Let R denote the set of all real numbers. Let | vedclass

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(A) For $x 
eq 0$,$f(x) = \frac{6x+\sin x}{2x+\sin x} = \frac{6 + \frac{\sin x}{x}}{2 + \frac{\sin x}{x}}$.As $x ightarrow 0$,$\frac{\sin x}{x} i

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$B, C, D$

Vedclass · Answer

(A) For $x 
eq 0$,$f(x) = \frac{6x+\sin x}{2x+\sin x} = \frac{6 + \frac{\sin x}{x}}{2 + \frac{\sin x}{x}}$.As $x ightarrow 0$,$\frac{\sin x}{x} ightarrow 1$,so $\lim_{x ightarrow 0} f(x) = \frac{6+1}{2+1} = \frac{7}{3}$.Since $f(0) = \frac{7}{3}$,$f$ is continuous at $x=0$.For $x$ near $0$,let $g(x) = \frac{\sin x}{x}$. Since $g(x)  0$.$f(x) = \frac{6 + (1-\epsilon)}{2 + (1-\epsilon)} = \frac{7-\epsilon}{3-\epsilon} = \frac{3(3-\epsilon) - 2 + 2\epsilon}{3-\epsilon} = 3 - \frac{2}{3-\epsilon} Since $f(x) $f'(x) = \frac{4(\sin x - x \cos x)}{(2x+\sin x)^2} = \frac{4 \cos x(	an x - x)}{(2x+\sin x)^2}$.For $x > 0$,$	an x > x$,so $f'(x) > 0$ when $\cos x > 0$.The local maxima occur when $f'(x)$ changes from positive to negative,i.e.,when $\cos x$ changes sign from positive to negative at $x = (2n+1)\frac{\pi}{2}$.In $[\pi, 6\pi]$,$\cos x$ changes sign at $x = \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$.Evaluating the behavior,options $B, C, D$ are correct.

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