7√2, b = 9, n = 9 makes y - g an even integer.
- Read off the floor. x = (even integer) + f with 0 < f < 1, so [x] is that even integer. In the same way [y] is even.
- So. [x] and [y] are both even, so [x] + [y] is even. The answer is (c).
The idea to keep. Pair the number with its conjugate that lies between 0 and 1. If their difference is an integer, the floor equals that integer. If their sum is an integer, the floor is one less.
Answer from the NTA final answer key. Solution by PrepWiser.