P2.3Quadratics and the discriminant

• M1 for \((x \pm 4)^2 \pm k\), \(k \ne 0\)
• A1 for \(p = -4\)
• A1 for \(q = 9\)

Worked solution: \(x^{2} - 8x + 25 = (x - 4)^2 - 16 + 25 = (x - 4)^2 + 9\)

• B1ft for \(x\)-coordinate \(4\)
• B1ft for \(y\)-coordinate \(9\)

Worked solution: The minimum occurs when \(x - 4 = 0\), so the minimum point is \((4, 9)\).

• B1* for equating \(x^2 + 3x = kx + k\) and rearranging to the given equation

Worked solution: At an intersection \(x^2 + 3x = kx + k\), so \(x^2 + 3x - kx - k = 0\), i.e. \(x^2 + (3 - k)x - k = 0\).

• M1 for \((3 - k)^2 - 4 \times 1 \times (-k)\)
• A1* for \(k^2 - 2k + 9\) with no errors seen

Worked solution: \(b^2 - 4ac = (3 - k)^2 - 4(1)(-k) = 9 - 6k + k^2 + 4k = k^2 - 2k + 9\).

• M1 for completing the square: \(k^2 - 2k + 9 = (k - 1)^2 + 8\)
• A1 for \((k - 1)^2 \ge 0\) so the discriminant is at least 8, hence positive
• A1* for a conclusion: discriminant \(\gt 0\) for all \(k\), so two distinct real roots and two distinct points of intersection

Worked solution: \(k^2 - 2k + 9 = (k - 1)^2 + 8\). Since \((k - 1)^2 \ge 0\), the discriminant is at least 8, so it is always positive.
So the quadratic always has two distinct real roots, and \(l\) meets \(C\) at two distinct points for every real \(k\).

• M1 for multiplying both sides by \(x^2 + 1\) (which is never zero)
• A1* for collecting terms to obtain the given equation

Worked solution: \(x^2 + 1 \gt 0\), so \(y(x^2 + 1) = x^2 + x + 1\), i.e. \(yx^2 + y - x^2 - x - 1 = 0\), which is \((y - 1)x^2 - x + (y - 1) = 0\).

• M1 for using the condition for real roots: \((-1)^2 - 4(y - 1)(y - 1) \ge 0\)
• A1 for \((y - 1)^2 \le \frac{1}{4}\) (or \(4y^2 - 8y + 3 \le 0\))
• M1 for solving: \(-\frac{1}{2} \le y - 1 \le \frac{1}{2}\) (or critical values \(\frac{1}{2}\), \(\frac{3}{2}\) with the inside region)
• A1 for \(\frac{1}{2} \le y \le \frac{3}{2}\)
• B1 for dealing with \(y = 1\), where the equation is linear (\(x = 0\)), and noting it lies in the interval; with a conclusion

Worked solution: Any point on \(C\) has a real \(x\)-coordinate, which is a root of \((y - 1)x^2 - x + (y - 1) = 0\).
If \(y \ne 1\) this is a quadratic, so its discriminant is non-negative: \(1 - 4(y - 1)^2 \ge 0\), so \((y - 1)^2 \le \frac{1}{4}\) and \(-\frac{1}{2} \le y - 1 \le \frac{1}{2}\), i.e. \(\frac{1}{2} \le y \le \frac{3}{2}\).
If \(y = 1\) the equation is \(-x = 0\), so \(x = 0\); and \(y = 1\) lies in the interval.
So \(\frac{1}{2} \le y \le \frac{3}{2}\) for every point on \(C\).

• M1 for substituting \(y = \frac{3}{2}\) or \(y = \frac{1}{2}\) into the equation in (a) and solving (a repeated root)
• A1 for \(\left(1, \frac{3}{2}\right)\) and \(\left(-1, \frac{1}{2}\right)\)

Worked solution: \(y = \frac{3}{2}\): \(\frac{1}{2}x^2 - x + \frac{1}{2} = 0 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1\).
\(y = \frac{1}{2}\): \(-\frac{1}{2}x^2 - x - \frac{1}{2} = 0 \Rightarrow (x + 1)^2 = 0 \Rightarrow x = -1\).
Greatest value at \(\left(1, \frac{3}{2}\right)\), least value at \(\left(-1, \frac{1}{2}\right)\).