Past Exam Questions

The equation of the line is given by y = mx - 2.

To find when the line is tangent to the curve y = x^2 - 2x + 7, equate the two equations:

\(x^2 - 2x + 7 = mx - 2\)

Rearrange to form a quadratic equation:

\(x^2 - (2 + m)x + 9 = 0\)

For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Substitute into the discriminant formula:

\((2 + m)^2 - 4 \times 1 \times 9 = 0\)

Simplify and solve for m:

\((2 + m)^2 = 36\)

\(2 + m = 6 \quad \text{or} \quad 2 + m = -6\)

\(m = 4 \quad \text{or} \quad m = -8\)

For m = 4, substitute back to find x:

\(x^2 - 6x + 9 = 0\)

\((x - 3)^2 = 0\)

\(x = 3\)

Substitute x = 3 into the line equation to find y:

\(y = 4 \times 3 - 2 = 10\)

Coordinates are (3, 10).

For m = -8, substitute back to find x:

\(x^2 + 6x + 9 = 0\)

\((x + 3)^2 = 0\)

\(x = -3\)

Substitute x = -3 into the line equation to find y:

\(y = -8 \times (-3) - 2 = 22\)

Coordinates are (-3, 22).

For the quadratic equation \(ax^2 + bx + c = 0\) to have no real roots, the discriminant must be less than zero.

The discriminant \(\Delta\) is given by \(b^2 - 4ac\).

Here, \(a = 4\), \(b = -24\), and \(c = p\).

So, \(\Delta = (-24)^2 - 4 \times 4 \times p\).

\(\Delta = 576 - 16p\).

For no real roots, \(576 - 16p < 0\).

Solving for \(p\):

\(576 < 16p\)

\(p > 36\)

(i) To find the value of \(c\), we need the line \(4y = x + c\) to be tangent to the curve \(y^2 = x + 3\). Substitute \(x = 4y - c\) into the curve equation:

\(y^2 = 4y - c + 3\)

\(y^2 - 4y + c - 3 = 0\)

This is a quadratic in \(y\). For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 1\), \(b = -4\), \(c = c - 3\). So:

\((-4)^2 - 4(1)(c - 3) = 0\)

\(16 - 4c + 12 = 0\)

\(28 - 4c = 0\)

\(4c = 28\)

\(c = 7\)

(ii) With \(c = 7\), substitute back to find \(P\):

\(y^2 = 4y - 7 + 3\)

\(y^2 - 4y + 4 = 0\)

\((y - 2)^2 = 0\)

\(y = 2\)

Substitute \(y = 2\) into \(x = 4y - c\):

\(x = 4(2) - 7 = 1\)

Coordinates of \(P\) are \((1, 2)\).

(i) To show that the curve and the line meet for all values of \(k\), equate the equations:

\(2x^2 - 3x + 1 = kx + k^2\)

Rearrange to form a quadratic equation:

\(2x^2 - (3+k)x + (1-k^2) = 0\)

For the curve and line to meet, the discriminant \(b^2 - 4ac\) must be non-negative:

\((3+k)^2 - 4 \times 2 \times (1-k^2) \geq 0\)

\(9 + 6k + k^2 - 8 + 8k^2 \geq 0\)

\(9k^2 + 6k + 1 \geq 0\)

\((3k + 1)^2 \geq 0\)

This is always true, hence the curve and line meet for all values of \(k\).

(ii) For the line to be tangent to the curve, the discriminant must be zero:

\((3+k)^2 - 4 \times 2 \times (1-k^2) = 0\)

\(9 + 6k + k^2 - 8 + 8k^2 = 0\)

\(9k^2 + 6k + 1 = 0\)

Solving for \(k\), we find \(k = -\frac{1}{3}\).

Substitute \(k = -\frac{1}{3}\) into the line equation:

\(y = -\frac{1}{3}x + \left(-\frac{1}{3}\right)^2 = -\frac{1}{3}x + \frac{1}{9}\)

Equate with the curve equation:

\(2x^2 - 3x + 1 = -\frac{1}{3}x + \frac{1}{9}\)

\(2x^2 - \frac{8}{3}x + \frac{8}{9} = 0\)

Solving gives \(x = \frac{2}{3}\).

Substitute \(x = \frac{2}{3}\) back into the line equation to find \(y\):

\(y = -\frac{1}{3} \times \frac{2}{3} + \frac{1}{9} = -\frac{1}{9}\)

Thus, the point of tangency is \(\left( \frac{2}{3}, -\frac{1}{9} \right)\).

