Question

For the function $f(x) = (\cos x) - x + 1, x \in \mathbb{R}$, find the correct relationship 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 $\left[\frac{\pi}{2}, \pi\right]$.

• A. Both (S1) and (S2) are correct
• B. Only (S1) is correct
• C. Both (S1) and (S2) are incorrect
• D. Only (S2) is correct

Answer 1

Answer: (B)

Only (S1) is correct

Answer 2

The function is:
$f(x) = \cos x - x + 1.$
The derivative is:
$f'(x) = -\sin x - 1.$
Since $-\sin x - 1 < 0$ for all $x \in \mathbb{R}$, $f(x)$ is strictly decreasing in $[0, \pi]$.
At $x = 0$,
$f(0) = \cos(0) - 0 + 1 = 2.$
At $x = \pi$,
f(\pi) = \cos(\pi) - \pi + 1 = -\pi < 0\.
By the intermediate value theorem, $f(x) = 0$ has exactly one root in $[0, \pi]$. Thus, (S1) is correct.
(S2) is incorrect because $f(x)$ is strictly decreasing in $[0, \pi]$.
Final Answer: Only (S1) is correct.