If \(y = ax^2 + bx\) has minima at \(x = 2\) and the minimum value is −12, then which of the following are correct?
(A) \(a = 3\)
(B) \(a = -3\)
(C) \(b = 12\)
(D) \(b = -12\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (D) only
At a minimum the derivative is zero:
\( \displaystyle y' = 2ax + b \)
\( \displaystyle y'(2) = 4a + b = 0 \)
\( \displaystyle b = -4a \)
The minimum value is y(2):
\( \displaystyle 4a + 2b = -12 \)
\( \displaystyle 4a - 8a = -12 \)
\( \displaystyle a = 3 \)
\( \displaystyle b = -4a = -12 \)
Check: \(y'' = 2a = 6 \gt 0\), so x = 2 is indeed a minimum.
So \(a = 3\) and \(b = -12\).
Checking the statements
(A) \(a = 3\) — Correct. \(4a - 8a = -12\) gives \(a = 3\).
(B) \(a = -3\) — Incorrect. \(a\) is 3; a negative a would give a maximum, not a minimum.
(C) \(b = 12\) — Incorrect. \(b = -4a = -12\).
(D) \(b = -12\) — Correct. \(b = -4a = -12\).