JEE Main 2022 · 29 July · evening · Single correct · +4 / −1
The number of elements in the set S={x ∈ ℝ: 2 cos ((x²+x)/6)=4ˣ+4⁻ˣ} is :
- (a) 1
- (b) 3
- (c) 0
- (d) infinite
Answer and solution
Answer (a) 1
Solution
Maths sir · 4 steps
The left side can never exceed 2 and the right side can never go below 2, so both must equal 2 together.
- Bound the left side. Cosine never exceeds 1, so 2 cos((x²+x)/6) ≤ 2 for every real x.
- Bound the right side. 4ˣ and 4⁻ˣ are positive and their product is 1. By AM–GM, their sum is at least 2√1 = 2, with equality only when 4ˣ = 4⁻ˣ, that is x = 0.
2 cos((x²+x)/6) ≤ 2 ≤ 4ˣ + 4⁻ˣ
- Force equality. The two sides can be equal only if both equal 2. The right side equals 2 only at x = 0. At x = 0 the left side is 2 cos 0 = 2, so it works.
- So. S = {0}, which has exactly 1 element. The answer is (a).
The idea to keep. When a bounded trig expression equals something like a + 1/a, compare their ranges. If they only meet at one value, both sides must take that value at the same x.
Answer from the NTA final answer key. Solution by PrepWiser.