Question:
If the maximum value of the function $f(x) = \alpha - 4x - x^2$ is 1, then the value of $\alpha$ is equal to
If the maximum value of the function $f(x) = \alpha - 4x - x^2$ is 1, then the value of $\alpha$ is equal to
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Alternatively, use differentiation: $f'(x) = -4 - 2x$. Setting $f'(x) = 0$ gives $x = -2$. Substitute this back into $f(x)$ and set equal to 1 to solve for $\alpha$.
Updated On: Jun 26, 2026
- -3
- 3
- -5
- 5
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
For a downward-opening parabola $f(x) = ax^2 + bx + c$ ($a < 0$), the maximum value occurs at the vertex where $x = -\frac{b}{2a}$.
Step 2: Detailed Explanation:
1. Identify coefficients from $f(x) = -x^2 - 4x + \alpha$:
$a = -1$, $b = -4$, $c = \alpha$.
2. Find the $x$-coordinate of the maximum:
\[ x = -\frac{b}{2a} = -\frac{-4}{2(-1)} = \frac{4}{-2} = -2 \]
3. Calculate the maximum value $f(-2)$:
\[ f(-2) = \alpha - 4(-2) - (-2)^2 = \alpha + 8 - 4 = \alpha + 4 \]
4. Given the maximum value is 1:
\[ \alpha + 4 = 1 \implies \alpha = -3 \]
Step 3: Final Answer:
The value of $\alpha$ is -3.
For a downward-opening parabola $f(x) = ax^2 + bx + c$ ($a < 0$), the maximum value occurs at the vertex where $x = -\frac{b}{2a}$.
Step 2: Detailed Explanation:
1. Identify coefficients from $f(x) = -x^2 - 4x + \alpha$:
$a = -1$, $b = -4$, $c = \alpha$.
2. Find the $x$-coordinate of the maximum:
\[ x = -\frac{b}{2a} = -\frac{-4}{2(-1)} = \frac{4}{-2} = -2 \]
3. Calculate the maximum value $f(-2)$:
\[ f(-2) = \alpha - 4(-2) - (-2)^2 = \alpha + 8 - 4 = \alpha + 4 \]
4. Given the maximum value is 1:
\[ \alpha + 4 = 1 \implies \alpha = -3 \]
Step 3: Final Answer:
The value of $\alpha$ is -3.
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