Question

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?

• A. \(F\) is strictly decreasing
• B. \(F\) is strictly increasing
• C. \(\lim_{x\to1}F(x)\) exists
• D. \(F'\left(\frac{3}{4}\right)=1\)

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept.
The function \(F(x)=\int_{1/2}^{x}\frac{1}{p(t)}\,dt\) is built from \(p(t)=t(t-1)(t-2)\). By the Fundamental Theorem of Calculus, \(F'(x)=\frac{1}{p(x)}\) wherever \(p(x)\neq0\). The sign of \(F'(x)\) tells us whether \(F\) increases or decreases, and the size of \(p(t)\) near \(t=1\) tells us whether the integral stays finite as \(x\to1\).

Step 2: Check the sign of \(p(t)\) on \(\left[\frac{1}{2},1\right)\).
For \(t\in\left(\frac{1}{2},1\right)\): \(t>0\), \(t-1<0\), \(t-2<0\). So \(p(t)=t(t-1)(t-2)\) is (positive)(negative)(negative), which is positive. So \(p(t)>0\) throughout \(\left(\frac{1}{2},1\right)\).

Step 3: Determine monotonicity of \(F\).
Since \(p(t)>0\) on \(\left(\frac{1}{2},1\right)\), we get \(F'(x)=\frac{1}{p(x)}>0\) there. A positive derivative on the whole interval means \(F\) is strictly increasing. This checks option (B).

Step 4: Check option (A).
\(F\) cannot be strictly decreasing, since we just showed \(F'(x)>0\). So (A) is FALSE.

Step 5: Check option (D).
\[ p\left(\frac{3}{4}\right)=\frac{3}{4}\left(\frac{3}{4}-1\right)\left(\frac{3}{4}-2\right)=\frac{3}{4}\left(-\frac{1}{4}\right)\left(-\frac{5}{4}\right)=\frac{15}{64} \] So \[ F'\left(\frac{3}{4}\right)=\frac{1}{p(3/4)}=\frac{64}{15}\neq1. \] So (D) is FALSE.

Step 6: Check option (C).
As \(t\to1^{-}\), \(p(t)\to0\). Near \(t=1\), \(p(t)=t(t-1)(t-2)\approx(1)(t-1)(-1)=(1-t)\), a simple zero. So \(\frac{1}{p(t)}\) behaves like \(\frac{1}{1-t}\) near \(t=1\), and \[ \int^{x}\frac{dt}{1-t}=-\ln(1-x)\to\infty \] as \(x\to1^{-}\). So \(F(x)\to\infty\) and the limit does NOT exist (it diverges). So (C) is FALSE.

Final Answer:
\(F\) is strictly increasing on \(\left[\frac{1}{2},1\right)\). \[ \boxed{\text{F is strictly increasing}} \]