(i) To find the values of \(k\) for which the line and curve meet at two distinct points, equate the equations:

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

Rearrange to form a quadratic equation:

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

For two distinct points, the discriminant \(b^2 - 4ac > 0\).

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

Calculate the discriminant:

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

\(16k^2 - 8k - 8 > 0\)

Factorize or solve the inequality:

\((k - 1)(k + \frac{1}{2}) > 0\)

Thus, \(k > 1\) or \(k < -\frac{1}{2}\).

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

\(16k^2 - 8k - 8 = 0\)

Solving gives \(k = 1\) and \(k = -\frac{1}{2}\).

For \(k = 1\), the line is \(y = 3x - 2\) and the curve is \(y = \frac{3}{2}x + 1\).

Equate the tangents:

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

Solve for \(x\):

\(x = \frac{2}{3}\)

Thus, the tangents meet on the x-axis at \(x = \frac{2}{3}\).

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).

(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 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 \((2, 8)\).

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 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:

\(k < 3.875\)

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 \(|k| < \sqrt{96}\).