If the equation ${x}^{3}-9{x}^{2}+24{x}+{k}=0$ has exactly one root in $(2,4)$, then $k$ lies in the interval
- $(- 20, -16)$
- $($16, $\infty)$
- $(-\infty, -20)$
-
None of these
For f(x) = x^3 - 9x^2 + 24x + k, the derivative is 3(x - 2)(x - 4), so f decreases on (2,4). Its endpoint values are f(2) = k + 20 and f(4) = k + 16, which must have opposite signs for one interior root. Thus -20 < k < -16.
Let f(x) = x^3 - 9x^2 + 24x + k, which has critical points where f'(x) = 3x^2 - 18x + 48 = 0, giving x = 4 and x = 4 is a double root, meaning the function is strictly increasing everywhere except at x = 4. Testing values, f(2) = 8 - 36 + 48 + k = 20 + k and f(4) = 64 - 144 + 96 + k = 16 + k. The solution provided claims k lies in the interval (-20, -16), but if we take k = -18 from this interval, f(2) = 2 and f(4) = -2, meaning f(x) goes from positive to negative across the interval (2, 4) and has exactly one root there. Wait, let us check the right boundary of the interval: f(4) = 16 + k, which for k = -16 is 0, making x=4 a root, but for k in (-20, -16), f(2) and f(4) are both positive if k > -16. The actual condition for exactly one root in (2, 4) requires f(2)f(4) < 0, leading to (20 + k)(16 + k) < 0, which means k must be in (-20, -16). Thus, k lies in the interval (-20, -16).