Past Exam Questions

To find the value of \(m\) for which the line is tangent to the curve, we set the equations equal: \(x^2 - 4x + 4 = mx\).

Rearrange to form a quadratic equation: \(x^2 - (4 + m)x + 4 = 0\).

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

Here, \(a = 1\), \(b = -(4 + m)\), \(c = 4\).

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

Simplify: \((4 + m)^2 - 16 = 0\).

\((4 + m)^2 = 16\).

\(4 + m = \\pm 4\).

Solving gives \(m = 0\) or \(m = -8\).

Since \(m\) must be non-zero, \(m = -8\).

Substitute \(m = -8\) back into the quadratic: \(x^2 + 4x + 4 = 0\).

Factor: \((x + 2)^2 = 0\).

\(x = -2\).

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

\(y = (-2)^2 - 4(-2) + 4 = 4 + 8 + 4 = 16\).

Thus, the coordinates are \((-2, 16)\).

(i) To find the value of \(k\), we equate the line and the curve:

\(\frac{x^2}{4} = \frac{x}{k} + k\)

Rearranging gives:

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

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

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

Substituting \(a = k\), \(b = -4\), \(c = -4k^2\), we get:

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

\(16 + 16k^3 = 0\)

\(k^3 = -1\)

\(k = -1\)

(ii) Substitute \(k = -1\) into the line equation:

\(y = -x - 1\)

Substitute into the curve equation \(4y = x^2\):

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

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

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

\(x = -2\)

Substitute \(x = -2\) back into \(y = -x - 1\):

\(y = -(-2) - 1 = 1\)

Thus, the coordinates of \(P\) are \((-2, 1)\).

To find the intersection points, substitute x = k - 2y into the curve equation xy = 6:

\((k - 2y)y = 6\)

\(2y^2 - ky + 6 = 0\)

This is a quadratic in y. For two distinct points, the discriminant must be positive:

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

Here, \(a = 2\), \(b = -k\), \(c = 6\).

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

\(k^2 - 48 > 0\)

\(k^2 > 48\)

\(k < -\sqrt{48} \text{ and } k > \sqrt{48}\)

To find the points of intersection, set the equations equal: \(3x - 2k = x^2 - kx + 2\).

Rearrange to form a quadratic equation: \(x^2 - kx - 3x + 2 + 2k = 0\).

Simplify to: \(x^2 - (k+3)x + (2 + 2k) = 0\).

Use the discriminant \(b^2 - 4ac\) to determine if there are real solutions: \(b = -(k+3), a = 1, c = 2 + 2k\).

Calculate the discriminant: \((k+3)^2 - 4(1)(2 + 2k)\).

Simplify: \((k+3)^2 - 8 - 8k = (k-1)^2\).

Since \((k-1)^2 \geq 0\) for all \(k\), the discriminant is non-negative, ensuring real solutions.

Hence, the line and the curve meet for all values of \(k\).

To find the value of k for which the line y = 6x + k is tangent to the curve y = 7/√x, we need to equate the derivatives and solve for k.

The derivative of y = 7/√x is:

\(\frac{d}{dx} \left( 7x^{-1/2} \right) = -\frac{7}{2}x^{-3/2}\)

The derivative of y = 6x + k is:

\(\frac{d}{dx} (6x + k) = 6\)

Setting the derivatives equal gives:

\(-\frac{7}{2}x^{-3/2} = 6\)

Solving for x:

\(x = \frac{49}{144}\)

Substitute x back into the curve equation to find y:

\(y = \frac{49}{12}\)

Substitute x and y into the line equation:

\(\frac{49}{12} = 6 \left( \frac{49}{144} \right) + k\)

Solve for k:

\(k = \frac{49}{24} \text{ or } 2.04\)

(i) The equation of a line with gradient \(m\) passing through the point \((2, 0)\) is given by the point-slope form:

\(y - 0 = m(x - 2)\)

Thus, the equation is \(y = m(x - 2)\).

