Let f(x)=2+cos x for all real x.STATEMENT-1: | vedclass

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(B) Given $f(x)=2+\cos x$ for all $x \in \mathbb{R}$.Statement-$1$: We need to check if there exists $c \in [t, t+\pi]$ such that $f'(c)=0$.$f'(x) = -

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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) Given $f(x)=2+\cos x$ for all $x \in \mathbb{R}$.Statement-$1$: We need to check if there exists $c \in [t, t+\pi]$ such that $f'(c)=0$.$f'(x) = -\sin x$.For $f'(c)=0$,we need $\sin c = 0$,which implies $c = n\pi$ for some integer $n$.In any interval of length $\pi$,such as $[t, t+\pi]$,there is always at least one multiple of $\pi$. For example,if $t=0.1$,the interval is $[0.1, 3.24]$,which contains $\pi \approx 3.14$. Thus,Statement-$1$ is True.Statement-$2$: $f(t) = 2+\cos t$ and $f(t+2\pi) = 2+\cos(t+2\pi) = 2+\cos t$. Thus,$f(t)=f(t+2\pi)$ is True.However,Statement-$2$ (the periodicity of $f$) does not imply the existence of a root of the derivative in every interval of length $\pi$. The existence of a root in $[t, t+\pi]$ is a property of the sine function's zeros,not the periodicity of $f$. Therefore,Statement-$2$ is not the correct explanation for Statement-$1$.

Let $f(x)=2+\cos x$ for all real $x$.
$STATEMENT-1$: For each real $t$,there exists a point $c$ in $[t, t+\pi]$ such that $f^{\prime}(c)=0$. because
$STATEMENT-2$: $f(t)=f(t+2\pi)$ for each real $t$.

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Let $f(x)=2+\cos x$ for all real $x$.$STATEMENT-1$: For each real $t$,there exists a point $c$ in $[t, t+\pi]$ such that $f^{\prime}(c)=0$. because$STATEMENT-2$: $f(t)=f(t+2\pi)$ for each real $t$.