If f(x) = sin^2 x + x sin 2x log x,then f(x) | vedclass

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(B) Given $f(x) = \sin^2 x + x \sin 2x \log x$. We can rewrite $f(x)$ as $f(x) = x \left( \frac{\sin^2 x}{x} + \sin 2x \log x \right)$. This does not

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at least two roots in $(0, 2\pi]$

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(B) Given $f(x) = \sin^2 x + x \sin 2x \log x$. We can rewrite $f(x)$ as $f(x) = x \left( \frac{\sin^2 x}{x} + \sin 2x \log x \right)$. This does not simplify directly,so let us consider the function $g(x) = \sin^2 x \log x$. Then $g'(x) = 2 \sin x \cos x \log x + \sin^2 x \cdot \frac{1}{x} = \sin 2x \log x + \frac{\sin^2 x}{x}$. Thus,$f(x) = x g'(x)$. We observe that $g(1) = \sin^2(1) \log(1) = 0$. Also,$g(\pi) = \sin^2(\pi) \log(\pi) = 0$ and $g(2\pi) = \sin^2(2\pi) \log(2\pi) = 0$. By Rolle's Theorem,$g'(x) = 0$ has at least one root in $(1, \pi)$ and at least one root in $(\pi, 2\pi)$. Since $f(x) = x g'(x)$,$f(x) = 0$ implies $g'(x) = 0$ for $x \in (0, 2\pi]$. Therefore,$f(x) = 0$ has at least two roots in $(1, 2\pi) \subset (0, 2\pi]$.

If $f(x) = \sin^2 x + x \sin 2x \log x$,then $f(x) = 0$ has

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