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If the function f(x) = x2 - ax - 2 is strictly decreasing on (2, 3) than a lies in the interval. |
[4, ∞) [6, ∞) [-4, ∞) [-6, ∞) |
[6, ∞) |
$f(x) = x^2-ax-2$ $f'(x) = 2x-a$ $f'(x) < 0$ 2x - a < 0 2x < a → for f to be strictly decreasing, f'(x) < 0 x ∈ (2, 3) → from question 2x ∈ (4, 6) a > 6 $\Rightarrow [6, ∞)$ |