(ii) To find the values of \(m\) for which the line is tangent to the curve \(y = x^2 - 4x + 5\), equate the line equation \(y = mx - 2m\) to the curve:

\(x^2 - 4x + 5 = mx - 2m\)

Rearrange to form a quadratic equation:

\(x^2 - (4 + m)x + (5 + 2m) = 0\)

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

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

\((4 + m)^2 - 4(5 + 2m) = 0\)

\(m^2 - 4 = 0\)

Solving gives \(m = \pm 2\).

For \(m = 2\):

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

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

\(x = 3\)

Substitute \(x = 3\) into \(y = 2(x - 2)\):

\(y = 2(3 - 2) = 2\)

Point is (3, 2).

For \(m = -2\):

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

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

\(x = 1\)

Substitute \(x = 1\) into \(y = -2(x - 2)\):

\(y = -2(1 - 2) = 2\)

Point is (1, 2).

To find the value of \(k\) for which the line is tangent to the curve, we first eliminate \(x\) from the equations. Substitute \(x = k - 2y\) from the line equation into the curve equation:

\(y^2 + 2(k - 2y) = 13\)

Simplify to get:

\(y^2 - 4y + 2k = 13\)

\(y^2 - 4y + 2k - 13 = 0\)

This is a quadratic equation in \(y\). 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 \(c = 2k - 13\). So,

\((-4)^2 - 4(1)(2k - 13) = 0\)

\(16 - 8k + 52 = 0\)

\(68 = 8k\)

\(k = \frac{68}{8} = 8 \frac{1}{2}\)

Thus, the value of \(k\) for which the line is tangent to the curve is \(8 \frac{1}{2}\).

To find the values of \(k\) for which the line is tangent to the curve, we set the equations equal to each other:

\(kx + 6 = x^2 + 3x + 2k\)

Rearrange to form a quadratic equation:

\(x^2 + (3-k)x + (2k-6) = 0\)

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

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

Here, \(a = 1\), \(b = 3-k\), \(c = 2k-6\).

Substitute into the discriminant formula:

\((3-k)^2 - 4(1)(2k-6) = 0\)

Simplify:

\((3-k)^2 - 8k + 24 = 0\)

\(9 - 6k + k^2 - 8k + 24 = 0\)

\(k^2 - 14k + 33 = 0\)

Factorize the quadratic:

\((3-k)(11-k) = 0\)

Thus, \(k = 3\) or \(k = 11\).

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

\(mx + 4 = 3x^2 - 4x + 7\)

Rearrange to form a quadratic equation:

\(3x^2 - (4 + m)x + 3 = 0\)

For the line to intersect the curve at two distinct points, the discriminant of this quadratic equation must be greater than zero:

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

Here, \(a = 3\), \(b = -(4 + m)\), and \(c = 3\).

Calculate the discriminant:

\((4 + m)^2 - 4 \times 3 \times 3 > 0\)

\((4 + m)^2 - 36 > 0\)

\((4 + m)^2 > 36\)

Taking the square root of both sides:

\(|4 + m| > 6\)

This gives two inequalities:

\(4 + m > 6\) or \(4 + m < -6\)

Solving these:

\(m > 2\) or \(m < -10\)

Thus, the set of values of m for which the line intersects the curve at two distinct points is \(m > 2 \text{ or } m < -10\).

(i) Since the roots of the equation are \(-3\) and \(5\), we can express the quadratic as \((x + 3)(x - 5) = 0\).

Expanding this, we get:

\(x^2 - 5x + 3x - 15 = x^2 - 2x - 15\).

Thus, \(p = -2\) and \(q = -15\).

(ii) For the equation \(x^2 + px + q + r = 0\) to have equal roots, the discriminant must be zero.

The discriminant \(b^2 - 4ac = 0\).

