Question

For $t \in (0, \frac{1}{2})$, consider the quadratic equation

$p_t(x) = (t-t^2)x^2 - (1-t)x + t^2 = 0$

Let $\alpha_t$ and $\beta_t$ denote the roots of $p_t$. Then which of the following statements is FALSE?

• A. $p_t(x)$ has distinct roots for all $t \in (0, \frac{1}{2})$
• B. For infinitely many values of t, the corresponding root $\alpha_t$ is rational and for infinitely many values of t, $\alpha_t$ is irrational
• C. $\alpha_t$ and $\beta_t$ are real and positive, with both $\alpha_t$ and $\beta_t$ less than 1
• D. If $t \rightarrow (\frac{1}{2})^{-}$, then $|\alpha_t - \beta_t| \rightarrow 0$

Answer 1

Answer: (C)

$\alpha_t$ and $\beta_t$ are real and positive, with both $\alpha_t$ and $\beta_t$ less than 1

Answer 2

The given quadratic equation is $p_t(x) = (t-t^2)x^2 - (1-t)x + t^2 = 0$ for $t \in (0, \frac{1}{2})$.

The coefficients are $a = t(1-t)$, $b = -(1-t)$, $c = t^2$.

For $t \in (0, \frac{1}{2})$, $t > 0$ and $1-t > \frac{1}{2} > 0$, so $a = t(1-t) > 0$.

Statement (A): $p_t(x)$ has distinct roots for all $t \in (0, \frac{1}{2})$.

The discriminant is $D = b^2 - 4ac = (-(1-t))^2 - 4(t(1-t))(t^2) = (1-t)^2 - 4t^3(1-t)$.

Since $t \in (0, \frac{1}{2})$, $1-t \neq 0$. We can factor out $(1-t)$:

$D = (1-t)(1-t - 4t^3)$.

For $t \in (0, \frac{1}{2})$, $1-t > \frac{1}{2} > 0$.

Let $f(t) = 1-t-4t^3$. $f'(t) = -1-12t^2$. For $t \in (0, \frac{1}{2})$, $f'(t) < 0$, so $f(t)$ is strictly decreasing.

$f(0) = 1$, $f(\frac{1}{2}) = 1 - \frac{1}{2} - 4(\frac{1}{8}) = 1 - \frac{1}{2} - \frac{1}{2} = 0$.

Since $f(t)$ is strictly decreasing on $(0, \frac{1}{2})$ and $f(\frac{1}{2}) = 0$, we have $f(t) > 0$ for all $t \in (0, \frac{1}{2})$.

Thus, $D = (1-t)f(t) > 0$ for all $t \in (0, \frac{1}{2})$. The roots are distinct and real.

Statement (A) is TRUE.

Statement (C): $\alpha_t$ and $\beta_t$ are real and positive, with both $\alpha_t$ and $\beta_t$ less than 1.

From (A), the roots are real.

The sum of roots is $\alpha_t + \beta_t = \frac{-b}{a} = \frac{1-t}{t(1-t)} = \frac{1}{t}$.

The product of roots is $\alpha_t \beta_t = \frac{c}{a} = \frac{t^2}{t(1-t)} = \frac{t}{1-t}$.

For $t \in (0, \frac{1}{2})$, $t > 0$ and $1-t > 0$.

Sum of roots $\frac{1}{t}$. Since $t \in (0, \frac{1}{2})$, $\frac{1}{t} \in (2, \infty)$. So $\alpha_t + \beta_t > 2$.

Product of roots $\frac{t}{1-t}$. Since $t \in (0, \frac{1}{2})$, $1-t \in (\frac{1}{2}, 1)$.

As $t$ increases from $0$ to $\frac{1}{2}$, $\frac{t}{1-t}$ increases.

At $t \rightarrow 0^+$, $\frac{t}{1-t} \rightarrow 0$.

At $t \rightarrow (\frac{1}{2})^-$, $\frac{t}{1-t} \rightarrow \frac{1/2}{1/2} = 1$.

So $\alpha_t \beta_t \in (0, 1)$.

Since the sum and product of roots are positive, both roots must be positive.

We need to check if both roots are less than 1.

Consider the value of $p_t(1) = (t-t^2)(1)^2 - (1-t)(1) + t^2 = t - t^2 - 1 + t + t^2 = 2t - 1$.

For $t \in (0, \frac{1}{2})$, $2t \in (0, 1)$, so $2t-1 \in (-1, 0)$.

Thus $p_t(1) = 2t-1 < 0$ for all $t \in (0, \frac{1}{2})$.

Since the leading coefficient $a = t(1-t) > 0$ and $p_t(1) < 0$, the roots $\alpha_t$ and $\beta_t$ must lie on opposite sides of 1.

Let $\alpha_t < \beta_t$. Then $\alpha_t < 1$ and $\beta_t > 1$.

This contradicts the statement that both $\alpha_t$ and $\beta_t$ are less than 1.

Statement (C) is FALSE.

Statement (D): If $t \rightarrow (\frac{1}{2})^{-}$, then $|\alpha_t - \beta_t| \rightarrow 0$

The difference of roots is given by:

$|\alpha_t - \beta_t| = \frac{\sqrt{D}}{|a|} = \frac{\sqrt{(1-t)(1-t-4t^3)}}{t(1-t)} = \frac{\sqrt{1-t-4t^3}}{t\sqrt{1-t}}$.

As $t \rightarrow (\frac{1}{2})^{-}$, we have:

$1 - t - 4t^3 \rightarrow 1 - \frac{1}{2} - 4(\frac{1}{8}) = 0$

$1 - t \rightarrow \frac{1}{2}$

Thus, $|\alpha_t - \beta_t| \rightarrow \frac{\sqrt{0}}{\frac{1}{2}\sqrt{\frac{1}{2}}} = 0$.

Statement (D) is TRUE.

Since (A) and (D) are TRUE, and (C) is FALSE, the answer is (C).