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

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(B) Consider the function $h(x) = g(x) - f(x) = x - \sin x$.To check if $f(x) \leq g(x)$ for $x > 0$,we analyze $h(x)$.The derivative is $h'(x) = 1 -

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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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(B) Consider the function $h(x) = g(x) - f(x) = x - \sin x$.To check if $f(x) \leq g(x)$ for $x > 0$,we analyze $h(x)$.The derivative is $h'(x) = 1 - \cos x$.Since $\cos x \leq 1$ for all $x$,$h'(x) = 1 - \cos x \geq 0$.This means $h(x)$ is a non-decreasing function for $x \geq 0$.Since $h(0) = 0 - \sin(0) = 0$,it follows that $h(x) \geq 0$ for all $x \geq 0$.Thus,$x - \sin x \geq 0$,which implies $x \geq \sin x$ or $g(x) \geq f(x)$ for $x \in (0, \infty)$.Therefore,Statement-$1$ is true.Now consider Statement-$2$: For $x \in (0, \infty)$,$f(x) = \sin x \leq 1$ is true.Also,as $x \rightarrow \infty$,$g(x) = x \rightarrow \infty$ is true.However,the fact that $f(x) \leq 1$ and $g(x) \rightarrow \infty$ does not directly prove $f(x) \leq g(x)$ without comparing the growth rates or using the derivative test.Thus,Statement-$2$ is true,but it is not the direct logical explanation for Statement-$1$ because the inequality $x \geq \sin x$ holds for all $x > 0$ regardless of the limit at infinity.

Let $f(x) = \sin x$ and $g(x) = x$.
Statement-$1$: For $x \in (0, \infty)$,$f(x) \leq g(x)$.
Statement-$2$: For $x \in (0, \infty)$,$f(x) \leq 1$ but as $x \rightarrow \infty$,$g(x) \rightarrow \infty$.

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Let $f(x) = \sin x$ and $g(x) = x$.Statement-$1$: For $x \in (0, \infty)$,$f(x) \leq g(x)$.Statement-$2$: For $x \in (0, \infty)$,$f(x) \leq 1$ but as $x \rightarrow \infty$,$g(x) \rightarrow \infty$.