Inequalities

Solve the inequality (x^2 - 2x > 15).",
"marks": 4,
"solution": "Step 1: Rearrange the inequality to have zero on one side.
(x^2 - 2x - 15 > 0)

Step 2: Find the critical values by solving the corresponding equation (x^2 - 2x - 15 = 0).
Factorising gives: ((x-5)(x+3) = 0)
So, the critical values are (x = 5) and (x = -3).

Step 3: Sketch the parabola. Since the coefficient of (x^2) is positive (1), it is a U-shaped curve crossing the x-axis at -3 and 5.

Step 4: Identify the correct region. We want where (x^2 - 2x - 15 > 0), which is where the curve is above the x-axis. This occurs in two separate regions.

Final answer: (x < -3) or (x > 5)",
"examiner_commentary": "This candidate earns full marks. M1 is awarded for correctly rearranging the inequality. A second M1 is given for a correct method to find the critical values (factorising shown here). A1 is for both correct critical values. The final A1 is for the correct inequalities representing the two disjoint regions. A common error is to write (-3 > x > 5), which is mathematically impossible and scores zero for the final mark."

Find the set of values for (x) for which (12 + 5x - 2x^2 \geq 0).",
"marks": 4,
"solution": "Step 1: The inequality is already arranged with zero on one side. To make sketching easier, we can multiply by -1 and flip the inequality sign: (2x^2 - 5x - 12 \leq 0).

Step 2: Find the critical values by solving (2x^2 - 5x - 12 = 0).
Factorising gives: ((2x+3)(x-4) = 0)
So, the critical values are (x = 4) and (x = -\frac{3}{2}).

Step 3: Sketch the parabola for (2x^2 - 5x - 12). It is a U-shaped curve crossing the x-axis at -1.5 and 4.

Step 4: Identify the region. We want where (2x^2 - 5x -12 \leq 0$), which is where the curve is below or on the x-axis. This is the single region between the roots.$

Final answer: (-\frac{3}{2} \leq x \leq 4)",
"examiner_commentary": "Full credit is given. An M1 is awarded for the method to find roots. A1 for both correct roots. A further M1 is implied by the correct selection of the 'inside' region, ideally supported by a sketch. The final A1 is for the correct compound inequality, including the non-strict (≤) signs. Candidates who work with the original n-shaped graph and correctly identify the region above the axis would also receive full marks."

Find the range of values of (k) for which the equation (x^2 + kx + 9 = 0) has real roots.",
"marks": 5,
"solution": "Step 1: For an equation to have real roots, the discriminant must be greater than or equal to zero.
(b^2 - 4ac \geq 0)

Step 2: Identify (a, b, c) from the equation (x^2 + kx + 9 = 0).
(a=1, b=k, c=9)

Step 3: Substitute these into the discriminant condition.
(k^2 - 4(1)(9) \geq 0)
(k^2 - 36 \geq 0)

Step 4: Solve the new quadratic inequality for (k). The critical values are found by solving (k^2 - 36 = 0), which gives (k = 6) and (k = -6).

Step 5: Sketch the parabola for (y = k^2 - 36). It is a U-shaped curve. We need the region where (k^2 - 36 \geq 0), which is above or on the x-axis.

Final answer: (k \leq -6) or (k \geq 6)",
"examiner_commentary": "This is an excellent response showing a synoptic link. M1 is awarded for correctly stating the discriminant condition (b²-4ac ≥ 0). A1 for correctly substituting to get k² - 36 ≥ 0. M1 for finding the critical values of k. A1 for both correct values. The final A1 is for the correct disjoint inequalities. A common mistake is to forget the 'or equal to' part of the condition for 'real roots' (as opposed to 'distinct real roots')."