2019/Promo/RI/P1/Q8

1. Rearrange the inequality to compare with zero and simplify into a single fraction.

$\frac{2x^2 - x}{x^2 + 3x - 4} - 1 > 0$
$\frac{2x^2 - x - (x^2 + 3x - 4)}{x^2 + 3x - 4} > 0$
$\frac{x^2 - 4x + 4}{x^2 + 3x - 4} > 0$

2. Factorise the quadratic expressions in the numerator and denominator.

$\frac{(x - 2)^2}{(x + 4)(x - 1)} > 0$

3. Analyse the signs of the factors to establish conditions for the inequality to hold.

For the rational expression to be strictly positive, since the numerator $(x - 2)^2 \ge 0$ for all real $x$, we must ensure it is not zero:
$(x - 2)^2 \neq 0 \implies x \neq 2$
Consequently, the denominator must be strictly positive:
$(x + 4)(x - 1) > 0$

4. Solve the resulting quadratic inequality and combine with the restriction to state the final solution.

Solving $(x + 4)(x - 1) > 0$ yields:
$x < -4 \text{ or } x > 1$
Combining this with the restriction $x \neq 2$, the final solution is:
$x < -4 \text{ or } 1 < x < 2 \text{ or } x > 2$

1. Determine the key features of the hyperbola $y = 2 + \frac{a}{x}$.

The graph of $y = 2 + \frac{a}{x}$ has a vertical asymptote at $x = 0$ and a horizontal asymptote at $y = 2$.
Since $a > 0$, the branches occupy the top-right and bottom-left regions relative to the intersection of the asymptotes $(0,2)$.
Find the $x$-intercept by setting $y = 0$:
$2 + \frac{a}{x} = 0 \implies x = -\frac{a}{2}$

2. Determine the key features of the modulus graph $y = 2 - |x|$.

The graph of $y = 2 - |x|$ is an inverted V-shape.
Its maximum point (vertex) is at $(0, 2)$.
Find the $x$-intercepts by setting $y = 0$:
$2 - |x| = 0 \implies |x| = 2 \implies x = -2 \text{ or } 2$

3. Determine the points of intersection to ensure correct relative positions on the sketch.

Equate the two functions to find any points of intersection:
$2 + \frac{a}{x} = 2 - |x| \implies \frac{a}{x} = -|x|$
Since $a > 0$, the left side has the same sign as $x$. The right side is always negative or zero. Thus, intersection can only occur when $x < 0$.
For $x < 0$, $|x| = -x$:
$\frac{a}{x} = x \implies x^2 = a \implies x = -\sqrt{a}$
Since $1 < a < 2$, we have $1 < \sqrt{a} < \sqrt{2}$. Thus, the intersection point $x = -\sqrt{a}$ lies between $x = -1.414$ and $x = -1$.
Note that $-\sqrt{a} < -\frac{a}{2}$, so the intersection happens to the left of the hyperbola's $x$-intercept.
The two graphs are then sketched on the same axes incorporating these relative positions.

1. Simplify the inequality for $x < 0$ and find the critical boundary value.

From the graph sketched in (b)(i), there are no intersections for $x > 0$ since $y = 2 + \dfrac{a}{x} > 2$ and $y = 2 - |x| < 2$.

For $x < 0$, $|x| = -x$. The inequality becomes:
$2 + \dfrac{a}{x} < 2 - (-x)$
$\dfrac{a}{x} < x$
The intersection point (boundary) occurs where:
$\dfrac{a}{x} = x$
$x^2 = a$
Since $x < 0$, $x = -\sqrt{a}$.

2. Solve the rational inequality taking into account the sign of $x$.

Multiplying both sides of the inequality $\dfrac{a}{x} < x$ by $x$ (which is strictly negative, $x < 0$) reverses the inequality sign:
$a > x^2$
$x^2 < a$
$-\sqrt{a} < x < \sqrt{a}$
Since we are only considering the region where $x < 0$, we take the intersection of the conditions:
$-\sqrt{a} < x < 0$