JEE Main 2021 · 27 July · evening · Single correct · +4 / −1
Let f : R → R be defined as f(x + y) + f(x - y) = 2f(x)f(y),f( 1/2 ) = - 1. Then, the value of Σₖ₌₁²⁰ 1/(sin (k)sin (k + f(k))) is equal to :
- (a) cosec²(21) cos(20) cos(2)
- (b) sec²(1) sec(21) cos(20)
- (c) cosec²(1) cosec(21) sin(20)
- (d) sec²(21) sin(20) sin(2)
Answer and solution
Answer (c) cosec²(1) cosec(21) sin(20)
Solution
Maths sir · 5 steps
First find f at the integers. It is always 1, and then the sum telescopes with the cot trick.
- Find f(0). Put y = 0: 2f(x) = 2f(x)f(0). If f(0) = 0, then f(x) = 0 for every x, which fails f(1/2) = -1. So f(0) = 1.
- Find f(1). Put x = y = 1/2: f(1) + f(0) = 2f(1/2)² = 2. So f(1) = 1.
- All integers. Put x = k, y = 1: f(k+1) + f(k-1) = 2f(k). So f(0), f(1), f(2), … form an arithmetic progression with common difference f(1) - f(0) = 0. Hence f(k) = 1, and each term is 1/(sin k sin(k+1)).
1/(sin k sin(k+1)) = 1/(sin 1)·sin((k+1)-k)/(sin k sin(k+1)) = 1/(sin 1)(cot k - cot(k+1))
- Telescope. Adding from k = 1 to 20 leaves 1/(sin 1)(cot 1 - cot 21). Since cot 1 - cot 21 = sin(21-1)/(sin 1 sin 21), the sum is (sin 20)/(sin² 1 sin 21).
- So. The sum equals cosec²(1) cosec(21) sin(20), which is option (c).
The idea to keep. For a sum of 1/(sin a sin b) where b - a is fixed, write the numerator as sin(b-a) and divide by it. Each term becomes a difference of cots, and the sum telescopes.
Answer from the NTA final answer key. Solution by PrepWiser.