IAL 2023 June Q3 | 國際數學站

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Given that

\begin{align*} \frac{x + 2}{x + 4} \leqslant\,& \frac{x}{k(x - 1)}\\[2mm] \end{align*}

where $k$ is a positive constant,

(a) show that

\begin{align*} (x + 4)(x - 1)(px^2 + qx + r) \leqslant\,& 0\\[2mm] \end{align*}

where $p$ , $q$ and $r$ are expressions in terms of $k$ to be determined.

(3)

(b) Hence, or otherwise, determine the values for $x$ for which

\begin{align*} \frac{x + 2}{x + 4} \leqslant\,& \frac{x}{3(x - 1)}\\[2mm] \end{align*}

(4)

解答

(a)

Given the inequality:

\begin{align*} \frac{x+2}{x+4} \le &\,\frac{x}{k(x-1)} \end{align*}

Subtract the right-hand side to bring all terms to one side:

\begin{align*} \frac{x+2}{x+4} - \frac{x}{k(x-1)} \le &\,0\\[4mm] \frac{k(x+2)(x-1) - x(x+4)}{k(x+4)(x-1)} \le &\,0 \end{align*}

Since $k$ is a positive constant, multiplying by $k^2(x+4)^2(x-1)^2$ (which is always strictly positive for $x \neq -4, 1$ ) preserves the inequality. In practice, we multiply by $k(x+4)^2(x-1)^2$ :

\begin{align*} (x+4)(x-1) \big[ k(x+2)(x-1) - x(x+4) \big] \le &\,0\\[4mm] (x+4)(x-1) \big[ k(x^2 + x - 2) - (x^2 + 4x) \big] \le &\,0\\[4mm] (x+4)(x-1) \big[ kx^2 + kx - 2k - x^2 - 4x \big] \le &\,0\\[4mm] (x+4)(x-1) \big[ (k-1)x^2 + (k-4)x - 2k \big] \le &\,0 \end{align*}

This matches the required form $(x+4)(x-1)(px^2+qx+r) \le 0$ , where $p = k-1$ , $q = k-4$ , and $r = -2k$ .

(b)

Substitute $k = 3$ into the quadratic expression derived in part (a):

\begin{align*} (3-1)x^2 + (3-4)x - 2(3) = &\,2x^2 - x - 6 \end{align*}

So the inequality becomes:

\begin{align*} (x+4)(x-1)(2x^2 - x - 6) \le &\,0 \end{align*}

Factorise the quadratic expression:

\begin{align*} 2x^2 - x - 6 = &\,(2x+3)(x-2) \end{align*}

The critical values are $x = -4$ , $x = 1$ , $x = -\frac{3}{2}$ , and $x = 2$ . Arrange them in ascending order: $-4$ , $-\frac{3}{2}$ , $1$ , $2$ .

We test the sign of $(x+4)(x-1)(2x+3)(x-2)$ in the intervals defined by these critical values:

For $x < -4$ , the sign is $(-)(-)(-)(-) \implies (+)$
For $-4 < x < -\frac{3}{2}$ , the sign is $(+)(-)(-)(-) \implies (-)$
For $-\frac{3}{2} < x < 1$ , the sign is $(+)(-)(+)(-) \implies (+)$
For $1 < x < 2$ , the sign is $(+)(+)(+)(-) \implies (-)$
For $x > 2$ , the sign is $(+)(+)(+)(+) \implies (+)$

We require the expression to be $\le 0$ . The original denominators cannot be zero, so $x \neq -4$ and $x \neq 1$ . The values of $x$ satisfying the inequality are:

\begin{align*} -4 < x \le -\frac{3}{2} \quad \text{or} \quad 1 < x \le 2 \end{align*}