Here, \(a = 1\), \(b = p = -2\), and \(c = q + r = -15 + r\).

So, \((-2)^2 - 4(1)(-15 + r) = 0\).

\(4 - 4(-15 + r) = 0\).

\(4 + 60 - 4r = 0\).

\(64 = 4r\).

\(r = 16\).

(i) To find when the curve \(y = kx^2 + 1\) and the line \(y = kx\) have no common points, set the equations equal: \(kx^2 + 1 = kx\).

Rearrange to form a quadratic equation: \(kx^2 - kx + 1 = 0\).

For no common points, the discriminant must be less than zero: \(b^2 - 4ac < 0\).

Here, \(a = k\), \(b = -k\), \(c = 1\). So, \((-k)^2 - 4(k)(1) < 0\).

Simplify: \(k^2 - 4k < 0\).

Factor: \(k(k - 4) < 0\).

The solution to this inequality is \(0 < k < 4\).

(ii) For the line to be tangent to the curve, the discriminant must be zero: \(k^2 - 4k = 0\).

Factor: \(k(k - 4) = 0\).

Since \(k \neq 0\), \(k = 4\).

Substitute \(k = 4\) into the quadratic equation: \(4x^2 - 4x + 1 = 0\).

Factor: \((2x - 1)^2 = 0\).

Solve for \(x\): \(x = \frac{1}{2}\).

Substitute \(x = \frac{1}{2}\) into \(y = 4x\): \(y = 4 \times \frac{1}{2} = 2\).

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

To find the intersection points, set the equations equal: \(\frac{9}{2-x} = x + k\).

Multiply both sides by \(2-x\) to eliminate the fraction: \(9 = (x + k)(2-x)\).

Expand and rearrange: \(x^2 - 2x + kx - 2k + 9 = 0\).

This simplifies to \(x^2 + (k-2)x + (9-2k) = 0\).

For two distinct points, the discriminant \(b^2 - 4ac\) must be greater than zero.

Here, \(a = 1\), \(b = k-2\), \(c = 9-2k\).

Calculate the discriminant: \((k-2)^2 - 4(1)(9-2k) > 0\).

Simplify: \(k^2 - 4k + 4 - 36 + 8k > 0\).

Further simplify: \(k^2 + 4k - 32 > 0\).

Factorize: \((k+8)(k-4) > 0\).

The solution to this inequality is \(k < -8\) or \(k > 4\).

To find the values of \(k\) for which the line \(y + kx = 12\) is a tangent to the curve \(y = 2x^2 - 8x + 14\), we first express the line equation in terms of \(y\):

\(y = -kx + 12\)

Set the expressions for \(y\) equal to each other:

\(2x^2 - 8x + 14 = -kx + 12\)

Rearrange to form a quadratic equation:

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

For the line to be tangent to the curve, the quadratic must have exactly one solution, which occurs when the discriminant is zero:

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

Here, \(a = 2\), \(b = k-8\), and \(c = 2\). Substitute these into the discriminant formula:

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

Simplify:

\((k-8)^2 = 16\)

Take the square root of both sides:

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

Solve for \(k\):

\(k = 12 \quad \text{or} \quad k = 4\)

To find the set of values of k for which the line does not intersect the curve, we substitute y from the line equation into the curve equation:

\(y = \frac{x + k}{2}\)

Substitute into the curve equation:

\(\frac{x + k}{2} = x^2 - 4x + 7\)

Multiply through by 2 to eliminate the fraction:

\(x + k = 2x^2 - 8x + 14\)

Rearrange to form a quadratic equation:

\(2x^2 - 9x + 14 - k = 0\)

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

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

Here, a = 2, b = -9, c = 14 - k. Calculate the discriminant:

\((-9)^2 - 4 \times 2 \times (14 - k) < 0\)

\(81 - 8(14 - k) < 0\)

Simplify:

\(81 - 112 + 8k < 0\)

\(8k < 31\)

\(k < \frac{31}{8}\)

Thus, the set of values of k for which the line does not intersect the curve is:

\(k < 3.875\)

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

Substitute these values into the discriminant condition:

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

\(k^2 - 64 < 0\)

\(k^2 < 64\)

This implies \(-8 < k < 8\).

