If y = x sin x and frac{frac{dy}{dx} - frac{y | vedclass

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(C) Given $y = x \sin x$. Differentiating with respect to $x$, we get $\frac{dy}{dx} = x \cos x + \sin x$. Now, consider the expression $\frac{\frac{d

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(C) Given $y = x \sin x$. Differentiating with respect to $x$, we get $\frac{dy}{dx} = x \cos x + \sin x$. Now, consider the expression $\frac{\frac{dy}{dx} - \frac{y}{x}}{x \frac{dy}{dx} - y}$. Substituting $\frac{dy}{dx} = x \cos x + \sin x$ and $y = x \sin x$: Numerator: $\frac{dy}{dx} - \frac{y}{x} = (x \cos x + \sin x) - \frac{x \sin x}{x} = x \cos x + \sin x - \sin x = x \cos x$. Denominator: $x \frac{dy}{dx} - y = x(x \cos x + \sin x) - x \sin x = x^2 \cos x + x \sin x - x \sin x = x^2 \cos x$. Thus, the expression becomes $\frac{x \cos x}{x^2 \cos x} = \frac{1}{x}$. Given that at $x = \alpha$, the expression equals $1$, we have $\frac{1}{\alpha} = 1$, which implies $\alpha = 1$.

If $y = x \sin x$ and $\frac{\frac{dy}{dx} - \frac{y}{x}}{x \frac{dy}{dx} - y}$ at $x = \alpha$ is $1$, then $\alpha =$

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Assertion: For $x < 0$, $\frac{d^2}{d x^2}(\log |x|) = \frac{1}{|x|^2}$.
Reason: For $x < 0$, $|x| = -x$.

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