Question

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

• A. 0
• B. \(\log 2\)
• C. 1
• D. \(\log 3\)

Hint:

When a tangent to a family of curves is fixed at a point for all parameter values, the point’s coordinates often make the derivative condition independent of the parameter.

Answer 1

The Correct Option is C

Step 1: Set up the condition for the tangent.
The curve is \( y = x^n \log x \), and \( y = x - 1 \) is the tangent at \((\alpha, \beta)\) for all \( n \). This means: - The point \((\alpha, \beta)\) lies on the curve: \[ \beta = \alpha^n \log \alpha \quad (1) \] - The line \( y = x - 1 \) is the tangent, so the slope of the curve at \( x = \alpha \) must equal the slope of the line \( y = x - 1 \), which is 1, and the point must satisfy the line’s equation: \[ \beta = \alpha - 1 \quad (2) \] - Compute the derivative of the curve to find the slope: \[ y = x^n \log x \implies \frac{dy}{dx} = \frac{d}{dx} (x^n \log x) \] Use the product rule: \[ \frac{dy}{dx} = x^n \cdot \frac{1}{x} + \log x \cdot n x^{n-1} = x^{n-1} + n x^{n-1} \log x = x^{n-1} (1 + n \log x) \] At \( x = \alpha \), the slope must be 1: \[ \alpha^{n-1} (1 + n \log \alpha) = 1 \quad (3) \]
Step 2: Solve the equations.
From (2), \(\beta = \alpha - 1\). Substitute into (1): \[ \alpha - 1 = \alpha^n \log \alpha \quad (4) \] Equation (3) must hold for all \( n \), so \( 1 + n \log \alpha = \frac{1}{\alpha^{n-1}} \). This must be independent of \( n \). Test possible values for \(\alpha\). Assume \( 1 + n \log \alpha \) is constant for all \( n \), which implies the coefficient of \( n \) must be zero: \[ \log \alpha = 0 \implies \alpha = 1 \] But if \( \alpha = 1 \), from (4): \[ 1 - 1 = 1^n \log 1 \implies 0 = 0 \] This holds, but check (3): \[ 1^{n-1} (1 + n \log 1) = 1 \cdot (1 + 0) = 1 \] This satisfies the slope condition. So, \( \alpha = 1 \), \( \beta = 1 - 1 = 0 \). Thus, the point is \((1, 0)\), and: \[ \alpha + \beta = 1 + 0 = 1 \] This matches option (3).
Step 3: Verify with the point \((1, 0)\).
The curve at \( x = 1 \): \( y = 1^n \log 1 = 0 \), so \((1, 0)\) is on the curve. The line \( y = x - 1 \) at \( x = 1 \): \( y = 1 - 1 = 0 \), so it passes through \((1, 0)\). The slope condition already holds for all \( n \), confirming \((1, 0)\) is the fixed point. Final Answer: \[ \boxed{3} \]