Question

If $y = \alpha \log x + \beta x^2 - x$ has extreme values at $x = -1$ and $x = 1$, then $\alpha$ and $\beta$ are respectively ______.

• A. 0 and $\frac{1}{2}$
• B. 0 and $-\frac{1}{2}$
• C. $-\frac{1}{2}$ and $\frac{1}{2}$
• D. $\frac{1}{2}$ and $\frac{1}{2}$

Hint:

Extreme values always mean $\frac{dy}{dx} = 0$. When a problem gives multiple conditions but options are limited, quickly test just ONE boundary condition against the options to find the unique match instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Extreme values occur where the first derivative $y'$ is equal to zero.

Step 2: Formula Application:
$y' = \frac{\alpha}{x} + 2\beta x - 1$.

Step 3: Explanation:
At $x = 1$, $y' = \alpha + 2\beta - 1 = 0 \implies \alpha + 2\beta = 1$. At $x = -1$, $y' = -\alpha - 2\beta - 1 = 0 \implies \alpha + 2\beta = -1$. Adding the two equations: $0 = 0$, which indicates the values must satisfy both. However, for $x=-1$ to be in the domain, $\log x$ usually implies $\alpha=0$ (since $\log$ is undefined for negative $x$ in real numbers). If $\alpha=0$, then $2\beta = 1 \implies \beta = 1/2$.

Step 4: Final Answer:
$\alpha = 0$ and $\beta = \frac{1}{2}$.