For the function f(x) = (cosx) – x + 1, x ∈ R , between the following two statements (S1) f(x) = 0 for only one value

Question

For the function \(f(x)=(\cos x)-x+1, \ x \in \mathbb{R}\), between the following two statements

(S1) \( f(x)=0\) for only one value of x is \([0, \pi]\).

(S2) \(f(x)\) is decreasing in \(\left[0, \frac{\pi}{2}\right]\) and increasing in \(\left[\frac{\pi}{2}, \pi\right]\).

(1) Both (S1) and (S2) are correct

(2) Only (S1) is correct

(3) Both (S1) and (S2) are incorrect

(4) Only (S2) is correct

Answer 1

Correct option is : (2) Only (S1) is correct 

\(f(x)=\cos x-x+1\)

\(f^{\prime}(x)=-\sin x-1\)

\(\mathrm{f}\) is decreasing \(\forall \ \mathrm{x} \in \mathrm{R}\)

\(\mathrm{f}(\mathrm{x})=0\)

\(f(0)=2, f(\pi)=-\pi\)

\(\mathrm{f}\) is strictly decreasing in \([0, \pi]\) and \(\mathrm{f}(0) . \mathrm{f}(\pi)<0\)

\(\Rightarrow\) only one solution of \(\mathrm{f}(\mathrm{x})=0\)

\(\mathrm{S} 1\) is correct and \(\mathrm{S} 2\) is incorrect.