Applications of Derivatives
Application of Derivatives
nta_pyq_2025_apr
Grade 12
Question:
Let $(2, 3)$ be the largest open interval in which the function $f(x) = 2\log_e(x-2) - x^2 + ax + 1$ is strictly increasing and $(b, c)$ be the largest open interval, in which the function $g(x) = (x-1)^3(x+2-a)^2$ is strictly decreasing. Then $100(a+b-c)$ is equal to:
Step-by-Step Solution
Key Concept: For $f$: compute $f'(x) = \tfrac{2}{x-2} - 2x + a$; since $f''(x) < 0$, $f'$ is decreasing, so the right endpoint of the monotone interval is where $f'(x) = 0$. Use $f'(3) \geq 0$ to get $a_{\min} = 4$. For $g$ with $a=4$: factor $g'(x)$ and apply the wavy-curve method.
$f'(x) = \dfrac{2}{x-2} - 2x + a$. Since $f''(x) = \dfrac{-2}{(x-2)^2} - 2 < 0$, $f'$ is strictly decreasing. For the interval to be exactly $(2,3)$, we need $f'(3) = 0$:
$$\frac{2}{1} - 6 + a = 0 \Rightarrow a = 4.$$
With $a = 4$: $g(x) = (x-1)^3(x-2)^2$.
$$g'(x) = (x-1)^2(x-2)[2(x-1) + 3(x-2)] = (x-1)^2(x-2)(5x-8).$$
$g'(x) < 0$ when $(x-2)(5x-8) < 0$, i.e., $x \in \left(\tfrac{8}{5}, 2\right)$. So $b = \tfrac{8}{5}$, $c = 2$.
$$100(a+b-c) = 100\!\left(4 + \frac{8}{5} - 2\right) = 100 \times \frac{18}{5} = 360.$$
Correct Answer: 2