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(C) Given the family of curves $y = x^n \log x$. For the tangent $y = x - 1$ to be common to all curves at a fixed point $(\alpha, \beta)$,the slope o

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(C) Given the family of curves $y = x^n \log x$. For the tangent $y = x - 1$ to be common to all curves at a fixed point $(\alpha, \beta)$,the slope of the tangent must be $1$ (since the slope of $y = x - 1$ is $1$). Calculating the derivative: $\frac{dy}{dx} = n x^{n-1} \log x + x^n \cdot \frac{1}{x} = x^{n-1} (n \log x + 1)$. At the point of tangency $(\alpha, \beta)$,the slope is $1$: $\alpha^{n-1} (n \log \alpha + 1) = 1$. For this to hold for all $n$,we test the condition at $\alpha = 1$: $1^{n-1} (n \log 1 + 1) = 1 \cdot (0 + 1) = 1$. This is independent of $n$. When $\alpha = 1$,the value of $y$ on the curve is $\beta = 1^n \log 1 = 0$. Thus,the fixed point is $(1, 0)$. Therefore,$\alpha + \beta = 1 + 0 = 1$.

Let $n \in (0, \infty)$. If all the curves $y = x^n \log x$ for distinct values of $n$ always have $y = x - 1$ as the tangent drawn at a fixed point $(\alpha, \beta)$,then $\alpha + \beta =$

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