Consider the polynomial \[ p(t)=t(t-1)(t-2) \] and define \[ F(x)=\int_{1/2}^{x}\frac{1}{p(t)}\,dt,\qquad x\in\left[\frac{1}{2},1\right). \] Which of the following statements is correct?
Show Hint
Check the sign of p(t) on the interval first: this fixes the sign of F'(x)=1/p(t), which tells you if F increases or decreases.
Step 1: What FTC gives us.
Since $F(x)=\int_{1/2}^{x}\frac{1}{p(t)}\,dt$ with $p(t)=t(t-1)(t-2)$, the Fundamental Theorem of Calculus tells us $F'(x)=\frac{1}{p(x)}$ at every point where $p(x)\neq0$. So we just need to track the sign and size of $p$ on the interval $[1/2,1)$.
Step 2: Sign check using a test point.
Take a test point $t=0.75$ inside $(1/2,1)$: $t=0.75>0$, $t-1=-0.25<0$, $t-2=-1.25<0$. Multiplying a positive number by two negative numbers gives a positive result, so $p(0.75)>0$. Since $p$ has no root inside the open interval $(1/2,1)$ (its roots are exactly $0,1,2$), the sign of $p$ cannot flip inside this interval, so $p(t)>0$ for every $t$ in $(1/2,1)$.
Step 3: Read off monotonicity.
Because $p(t)>0$ throughout $(1/2,1)$, we get $F'(x)=1/p(x)>0$ there too. A function with a positive derivative on an interval is strictly increasing on it, so option (B) holds and option (A) is ruled out.
Step 4: Test the numeric claim in option (D).
At $x=3/4$: \[ p(3/4)=\frac{3}{4}\cdot\left(-\frac{1}{4}\right)\cdot\left(-\frac{5}{4}\right)=\frac{15}{64} \] so $F'(3/4)=64/15$, about $4.27$, not $1$. Option (D) fails.
Step 5: Test whether the limit at $x=1$ exists.
Write $p(t)=t(t-1)(t-2)$ and let $t\to1^{-}$. Both $t$ and $(t-2)$ tend to the finite nonzero values $1$ and $-1$, while $(t-1)\to0^{-}$. So near $t=1$, $p(t)$ behaves like a constant times $(t-1)$, a simple zero. This makes $1/p(t)$ blow up like $1/(1-t)$, and integrating $1/(1-t)$ up to $x\to1$ gives a log term that diverges to infinity. So $F(x)\to\infty$, meaning the limit in option (C) does not exist as a finite number.
Step 6: Conclude.
Only the monotonicity claim survives all checks.
\[ \boxed{\text{F is strictly increasing}} \]
