The diagram shows the graphs of $y = |a - bx|$ and $y = |mx - 6|$ where $a$, $b$ and $m$ are positive constants. The graph of $y = |a - bx|$ meets the $x$-axis at the point $\left(\frac{9}{4}, 0\right)$. The solution set of the inequality $|mx - 6| < |a - bx|$ is given as $\{x : x \in \mathbb{R}, x < 1 \text{ or } x > 3\}$. Find the values $a$, $b$ and $m$. Given that $k$ is a constant and $k > 1$, solve the inequality $\frac{kx^2 + 1}{x^2 + (k+1)x + k} > -1$ giving your answer in terms of $k$.

1. Express $a$ in terms of $b$ using the $x$-intercept of $y = |a - bx|$.

Given that the graph of $y = |a - bx|$ meets the $x$-axis at $\left(\frac{9}{4}, 0\right)$:
$\left| a - b\left(\frac{9}{4}\right) \right| = 0$
$a = \frac{9}{4}b = 2.25b$

2. Set up modulus equations using the boundary values of the inequality's solution.

The solution set for the inequality $|mx - 6| < |a - bx|$ is bounded by $x = 1$ and $x = 3$.
This means the graphs intersect at these exact $x$-coordinates.
At $x = 1$:
$|m(1) - 6| = |a - b(1)| \implies |m - 6| = |a - b|$
At $x = 3$:
$|m(3) - 6| = |a - b(3)| \implies |3m - 6| = |a - 3b|$

3. Remove the absolute values by determining the signs of the inner expressions.

Since $a = 2.25b$ and $b > 0$:
$a - b = 1.25b > 0 \implies |a - b| = 1.25b$
$a - 3b = -0.75b < 0 \implies |a - 3b| = 0.75b$

From the graph description, the intersection at $x=3$ is to the left of the vertex of $y = |mx - 6|$.
The vertex is at $x = \frac{6}{m}$, so $\frac{6}{m} > 3 \implies m < 2$.
Thus, $m - 6 < 0$ and $3m - 6 < 0$, giving:
$|m - 6| = 6 - m$
$|3m - 6| = 6 - 3m$

4. Solve the resulting linear equations for $a$, $b$, and $m$.

We now have a system of linear equations:
$6 - m = 1.25b \quad \text{--- (1)}$
$6 - 3m = 0.75b \quad \text{--- (2)}$
From (1), $m = 6 - 1.25b$. Substituting this into (2):
$6 - 3(6 - 1.25b) = 0.75b$
$-12 + 3.75b = 0.75b$
$3b = 12 \implies b = 4$

Substitute $b = 4$ into the expression for $m$:
$m = 6 - 1.25(4) = 1$
Substitute $b = 4$ into the expression for $a$:
$a = 2.25(4) = 9$

1. Rearrange the inequality to compare with zero and express it as a single fraction.

$\frac{kx^2 + 1}{x^2 + (k+1)x + k} + 1 > 0$
$\frac{kx^2 + 1 + x^2 + (k+1)x + k}{x^2 + (k+1)x + k} > 0$
$\frac{(k+1)x^2 + (k+1)x + (k+1)}{x^2 + (k+1)x + k} > 0$

2. Factorise the numerator and the denominator.

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

3. Analyse the sign of the numerator to simplify the inequality.

Since $k > 1$, we have $k + 1 > 0$.

For the quadratic factor $x^2 + x + 1$, the discriminant is:
$\Delta = 1^2 - 4(1)(1) = -3 < 0$
Since the coefficient of $x^2$ is positive ($1 > 0$) and $\Delta < 0$, $x^2 + x + 1 > 0$ for all real $x$.

Therefore, the entire numerator $(k+1)(x^2 + x + 1)$ is strictly positive. For the fraction to be positive, the denominator must also be positive:
$(x + k)(x + 1) > 0$

4. Solve the resulting quadratic inequality for the denominator.

The critical values for the denominator are $x = -k$ and $x = -1$.
Since $k > 1$, it follows that $-k < -1$.

The quadratic inequality $(x + k)(x + 1) > 0$ therefore represents the regions outside the roots:
$x < -k \text{ or } x > -1$