Let f(x) = sin x and g(x) = x.Statement 1: f( | vedclass

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(C) Consider the function $h(x) = g(x) - f(x) = x - \sin x$ for $x \in (0, \infty)$.Taking the derivative,$h'(x) = 1 - \cos x$.Since $-1 \le \cos x \l

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Statement $1$ is true,Statement $2$ is true,Statement $2$ is not a correct explanation for Statement $1$.

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(C) Consider the function $h(x) = g(x) - f(x) = x - \sin x$ for $x \in (0, \infty)$.Taking the derivative,$h'(x) = 1 - \cos x$.Since $-1 \le \cos x \le 1$,we have $h'(x) = 1 - \cos x \ge 0$ for all $x$.Since $h'(x) \ge 0$ and $h(0) = 0 - \sin(0) = 0$,the function $h(x)$ is non-decreasing and $h(x) \ge 0$ for all $x \in (0, \infty)$.Thus,$x \ge \sin x$,which means $g(x) \ge f(x)$ is true. So,Statement $1$ is true.For Statement $2$,we know that $\sin x \le 1$ for all $x \in (0, \infty)$ and $\lim_{x 	o \infty} x = \infty$. Both parts of Statement $2$ are true.However,the fact that $\sin x \le 1$ and $x 	o \infty$ does not directly prove $\sin x \le x$ for all $x > 0$ without considering the behavior of the difference $x - \sin x$. Therefore,Statement $2$ is not the correct explanation for Statement $1$.

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 $x \to \infty$.

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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 $x \to \infty$.