Question:
Let \( f(x) = x^2 + ax + \beta \). If \( f \) has a local minimum at \( (2,6) \), then \( f(0) \) is equal to
Let \( f(x) = x^2 + ax + \beta \). If \( f \) has a local minimum at \( (2,6) \), then \( f(0) \) is equal to
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Use both derivative condition and point value for extrema problems.
Updated On: Apr 21, 2026
- \(10 \)
- \(-6 \)
- \(8 \)
- \(-8 \)
- \(6 \)
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The Correct Option is A
Solution and Explanation
Concept:
At minimum:
\[
f'(x)=0
\]
Step 1: Differentiate.
\[ f'(x) = 2x + a \] \[ f'(2)=0 \Rightarrow 4 + a = 0 \Rightarrow a=-4 \]
Step 2: Use point condition.
\[ f(2)=6 \Rightarrow 4 -8 + \beta = 6 \Rightarrow \beta = 10 \]
Step 3: Find \( f(0) \).
\[ f(0)=\beta=10 \]
Step 1: Differentiate.
\[ f'(x) = 2x + a \] \[ f'(2)=0 \Rightarrow 4 + a = 0 \Rightarrow a=-4 \]
Step 2: Use point condition.
\[ f(2)=6 \Rightarrow 4 -8 + \beta = 6 \Rightarrow \beta = 10 \]
Step 3: Find \( f(0) \).
\[ f(0)=\beta=10 \]
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