P2.7Curve sketching and the modulus function

(a) Answer: A cubic with a positive leading coefficient (rising from bottom left to top right) which crosses the \(x\)-axis at \((1, 0)\), touches the \(x\)-axis at \((-1, 0)\), and meets the \(y\)-axis at \((0, -1)\).

• B1 for the correct shape: a cubic with a positive leading coefficient (rising from bottom left to top right)
• B1 for the correct behaviour at the \(x\)-axis: it crosses the \(x\)-axis at \((1, 0)\); touches the \(x\)-axis at \((-1, 0)\)
• B1 for the \(y\)-intercept \((0, -1)\)

Worked solution: \(y = (x + 1)^2(x - 1)\). Expanding would give a leading term \(x^3\), so the curve is a cubic with a positive leading coefficient.
Roots: crosses the \(x\)-axis at \((1, 0)\); touches the \(x\)-axis at \((-1, 0)\).
When \(x = 0\), \(y = -1\), so the curve meets the \(y\)-axis at \((0, -1)\).

(b) Answer: \(x \lt -1\) or \(-1 \lt x \lt 1\)

• M1 for using the \(x\)-intercepts as critical values and choosing regions consistent with their sketch
• A1 for \(x \lt -1\) or \(-1 \lt x \lt 1\)

Worked solution: From the sketch, the curve is below the \(x\)-axis when \(x \lt -1\) or \(-1 \lt x \lt 1\).

(a) Answer: A quartic with a negative leading coefficient (with both ends going down) which crosses the \(x\)-axis at \((-4, 0)\) and \((-3, 0)\), touches the \(x\)-axis at \((1, 0)\), and meets the \(y\)-axis at \((0, -12)\).

• B1 for the correct shape: a quartic with a negative leading coefficient (with both ends going down)
• B1 for the correct behaviour at the \(x\)-axis: it crosses the \(x\)-axis at \((-4, 0)\) and \((-3, 0)\); touches the \(x\)-axis at \((1, 0)\)
• B1 for all \(x\)-intercepts labelled correctly (as coordinates or values)
• B1 for the \(y\)-intercept \((0, -12)\)

Worked solution: \(y = -(x + 4)(x + 3)(x - 1)^2\). Expanding would give a leading term \(-x^4\), so the curve is a quartic with a negative leading coefficient.
Roots: crosses the \(x\)-axis at \((-4, 0)\) and \((-3, 0)\); touches the \(x\)-axis at \((1, 0)\).
When \(x = 0\), \(y = -12\), so the curve meets the \(y\)-axis at \((0, -12)\).

(b) Answer: \(x \le -4\) or \(x \ge -3\)

• M1 for using the \(x\)-intercepts as critical values and choosing regions consistent with their sketch
• A1 for \(x \le -4\) or \(x \ge -3\)

Worked solution: From the sketch, the curve is below the \(x\)-axis (or on it) when \(x \le -4\) or \(x \ge -3\).

(a) Answer: \(x = -1\): \(|{-5}| = 5\) but \(x - 4 = -5\). \(x = \frac{7}{3}\): \(\left|\frac{5}{3}\right| = \frac{5}{3}\) but \(x - 4 = -\frac{5}{3}\).

• M1 for substituting at least one value into both sides
• A1 for both values shown to fail, with a comment

Worked solution: \(x = -1\): LHS \(= |{-5}| = 5\), RHS \(= -5\). \(x = \frac{7}{3}\): LHS \(= \left|\frac{14}{3} - 3\right| = \frac{5}{3}\), RHS \(= -\frac{5}{3}\). Neither value satisfies the equation.

(b) Answer: The V-shaped graph has vertex \(\left(\frac{3}{2}, 0\right)\) and arms of gradient \(\pm 2\). For \(x \lt 4\) the line is below the \(x\)-axis while \(|2x - 3| \ge 0\); for \(x \ge \frac{3}{2}\) the right arm starts above the line and rises faster (gradient 2 against 1). So the graphs never meet.

• B1 for the graph of \(y = |2x - 3|\): V shape with vertex \(\left(\frac{3}{2}, 0\right)\), arms of gradient \(-2\) and \(2\)
• M1 for comparing: where \(x - 4 \lt 0\) (i.e. \(x \lt 4\)) the line is below the graph, since \(|2x - 3| \ge 0\)
• A1 for completing the argument for large \(x\): the right arm \(y = 2x - 3\) is above the line at \(x = 4\) (5 against 0) and has the greater gradient, so there are no intersections

Worked solution: \(y = |2x - 3|\) is a V shape with vertex \(\left(\frac{3}{2}, 0\right)\), gradient \(-2\) to the left and \(2\) to the right, and it is never negative.
The line \(y = x - 4\) is negative for all \(x \lt 4\), so it cannot meet the V there.
For \(x \ge 4\) (on the right arm), \(2x - 3 - (x - 4) = x + 1 \gt 0\), so the arm is always above the line.
So the graphs never intersect and the equation has no real solutions.

(c) Answer: \(k \lt \frac{3}{2}\)

• M1 for considering the line through the vertex \(\left(\frac{3}{2}, 0\right)\): \(k = \frac{3}{2}\)
• M1 for showing that the solutions of the two branch equations, \(x = 3 - k\) and \(x = \frac{3 + k}{3}\), are valid (and distinct) exactly when \(k \lt \frac{3}{2}\)
• A1 for \(k \lt \frac{3}{2}\)

Worked solution: The line \(y = x - k\) has gradient 1, which lies between the arm gradients \(-2\) and \(2\). It meets both arms (at two distinct points) exactly when it passes above the vertex, i.e. when \(\frac{3}{2} - k \gt 0\).
Check: \(2x - 3 = x - k\) gives \(x = 3 - k\), valid (\(x \gt \frac{3}{2}\)) when \(k \lt \frac{3}{2}\); \(3 - 2x = x - k\) gives \(x = \frac{3 + k}{3}\), valid (\(x \lt \frac{3}{2}\)) when \(k \lt \frac{3}{2}\).
So there are two distinct solutions when \(k \lt \frac{3}{2}\).