To find the set of values of \(k\) for which the line does not meet the curve, we need to solve the system of equations:

1. \(y = 2x + \frac{12}{x}\)

2. \(y + x = k\)

Substitute \(y = k - x\) from the second equation into the first equation:

\(k - x = 2x + \frac{12}{x}\)

Rearrange to form a quadratic equation:

\(2x + \frac{12}{x} = k - x\)

\(2x^2 + 12 = kx - x^2\)

\(3x^2 - kx + 12 = 0\)

For the line not to meet the curve, the quadratic equation must have no real solutions. This occurs when the discriminant \(b^2 - 4ac\) is less than zero.

Here, \(a = 3\), \(b = -k\), \(c = 12\).

The discriminant is:

\((-k)^2 - 4 \times 3 \times 12\)

\(k^2 - 144 \lt 0\)

Solving \(k^2 - 144 \lt 0\) gives:

\(-12 \lt k \lt 12\)

To find the values of \(b\) for which the line meets the curve, set the equations equal to each other:

\(x + 1 = x^2 + bx + 5\)

Rearrange to form a quadratic equation:

\(x^2 + (b-1)x + 4 = 0\)

For the line to meet the curve, the quadratic must have real roots. This requires the discriminant to be non-negative:

\((b-1)^2 - 4 \times 1 \times 4 \geq 0\)

\((b-1)^2 - 16 \geq 0\)

\((b-1)^2 \geq 16\)

Taking square roots gives:

\(b-1 \geq 4\) or \(b-1 \leq -4\)

Solving these inequalities:

\(b \geq 5\) or \(b \leq -3\)

Thus, the set of values for \(b\) is \(b \geq 5\) or \(b \leq -3\).

(i) To find when the curve \(y = x^2 - 6x + k\) lies entirely above the \(x\)-axis, the quadratic must have no real roots. This occurs when the discriminant \(b^2 - 4ac < 0\). Here, \(a = 1\), \(b = -6\), and \(c = k\). The discriminant is \((-6)^2 - 4(1)(k) = 36 - 4k\). For the curve to lie above the \(x\)-axis, \(36 - 4k < 0\), which simplifies to \(k > 9\).

(ii) For the line \(y + 2x = 7\) to be tangent to the curve, it must touch the curve at exactly one point. Substitute \(y = 7 - 2x\) into the curve equation: \(7 - 2x = x^2 - 6x + k\). Rearrange to form a quadratic: \(x^2 - 4x + (k - 7) = 0\). For tangency, the discriminant must be zero: \((-4)^2 - 4(1)(k - 7) = 0\). Simplifying gives \(16 - 4k + 28 = 0\), leading to \(44 = 4k\), so \(k = 11\).

To find the points of intersection, set the equations equal: \(ax + 3a = -\frac{2}{x}\).

Multiply through by \(x\) to eliminate the fraction: \(ax^2 + 3ax + 2 = 0\).

This is a quadratic equation in \(x\): \(ax^2 + 3ax + 2 = 0\).

For the line and curve to intersect at two distinct points, the discriminant of the quadratic must be greater than zero: \(b^2 - 4ac > 0\).

Here, \(a = a\), \(b = 3a\), \(c = 2\).

Calculate the discriminant: \((3a)^2 - 4(a)(2) > 0\).

Simplify: \(9a^2 - 8a > 0\).

Factorize: \(a(9a - 8) > 0\).

The critical points are \(a = 0\) and \(a = \frac{8}{9}\).

Test intervals around the critical points to determine where the inequality holds:

- For \(a < 0\), \(a(9a - 8) > 0\) is true.

- For \(0 < a < \frac{8}{9}\), \(a(9a - 8) < 0\) is false.

- For \(a > \frac{8}{9}\), \(a(9a - 8) > 0\) is true.

Thus, the set of values for \(a\) is \(a < 0\) or \(a > \frac{8}{9}\).

To find the values of \(k\) for which the quadratic equation \(2x^2 + 3kx + k = 0\) has distinct real roots, we need to ensure that the discriminant is greater than zero.

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\).

Here, \(a = 2\), \(b = 3k\), and \(c = k\).

Thus, the discriminant is:

\(\Delta = (3k)^2 - 4 \times 2 \times k = 9k^2 - 8k\).

For distinct real roots, \(\Delta > 0\):

\(9k^2 - 8k > 0\).

Factorizing gives:

\(k(9k - 8) > 0\).

The critical points are \(k = 0\) and \(k = \frac{8}{9}\).

Using a sign chart or test intervals, we find that the inequality holds for:

\(k < 0\) or \(k > \frac{8}{9}\).

Thus, the set of values of \(k\) for which the equation has distinct real roots is \(k < 0, k > \frac{8}{9}\).