To find the values of k for which the line intersects the curve at two distinct points, we set the equations equal to each other:

\(kx - 4 = x^2 - 2x\)

Rearrange to form a quadratic equation:

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

For the line to intersect the curve at two distinct points, the discriminant of this quadratic equation must be greater than zero. The discriminant \(\Delta\) is given by:

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

Here, \(a = 1\), \(b = -(2 + k)\), and \(c = 4\). Thus,

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

\(\Delta = (2 + k)^2 - 16\)

For two distinct points, \(\Delta > 0\):

\((2 + k)^2 > 16\)

Solving this inequality:

\(2 + k > 4\) or \(2 + k < -4\)

\(k > 2\) or \(k < -6\)

To find the values of k for which the line y = 4x + k does not intersect the curve y = x^2, we set the equations equal to each other:

\(x^2 = 4x + k\)

Rearrange to form a quadratic equation:

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

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

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

Here, a = 1, b = -4, and c = -k. Substitute these into the discriminant:

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

Simplify:

\(16 + 4k < 0\)

Solve for k:

\(4k < -16\)

\(k < -4\)

Thus, the set of values for k is k < -4.

To find the value of c, we need the line y = 2x + c to be tangent to the curve y² = 4x. This means they intersect at exactly one point.

Substitute y = 2x + c into y² = 4x:

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

Expand and simplify:

\(4x^2 + 4xc + c^2 = 4x\)

Rearrange to form a quadratic equation:

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

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

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

Here, \(a = 4\), \(b = 4c - 4\), \(c = c^2\).

Calculate the discriminant:

\((4c - 4)^2 - 4 \times 4 \times c^2 = 0\)

\(16c^2 - 32c + 16 - 16c^2 = 0\)

\(-32c + 16 = 0\)

Solve for c:

\(-32c = -16\)

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

To find the set of values of \(k\) for which the line \(l\) does not intersect the curve, substitute \(y = k - 2x\) from the line equation into the curve equation \(xy = 12\).

This gives:

\(x(k - 2x) = 12\)

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

Rearrange to form a quadratic equation:

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

For the line not to intersect the curve, the quadratic equation must have no real roots. This occurs when the discriminant \(b^2 - 4ac < 0\).

Here, \(a = 2\), \(b = -k\), and \(c = 12\).

Calculate the discriminant:

\((-k)^2 - 4(2)(12) < 0\)

\(k^2 - 96 < 0\)

Thus, \(k^2 < 96\).

Taking the square root of both sides gives:

\(-\sqrt{96} < k < \sqrt{96}\)

Therefore, the set of values of \(k\) for which the line does not intersect the curve is \(|k| < \sqrt{96}\).

Substitute \(a = -\frac{7}{2}\) into the line equation, giving \(y = x + \frac{7}{2}\).

Set the curve equal to the line: \(4x^2 - kx + \frac{1}{2}k^2 = x + \frac{7}{2}\).

Rearrange to form a quadratic equation: \(4x^2 - (k+1)x + \left(\frac{1}{2}k^2 - \frac{7}{2}\right) = 0\).

For the line to be tangent, the discriminant must be zero: \((k+1)^2 - 4 \times 4 \times \left(\frac{1}{2}k^2 - \frac{7}{2}\right) = 0\).

Simplify: \((k+1)^2 - 2k^2 + 14 = 0\).

Further simplify to: \(7k^2 - 2k - 57 = 0\).

Factor or use the quadratic formula to solve: \((k-3)(7k+19) = 0\).

Thus, \(k = 3\) or \(k = \frac{19}{7}\).