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If $y=x \sin x$ and $\frac{\frac{d y}{d x}-\frac{y}{x}}{x \frac{d y}{d x}-y}$ at $x=\alpha$ is 1 , then $\alpha=$
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$y=x \sin x$
$\frac{d y}{d x}=x \cos x+\sin x$
$\begin{aligned} & \therefore \frac{\frac{d y}{d x}-\frac{y}{x}}{x \frac{d y}{d x}-y}=\frac{x \cos x+\sin x-\sin x}{x(x \cos x+\sin x)-x \sin x} \\ & =\frac{\cos x}{x \cos x+\sin x-\sin x}=\frac{1}{x} \\ & \therefore \quad \frac{1}{\alpha}=1 \Rightarrow \alpha=1 .\end{aligned}$
$\frac{d y}{d x}=x \cos x+\sin x$
$\begin{aligned} & \therefore \frac{\frac{d y}{d x}-\frac{y}{x}}{x \frac{d y}{d x}-y}=\frac{x \cos x+\sin x-\sin x}{x(x \cos x+\sin x)-x \sin x} \\ & =\frac{\cos x}{x \cos x+\sin x-\sin x}=\frac{1}{x} \\ & \therefore \quad \frac{1}{\alpha}=1 \Rightarrow \alpha=1 .\end{aligned}$
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