Let f(x) = x sin pi x,x 0. Then for all nat | vedclass

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(B) Given $f(x) = x \sin \pi x$.Taking the derivative with respect to $x$,we get:$f^{\prime}(x) = \sin \pi x + \pi x \cos \pi x$.To find where $f^{\pr

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$(B, C)$

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(B) Given $f(x) = x \sin \pi x$.Taking the derivative with respect to $x$,we get:$f^{\prime}(x) = \sin \pi x + \pi x \cos \pi x$.To find where $f^{\prime}(x)$ vanishes,set $f^{\prime}(x) = 0$:$\sin \pi x + \pi x \cos \pi x = 0$$\sin \pi x = -\pi x \cos \pi x$$	an \pi x = -\pi x$.Consider the graphs of $y = 	an \pi x$ and $y = -\pi x$.For any natural number $n$,the interval $(n, n+1)$ corresponds to the branch of $	an \pi x$ that increases from $-\infty$ to $+\infty$.The line $y = -\pi x$ is a straight line with a negative slope passing through the origin.In the interval $(n, n+1)$,the function $y = 	an \pi x$ covers all real values from $-\infty$ to $+\infty$ exactly once.Since the value of $-\pi x$ in the interval $(n, n+1)$ lies between $-\pi(n+1)$ and $-\pi n$,the line $y = -\pi x$ intersects the branch of $	an \pi x$ exactly once in the interval $(n, n+1)$.Specifically,for $n \geq 1$,the intersection occurs in the interval $\left(n + \frac{1}{2}, n+1ight)$ because $	an \pi x$ is negative in $\left(n, n+\frac{1}{2}ight)$ and positive in $\left(n+\frac{1}{2}, n+1ight)$,while $-\pi x$ is always negative.Thus,$f^{\prime}(x)$ vanishes at a unique point in $(n, n+1)$,and this point lies in $\left(n+\frac{1}{2}, n+1ight)$.Therefore,statements $(B)$ and $(C)$ are correct.

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