Question

Let $(2\alpha,\alpha)$ be the largest interval in which the function \[ f(t)=\frac{|t+1|}{t^2},\; t<0 \] is strictly decreasing. Then the local maximum value of the function \[ g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha,\; x>2 \] is
 

Hint:

First determine parameters using monotonicity, then substitute into the second function.

Answer 1

Correct Answer: 2

Step 1: Simplifying $f(t)$.
For $t<0$, 
|t+1|=-(t+1), & t<-1  t+1, & -1
Step 2: Checking monotonicity. 
Differentiate in both intervals. 
For $t<-1$: \[ f'(t)=\frac{t+2}{t^3} \Rightarrow f'(t)<0 \text{ for } (-2,-1) \] For -10  Thus, the largest decreasing interval is \[ (2\alpha,\alpha)=(-2,-1) \Rightarrow \alpha=-1 \] 
Step 3: Maximizing $g(x)$. 
Substitute $\alpha=-1$: \[ g(x)=2\log(x-2)-x^2+4x+1 \] Differentiate: \[ g'(x)=\frac{2}{x-2}-2x+4 \] Set $g'(x)=0$: \[ \frac{2}{x-2}=2x-4 \Rightarrow x=3 \] Step 4: Local maximum value. 
\[ g(3)=2\log 1-9+12+1=2 \]