Question ID: #406
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:
- (1) 280
- (2) 360
- (3) 420
- (4) 160
Solution:
$$ f'(x) = \frac{2}{x-2} – 2x + a $$
For $f(x)$ to be increasing in $(2, 3)$, $f'(x) \ge 0$ on this interval.
At the boundary $x=3$:
$$ f'(3) \ge 0 \Rightarrow \frac{2}{3-2} – 6 + a \ge 0 \Rightarrow 2 – 6 + a \ge 0 \Rightarrow a \ge 4 $$
Taking $a=4$:
$$ f'(x) = \frac{2}{x-2} – 2x + 4 = \frac{-2(x-1)(x-3)}{x-2} $$
Sign scheme gives increasing in $(2, 3)$. Thus, $a=4$.
Substitute $a=4$ into $g(x)$:
$$ g(x) = (x-1)^3(x+2-4)^2 = (x-1)^3(x-2)^2 $$
$$ g'(x) = 3(x-1)^2(x-2)^2 + (x-1)^3 \cdot 2(x-2) $$
$$ g'(x) = (x-1)^2(x-2) [3(x-2) + 2(x-1)] $$
$$ g'(x) = (x-1)^2(x-2) (5x-8) $$
For decreasing, $g'(x) < 0$: $$ (x-1)^2(x-2)(5x-8) < 0 $$ Critical points are $1, 1.6 (\text{i.e., } 8/5), 2$. Since $(x-1)^2 > 0$, we check $(x-2)(5x-8) < 0$. $$ \frac{8}{5} < x < 2 \Rightarrow x \in (1.6, 2) $$ Thus, $b = 1.6, c = 2$.
$$ 100(a+b-c) = 100(4 + 1.6 – 2) $$
$$ = 100(3.6) = 360 $$
Ans. (2)
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