The set of all real values of a for which the function \(f : R \to R\), \(f(x) = x^{3} + ax^{2} + 12x + 1\), is one-one is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \([-6, 6]\)
The derivative is
\(f'(x) = 3x^{2} + 2ax + 12\)
If \(f'(x) \lt 0\) somewhere, the cubic rises, falls and rises again, so some value is taken at three points and f is not one-one. So f is one-one exactly when \(f'(x) \ge 0\) for all x, which for this upward quadratic means its discriminant is not positive:
\((2a)^{2} - 4 \times 3 \times 12 \le 0\)
\(4a^{2} \le 144\)
\(a^{2} \le 36\)
\(-6 \le a \le 6\)
At the end values f is still one-one. For \(a = 6\),
\(f'(x) = 3(x + 2)^{2}\)
which is zero only at \(x = -2\) and positive elsewhere, so f is still strictly increasing. The case \(a = -6\) is the same with \(3(x - 2)^{2}\).
Why the other options are incorrect
Option (2) \((-6, 6)\) — Incorrect. It leaves out \(a = \pm 6\), where \(f'(x) = 3(x \pm 2)^{2}\) and f is still increasing.
Option (3) \([-3, 3]\) — Incorrect. It leaves out values such as \(a = 5\), where \(f'(x) = 3x^{2} + 10x + 12\) has discriminant \(100 - 144 \lt 0\), so f is one-one.
Option (4) \((-\infty, -6] \cup [6, \infty)\) — Incorrect. For \(a = 7\), \(f'(x) = 3x^{2} + 14x + 12\) has discriminant \(196 - 144 \gt 0\), so f falls between two turning points and is not one-one.