For the function f(x) = cos x - x + 1, x in R | vedclass

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(B) Given $f(x) = \cos x - x + 1$.Find the derivative: $f'(x) = -\sin x - 1$.Since $-1 \le \sin x \le 1$,we have $f'(x) = -(\sin x + 1) \le 0$ for all

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Only $(S1)$ is correct

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(B) Given $f(x) = \cos x - x + 1$.Find the derivative: $f'(x) = -\sin x - 1$.Since $-1 \le \sin x \le 1$,we have $f'(x) = -(\sin x + 1) \le 0$ for all $x \in R$.Specifically,$f'(x) Thus,$f(x)$ is a strictly decreasing function on $R$.For $(S1)$: Evaluate at endpoints of $[0, \pi]$: $f(0) = \cos(0) - 0 + 1 = 2$ and $f(\pi) = \cos(\pi) - \pi + 1 = -1 - \pi + 1 = -\pi$.Since $f(0) = 2 > 0$ and $f(\pi) = -\pi For $(S2)$: Since $f'(x) \le 0$ for all $x$,$f(x)$ is decreasing on the entire interval $[0, \pi]$. Therefore,the statement that it is increasing in $[\frac{\pi}{2}, \pi]$ is incorrect. So,$(S2)$ is incorrect.Conclusion: Only $(S1)$ is correct.

For the function $f(x) = \cos x - x + 1, x \in R$,consider the following two statements:
$(S1)$ $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.
$(S2)$ $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.

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For the function $f(x) = \cos x - x + 1, x \in R$,consider the following two statements:$(S1)$ $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.$(S2)$ $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.