To find when the curve and the line do not meet, we set the equations equal to each other:

\(kx^2 - 3x = x - k\)

Rearrange to form a quadratic equation:

\(kx^2 - 4x + k = 0\)

For the curve and line not to meet, the discriminant of the quadratic must be less than zero:

\(b^2 - 4ac < 0\)

Here, \(a = k\), \(b = -4\), and \(c = k\). Calculate the discriminant:

\((-4)^2 - 4(k)(k) < 0\)

\(16 - 4k^2 < 0\)

Rearrange:

\(4k^2 > 16\)

\(k^2 > 4\)

Taking the square root of both sides gives:

\(k > 2 \text{ or } k < -2\)

To find the value of \(k\) for which the line \(y = 2x + k\) is a tangent to the curve \(y = 2x^2 - 6x + 5\), we equate the two equations:

\(2x^2 - 6x + 5 = 2x + k\)

Rearrange to form a quadratic equation:

\(2x^2 - 8x + 5 - k = 0\)

For the line to be tangent to the curve, the discriminant of this quadratic equation must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 2\), \(b = -8\), and \(c = 5 - k\).

Calculate the discriminant:

\((-8)^2 - 4(2)(5 - k) = 0\)

\(64 - 8(5 - k) = 0\)

\(64 - 40 + 8k = 0\)

\(24 + 8k = 0\)

\(8k = -24\)

\(k = -3\)

Thus, the value of \(k\) is \(-3\).

To show that \(4c = m^2 - 12m + 16\), we start by equating the line and the curve: \(mx + c = 6x - x^2 - 5\).

Rearrange to form a quadratic equation: \(x^2 - (6-m)x + (c+5) = 0\).

For the line to be tangent to the curve, the discriminant of this quadratic must be zero: \(b^2 - 4ac = 0\).

Here, \(a = 1\), \(b = -(6-m)\), and \(c = c+5\).

Calculate the discriminant: \((6-m)^2 - 4(1)(c+5) = 0\).

Expand and simplify: \(36 - 12m + m^2 - 4c - 20 = 0\).

Rearrange to find \(4c = m^2 - 12m + 16\).

To find the point of tangency, equate the line and curve equations:

\(kx - k = -\frac{1}{2x}\)

Multiply through by \(2x\) to clear the fraction:

\(2kx^2 - 2kx = -1\)

Rearrange to form a quadratic equation:

\(2kx^2 - 2kx + 1 = 0\)

For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 2k\), \(b = -2k\), \(c = 1\).

\((-2k)^2 - 4(2k)(1) = 0\)

\(4k^2 - 8k = 0\)

Factor out \(4k\):

\(4k(k - 2) = 0\)

Since \(k\) is positive, \(k = 2\).

Substitute \(k = 2\) back into the line equation:

\(y = 2x - 2\)

Equate with the curve equation:

\(2x - 2 = -\frac{1}{2x}\)

Multiply through by \(2x\):

\(4x^2 - 4x + 1 = 0\)

\((2x - 1)^2 = 0\)

\(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into the line equation:

\(y = 2 \times \frac{1}{2} - 2 = -1\)

The point of tangency is \(\left( \frac{1}{2}, -1 \right)\).

To find the values of m for which the line y = mx is a tangent to the curve y = 2x^2 - 4x + 8, we equate the two equations:

\(mx = 2x^2 - 4x + 8\)

Rearrange to form a quadratic equation:

\(2x^2 - (4 + m)x + 8 = 0\)

For the line to be a tangent, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 2\), \(b = -(4 + m)\), \(c = 8\).

Substitute into the discriminant formula:

\((-(4 + m))^2 - 4 \times 2 \times 8 = 0\)

\((4 + m)^2 - 64 = 0\)

\((4 + m)^2 = 64\)

Take the square root of both sides:

\(4 + m = 8\) or \(4 + m = -8\)

Solve for \(m\):

\(m = 4\) or \(m = -12\)

To find when the line does not intersect the curve, set the equations equal to each other:

\(x^2 - 4x + c = 2x - 7\)

Rearrange to form a quadratic equation:

\(x^2 - 6x + c + 7 = 0\)

For the line not to intersect the curve, the quadratic equation must have no real roots. This occurs when the discriminant is less than zero:

\(b^2 - 4ac < 0\)

Here, \(a = 1\), \(b = -6\), and \(c = c + 7\). Calculate the discriminant:

\((-6)^2 - 4(1)(c + 7) < 0\)

\(36 - 4(c + 7) < 0\)

\(36 - 4c - 28 < 0\)

\(8 - 4c < 0\)

\(8 < 4c\)

\(c > 2\)

To find the set of values of \(p\) for which the equation \(f(x) = p\) has no real roots, we consider the equation:

\(2x^2 - 6x + 5 - p = 0\)

This is a quadratic equation in the form \(ax^2 + bx + c = 0\) where \(a = 2\), \(b = -6\), and \(c = 5 - p\).

