The equation of the tangent at x = frac{pi}{2 | vedclass

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(B) Given the curve $y = x \sin x$. At $x = \frac{\pi}{2}$,$y = \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2} 	imes 1 = \frac{\pi}{2}$. So,the po

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$x - y = 0$

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(B) Given the curve $y = x \sin x$. At $x = \frac{\pi}{2}$,$y = \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2} 	imes 1 = \frac{\pi}{2}$. So,the point of contact is $(\frac{\pi}{2}, \frac{\pi}{2})$. Now,differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x \sin x) = x \cos x + \sin x$. Calculate the slope $m$ at $x = \frac{\pi}{2}$: $m = \left. \frac{dy}{dx} ight|_{x = \frac{\pi}{2}} = \frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) = \frac{\pi}{2}(0) + 1 = 1$. The equation of the tangent is given by $y - y_1 = m(x - x_1)$: $y - \frac{\pi}{2} = 1(x - \frac{\pi}{2})$. $y - \frac{\pi}{2} = x - \frac{\pi}{2}$. $x - y = 0$.

The equation of the tangent at $x = \frac{\pi}{2}$ to the curve $y = x \sin x$ is

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