JEE Main 2024 · 8 April · morning · Single correct · +4 / −1
For the function f(x)=(cos x)-x+1, x ∈ ℝ, between the following two statements
(S1) f(x)=0 for only one value of x in [0, π].
(S2) f(x) is decreasing in [0, π/2] and increasing in [π/2, π].
- (a) Both (S1) and (S2) are incorrect.
- (b) Only (S1) is correct.
- (c) Only (S2) is correct.
- (d) Both (S1) and (S2) are correct.
Answer and solution
Answer (b) Only (S1) is correct.
Solution
Maths sir · 4 steps
Find the sign of f'(x) first, and both statements settle themselves.
- Differentiate. Differentiate term by term. The derivative of cos x is -sin x, and the derivative of -x + 1 is -1.
f'(x) = -sin x - 1 = -(1 + sin x) ≤ 0
- Check S2. On [0, π] we have sin x ≥ 0, so f'(x) ≤ -1 < 0 throughout. That makes f strictly decreasing on all of [0, π]. It does not turn and increase on [π/2, π], so (S2) is incorrect.
- Check S1. At the ends, f(0) = 1 - 0 + 1 = 2 > 0 and f(π) = -1 - π + 1 = -π < 0. The function is continuous and changes sign, so it has at least one zero. Because it is strictly decreasing, it cannot cross zero a second time. So (S1) is correct.
- So. Only (S1) is correct, which is option (b).
The idea to keep. To count roots in an interval, check the sign at both ends and check monotonicity. A sign change plus a strictly monotonic function gives exactly one root.
Answer from the NTA final answer key. Solution by PrepWiser.