Quadratic and two-variable inequalities

(a) $-3 < x < 3$

1. $x^2 - 9 = (x + 3)(x - 3)$ is zero at $x = \pm 3$.
2. The graph of $y = x^2 - 9$ is below the $x$-axis between the roots, so $-3 < x < 3$.

• M1 Finding the critical values $\pm 3$.
• A1 The correct answer, $-3 < x < 3$.

(b) $x \le -4$ or $x \ge 4$

1. $x^2 = 16$ at $x = \pm 4$. $x^2$ is at least 16 outside the roots: $x \le -4$ or $x \ge 4$.

• B1 $x \le -4$ or $x \ge 4$.

(a) 2 < x < 5

1. Factorise: $x^{2}$ $-$ 7x + 10 = (x $-$ 2)(x $-$ 5).
2. The upward-opening quadratic is negative between its roots, with the roots excluded because the inequality is strict.

• B1 Correct answer: 2 < x < 5

(a) $-\frac{1}{2} \le x \le 3$

1. Rearrange and factorise.
   $$2x^2 - 5x - 3 \le 0 \Rightarrow (2x + 1)(x - 3) \le 0$$
2. Critical values $-\frac{1}{2}$ and $3$; the solution is between them: $-\frac{1}{2} \le x \le 3$.

• M1 Rearranging to $2x^2 - 5x - 3 \le 0$.
• M1 Factorising to $(2x + 1)(x - 3)$ and finding the critical values.
• A1 $-\frac{1}{2} \le x \le 3$.

(b) $0, 1, 2, 3$

1. Integers from $-\frac{1}{2}$ to 3 inclusive: $0, 1, 2, 3$.

• B1 The correct answer, $0, 1, 2, 3$.

(a) $-15$

1. $$x^{2}+3x-24$$
2. $$10\times (-6+3)/2$$
3. Expand and rearrange: $x^{2}$ + 3x $-$ 24 $\le$ 0.
4. The boundary roots are ($-3$ ± $\sqrt{105}$)/2, approximately $-6.623$ and 3.623. The upward-opening quadratic is non-positive between them.
5. The integer solutions are $-6$, $-5$, $-4$, $-3$, $-2$, $-1$, 0, 1, 2 and 3.
6. Their sum is 10($-6$ + 3)/2 = $-15.$

• P1 Establishing $x^{2}+3x-24$ or an equivalent valid method.
• P1 Establishing $10\times (-6+3)/2$ or an equivalent valid method.
• A1 Correct answer: $-15$

(a) $-2<x<5$

1. The boundary roots are −2 and 5; the upward-opening quadratic is negative between them.
   $$(x+2)(x-5)=0$$
2. Therefore $-2<x<5$.

• M1 The boundary roots are −2 and 5; the upward-opening quadratic is negative between them.
• A1 Correct answer: $-2<x<5$

(b) $6$

1. Both endpoints are excluded.
   $$5-(-2)-1$$
2. Therefore $6$.

• M1 Both endpoints are excluded.
• A1 Correct answer: $6$

(a) The largest feasible y for each x gives [(1, 6), (2, 5), (3, 5), (4, 4), (5, 2)]. The maximum is 37, attained at [(3, 5)].

1. Use y at least 1 to bound the positive integer x values.
   $$3x+1\le 17$$
2. For each x choose the smaller integer upper bound on y, then compare 4x+5y.
3. The largest feasible y for each x gives [(1, 6), (2, 5), (3, 5), (4, 4), (5, 2)]. The maximum is 37, attained at [(3, 5)].

• P1 Use y at least 1 to bound the positive integer x values.
• P1 For each x choose the smaller integer upper bound on y, then compare 4x+5y.
• A1 Correct answer: The largest feasible y for each x gives [(1, 6), (2, 5), (3, 5), (4, 4), (5, 2)]. The maximum is 37, attained at [(3, 5)].

(b) The coefficient of y in 4x+5y is positive, so increasing y increases the objective for that fixed x.

1. The coefficient of y in 4x+5y is positive, so increasing y increases the objective for that fixed x.

• C1 Correct conclusion with supporting reasoning: The coefficient of y in 4x+5y is positive, so increasing y increases the objective for that fixed x.

(a) $(5, 1)$

1. Intersect the boundary lines in pairs.
   $$y = 1 \text{ and } x + y = 6: (5, 1); \quad y = 2x \text{ and } x + y = 6: (2, 4); \quad y = 1 \text{ and } y = 2x: (0.5, 1)$$
2. The vertex with the largest $x$-coordinate is $(5, 1)$.

• P1 Finding at least two vertices correctly.
• A1 The correct answer, $(5, 1)$.

(b) $11$

1. A linear expression is greatest at a vertex of the region.
2. At $(5, 1)$: 11. At $(2, 4)$: 8. At $(0.5, 1)$: 2.
3. The greatest value is 11.

• P1 Evaluating $2x + y$ at the vertices.
• A1 The correct answer, $11$.

(a) $37$

1. Since y $\ge$ 1, the inequality 3x + y $\le$ 17 gives 1 $\le$ x $\le$ 5.
2. $$4\times 3+5\times 5$$
3. Since y $\ge$ 1, the inequality 3x + y $\le$ 17 gives 1 $\le$ x $\le$ 5.
4. For a fixed x, maximise y because its coefficient in 4x + 5y is positive. For x = 1, 2, 3, 4, 5, the largest permitted integer y is 6, 5, 5, 4, 2 respectively.
5. These pairs give 4x + 5y values 34, 33, 37, 36 and 30.
6. The greatest is 37, achieved by x = 3 and y = 5.

• P1 Since y $\ge$ 1, the inequality 3x + y $\le$ 17 gives 1 $\le$ x $\le$ 5.
• P1 Establishing $4\times 3+5\times 5$ or an equivalent valid method.
• A1 Correct answer: $37$