f′(x)=cosx−x+1f′(x)=−sinx−1 f is decreasing ∀x∈R f(x)=0f(0)=2,f(π)=−π f is strictly decreasing in [0,π] and f(0).f(π)<0 ⇒ only one solution of f(x)=0 S1 is correct and S2 is incorrect.
For the function f(x)=(cosx)−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 [0,2π] and increasing in [2π,π].
Held on 8 Apr 2024 · Verified 6 Jul 2026.
Both (S1) and (S2) are correct.
Both (S1) and (S2) are incorrect.
Only (S2) is correct.
Only (S1) is correct.
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