Given
\[
f(t)=\left|\frac{t+1}{t^2}\right|,\ (t<0)
\]
is strictly decreasing in the interval $(2\alpha,\alpha)$, then the maximum value of
\[
g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha
\]
is:
Show Hint
Correct Answer: 4
Solution and Explanation
Step 1: Remove the modulus in $f(t)$.
For $t<0$, we have: \[ f(t)=\frac{-(t+1)}{t^2} \] Step 2: Differentiate $f(t)$.
\[ f'(t)=\frac{t+2}{t^3} \] Step 3: Use the decreasing condition.
For $f(t)$ to be strictly decreasing: \[ f'(t)<0 \Rightarrow \frac{t+2}{t^3}<0 \] Since $t<0$, this gives: \[ -2Step 4: Compare intervals.
Given interval $(2\alpha,\alpha)=(-2,0)$: \[ \Rightarrow \alpha=-1 \] Step 5: Substitute $\alpha$ in $g(x)$.
\[ g(x)=2\log(x-2)-x^2+4x+1 \] Step 6: Differentiate $g(x)$.
\[ g'(x)=\frac{2}{x-2}-2x+4 \] Setting $g'(x)=0$: \[ \frac{2}{x-2}=2x-4 \Rightarrow x=3 \] Step 7: Find maximum value.
\[ g(3)=2\log 1-9+12+1=4 \]
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