Prove that the function given by $f(x) = \cos x$ is decreasing in $(0, \pi)$.
(N/A) Given the function $f(x) = \cos x$.
To determine the intervals of increase or decrease,we find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(\cos x) = -\sin x$.
For the function to be decreasing,we require $f'(x) < 0$.
In the interval $(0, \pi)$,the value of $\sin x$ is always positive (i.e.,$\sin x > 0$).
Therefore,$f'(x) = -\sin x < 0$ for all $x \in (0, \pi)$.
Since the derivative is negative throughout the interval $(0, \pi)$,the function $f(x) = \cos x$ is strictly decreasing in $(0, \pi)$.
To determine the intervals of increase or decrease,we find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(\cos x) = -\sin x$.
For the function to be decreasing,we require $f'(x) < 0$.
In the interval $(0, \pi)$,the value of $\sin x$ is always positive (i.e.,$\sin x > 0$).
Therefore,$f'(x) = -\sin x < 0$ for all $x \in (0, \pi)$.
Since the derivative is negative throughout the interval $(0, \pi)$,the function $f(x) = \cos x$ is strictly decreasing in $(0, \pi)$.
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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 $[\frac{\pi}{2}, \pi]$.
$(S1)$ $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.
$(S2)$ $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.
Let $f(x) = \sin x$ and $g(x) = x$.
Statement $1$: $f(x) \le g(x)$ for $x$ in $(0, \infty)$.
Statement $2$: $f(x) \le 1$ for $x$ in $(0, \infty)$ but $g(x) \to \infty$ as $x \to \infty$.
Statement $1$: $f(x) \le g(x)$ for $x$ in $(0, \infty)$.
Statement $2$: $f(x) \le 1$ for $x$ in $(0, \infty)$ but $g(x) \to \infty$ as $x \to \infty$.
DifficultAIEEE 2012
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