Show that $2x^{2}-2x+1$ is always positive for all real $x$. Hence, without using a calculator, solve $\displaystyle \frac{2x^{3}-2x^{2}+x}{(2x+1)(x-1)^{2}} \ge 0$. Using your result in part (ii), find the exact range of values of $x$ for which $\frac{2(\ln x)^{3}-2(\ln x)^{2}+\ln x}{(2\ln x+1)(\ln x-1)^{2}} \ge 0.$

1. Express the quadratic expression in completed square form.

$\begin{aligned} 2x^{2}-2x+1 &= 2\left(x^{2}-x\right)+1 \\ &= 2\left[\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\right]+1 \\ &= 2\left(x-\frac{1}{2}\right)^{2}-\frac{1}{2}+1 \\ &= 2\left(x-\frac{1}{2}\right)^{2}+\frac{1}{2} \end{aligned}$

2. Use the property of squares to conclude the expression is always positive.

For all real $x$, $\left(x-\frac{1}{2}\right)^{2} \ge 0$.

Thus, $2\left(x-\frac{1}{2}\right)^{2} + \frac{1}{2} \ge \frac{1}{2} > 0$.

Therefore, $2x^{2}-2x+1$ is always positive for all real $x$.

1. Factorise the numerator and simplify using the result from part (i).

$\frac{x(2x^2-2x+1)}{(2x+1)(x-1)^2} \ge 0$
From part (i), $2x^2 - 2x + 1 > 0$ for all real $x$. Since this quadratic factor is strictly positive, we can divide both sides by it without changing the direction of the inequality:
$\frac{x}{(2x+1)(x-1)^2} \ge 0$

2. Identify conditions for the inequality to be defined and further simplify.

For the rational expression to be defined, the denominator cannot be zero. This gives the restrictions $x \neq -\frac{1}{2}$ and $x \neq 1$.
Since $(x-1)^2 > 0$ for all $x \neq 1$, we can multiply both sides by $(x-1)^2$ (or reason that its sign doesn't change the inequality) to simplify the problem to:
$\frac{x}{2x+1} \ge 0$

3. Solve the simplified inequality and state the final solution incorporating the restrictions.

The critical values of $\frac{x}{2x+1}$ are $x = -\frac{1}{2}$ and $x = 0$.
Testing the intervals around these critical values:

• For $x < -\frac{1}{2}$, $\frac{x}{2x+1} > 0$
• For $-\frac{1}{2} < x < 0$, $\frac{x}{2x+1} < 0$
• For $x > 0$, $\frac{x}{2x+1} > 0$

Thus, the basic solution to $\frac{x}{2x+1} \ge 0$ is $x < -\frac{1}{2}$ or $x \ge 0$.
Notice that $x \neq -\frac{1}{2}$ is naturally maintained by the strict inequality.
However, we must also apply the restriction $x \neq 1$. This splits the region $x \ge 0$ into two parts:
$x < -\frac{1}{2} \text{ or } 0 \le x < 1 \text{ or } x > 1$

1. Solve the rational inequality from part (ii) to find the base intervals.

From part (ii), we need to solve $\displaystyle \frac{x(2x^2-2x+1)}{(2x+1)(x-1)^2} \ge 0$.

Since $2x^2-2x+1 > 0$ for all real $x$ (as shown in part (i)) and $(x-1)^2 > 0$ for $x \neq 1$, the sign of the expression depends entirely on $\frac{x}{2x+1}$. We therefore require:
$\frac{x}{2x+1} \ge 0 \quad \text{and} \quad x \neq 1$

Solving this using a number line or sign test yields the solution to part (ii):
$x < -\frac{1}{2} \quad \text{or} \quad 0 \le x < 1 \quad \text{or} \quad x > 1$

2. Apply the substitution $x \to \ln x$ to the intervals found in part (ii).

The expression in part (iii) is structurally identical to the one in part (ii) with the variable $x$ replaced by $\ln x$. By using the result directly from part (ii), we substitute $\ln x$ into the established inequality ranges:
$\ln x < -\frac{1}{2} \quad \text{or} \quad 0 \le \ln x < 1 \quad \text{or} \quad \ln x > 1$

3. Solve for $x$, ensuring the domain of the natural logarithm is respected.

Recall that $\ln x$ is only defined for $x > 0$.

For $\ln x < -\frac{1}{2}$:
$0 < x < e^{-1/2}$

For $0 \le \ln x < 1$:
$e^0 \le x < e^1 \implies 1 \le x < e$

For $\ln x > 1$:
$x > e^1 \implies x > e$

Combining these, the exact range of values for $x$ is:
$0 < x < e^{-1/2} \quad \text{or} \quad 1 \le x < e \quad \text{or} \quad x > e$