The curve f(x) = e^x sin x is defined in the | vedclass

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(B) Given: $f(x) = e^x \sin x$. The slope of the tangent is given by $m(x) = f'(x)$. $f'(x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)$. Let $g

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$\frac{\pi}{2}$

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(B) Given: $f(x) = e^x \sin x$. The slope of the tangent is given by $m(x) = f'(x)$. $f'(x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)$. Let $g(x) = f'(x) = e^x (\sin x + \cos x)$. To find the maximum slope,we find $g'(x) = 0$. $g'(x) = e^x (\sin x + \cos x) + e^x (\cos x - \sin x) = 2 e^x \cos x$. Setting $g'(x) = 0$,we get $2 e^x \cos x = 0$. Since $e^x 
eq 0$ for all $x$,we have $\cos x = 0$. In the interval $[0, 2 \pi]$,$x = \frac{\pi}{2}$ or $x = \frac{3 \pi}{2}$. Now,we check the second derivative $g''(x) = 2 e^x \cos x - 2 e^x \sin x = 2 e^x (\cos x - \sin x)$. At $x = \frac{\pi}{2}$,$g''(\frac{\pi}{2}) = 2 e^{\frac{\pi}{2}} (0 - 1) = -2 e^{\frac{\pi}{2}} At $x = \frac{3 \pi}{2}$,$g''(\frac{3 \pi}{2}) = 2 e^{\frac{3 \pi}{2}} (0 - (-1)) = 2 e^{\frac{3 \pi}{2}} > 0$,which indicates a local minimum. Thus,the slope is maximum at $x = \frac{\pi}{2}$.

The curve $f(x) = e^x \sin x$ is defined in the interval $[0, 2 \pi]$. The value of $x$ for which the slope of the tangent drawn to the curve at $x$ is maximum,is

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