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:

\(\Delta = b^2 - 4ac\)

For the quadratic equation to have no real roots, the discriminant must be less than zero:

\(\Delta < 0\)

Substituting the values of \(a\), \(b\), and \(c\) into the discriminant:

\(\Delta = (-6)^2 - 4 \times 2 \times (5 - p)\)

\(\Delta = 36 - 8(5 - p)\)

\(\Delta = 36 - 40 + 8p\)

\(\Delta = 8p - 4\)

Setting \(\Delta < 0\):

\(8p - 4 < 0\)

\(8p < 4\)

\(p < \frac{1}{2}\)

Thus, the set of values of \(p\) for which the equation \(f(x) = p\) has no real roots is \(p < \frac{1}{2}\).

To find the values of k for which the line and the curve intersect at two distinct points, we equate the two equations:

\(2x - k = x^2 + kx - 2\)

Rearrange to form a quadratic equation:

\(x^2 + (k - 2)x + (k - 2) = 0\)

For the quadratic to have two distinct solutions, the discriminant must be greater than zero:

\(b^2 - 4ac > 0\)

Here, \(a = 1\), \(b = k - 2\), and \(c = k - 2\). So,

\((k - 2)^2 - 4(k - 2) > 0\)

Simplify:

\((k - 2)^2 - 4(k - 2) = (k - 2)(k - 6) > 0\)

This inequality holds when:

\(k < 2 \text{ or } k > 6\)

First, rewrite the equation \(2x^2 - 10x + 8 = kx\) as a standard quadratic equation:

\(2x^2 - (10 + k)x + 8 = 0\).

For the quadratic equation \(ax^2 + bx + c = 0\) to have no real roots, the discriminant must be less than zero:

\(b^2 - 4ac < 0\).

Here, \(a = 2\), \(b = -(10 + k)\), and \(c = 8\).

Calculate the discriminant:

\((-(10 + k))^2 - 4 \times 2 \times 8 < 0\)

\((10 + k)^2 - 64 < 0\)

\(100 + 20k + k^2 - 64 < 0\)

\(k^2 + 20k + 36 < 0\).

To find the values of \(k\), solve the inequality:

Factor the quadratic:

\((k + 18)(k + 2) < 0\).

The critical points are \(k = -18\) and \(k = -2\).

Test intervals around these points to determine where the inequality holds:

- For \(k < -18\), both factors are negative, so the product is positive.

- For \(-18 < k < -2\), one factor is negative and the other is positive, so the product is negative.

- For \(k > -2\), both factors are positive, so the product is positive.

Thus, the solution is \(-18 < k < -2\).

To find the value of \(c\) where the line is tangent to the curve, we need to set the equations equal and solve for \(x\) and \(c\).

First, equate the line and curve equations:

\(2x + c = 8 - 2x - x^2\)

Rearrange to form a quadratic equation:

\(x^2 + 4x + c - 8 = 0\)

For the line to be tangent to the curve, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 1\), \(b = 4\), and \(c = c - 8\).

Substitute into the discriminant formula:

\(4^2 - 4(1)(c - 8) = 0\)

Simplify:

\(16 - 4c + 32 = 0\)

\(48 - 4c = 0\)

Solve for \(c\):

\(4c = 48\)

\(c = 12\)

To find the value of \(m\) and the coordinates of \(P\), we equate the line and the curve:

\(mx + 14 = \frac{12}{x} + 2\)

Multiply through by \(x\) to eliminate the fraction:

\(mx^2 + 14x = 12 + 2x\)

Rearrange to form a quadratic equation:

\(mx^2 + 12x - 12 = 0\)

Since the line is tangent to the curve, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Substitute \(a = m\), \(b = 12\), \(c = -12\):

\(12^2 - 4m(-12) = 0\)

\(144 + 48m = 0\)

\(48m = -144\)

\(m = -3\)

Substitute \(m = -3\) back into the equation to find \(x\):

\(-3x^2 + 12x - 12 = 0\)

Factor or solve for \(x\):

\(x = 2\)

Substitute \(x = 2\) into the original curve equation to find \(y\):

\(y = \frac{12}{2} + 2 = 6 + 2 = 8\)

Thus, the coordinates of \(P\) are \((2, 8)\).