If the function f(x) = x^2 - ax - 2 is strictly decreasing on (2, 3) then a lies in the interval. Option 1: [4, ∞) Option 2: [6, ∞) Option 3: [-4, ∞) Option 4: [-6, ∞)

Answer: [6, ∞). Solution: f'(x) = 2x - a ≤ 0 on (2,3) requires a ≥ 2x for all x (2,3), so a ≥ 6. Thus a [6, ∞).

Calculus

Application of Derivatives

Medium

If the function $f(x) = x^2 - ax - 2$ is strictly decreasing on $(2, 3)$ then $a$ lies in the interval.

$[4, \infty)$

$[6, \infty)$

$[-4, \infty)$

$[-6, \infty)$

✅ Correct Option: 2

$f'(x) = 2x - a \leq 0$ on $(2,3)$ requires $a \geq 2x$ for all $x \in (2,3)$ , so $a \geq 6$ . Thus $a \in [6, \infty)$ .

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(A) $f'(x) = 3x^2 - 12x + 15$ .

(B) The critical points are 3 and 1.

(D) Local minimum value is 19.

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Match the LIST-I with LIST-II

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