Past Exam Questions

Let \(y = \frac{1}{x^2}\). Then \(y^2 = \frac{1}{x^4}\).

Substitute into the equation:

\(18y^2 + y = 4\)

Rearrange to form a quadratic equation:

\(18y^2 + y - 4 = 0\)

Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 18\), \(b = 1\), \(c = -4\):

\(y = \frac{-1 \pm \sqrt{1^2 - 4 \times 18 \times (-4)}}{2 \times 18}\)

\(y = \frac{-1 \pm \sqrt{1 + 288}}{36}\)

\(y = \frac{-1 \pm \sqrt{289}}{36}\)

\(y = \frac{-1 \pm 17}{36}\)

\(y = \frac{16}{36} \text{ or } y = \frac{-18}{36}\)

\(y = \frac{4}{9} \text{ or } y = -\frac{1}{2}\)

Since \(y = \frac{1}{x^2}\), solve for \(x\):

For \(y = \frac{4}{9}\):

\(\frac{1}{x^2} = \frac{4}{9}\)

\(x^2 = \frac{9}{4}\)

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

For \(y = -\frac{1}{2}\), \(x^2\) cannot be negative, so ignore.

Thus, the real roots are \(x = 1.5\) and \(x = -1.5\).

Given the function \(f(x) = x^2 - 6x + c\), we need \(f(x) > 2\) for all \(x\).

Rearrange the inequality:

\(x^2 - 6x + c > 2\)

\(x^2 - 6x + c - 2 > 0\)

Complete the square for \(x^2 - 6x\):

\((x-3)^2 - 9\)

Substitute back:

\((x-3)^2 - 9 + c - 2 > 0\)

\((x-3)^2 + c - 11 > 0\)

Since \((x-3)^2 \geq 0\) for all \(x\), we need:

\(c - 11 > 0\)

Thus, \(c > 11\).

To solve the equation \(3x + 2 = \frac{2}{x - 1}\), follow these steps:

1. Multiply both sides by \(x - 1\) to eliminate the fraction:

\((3x + 2)(x - 1) = 2\)

2. Expand the left side:

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

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

3. Bring all terms to one side to form a quadratic equation:

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

4. Factor the quadratic equation:

\((3x - 4)(x + 1) = 0\)

5. Solve for \(x\) by setting each factor to zero:

\(3x - 4 = 0\) or \(x + 1 = 0\)

\(3x = 4\) gives \(x = \frac{4}{3}\)

\(x = -1\)

Thus, the solutions are \(x = -1\) and \(x = \frac{4}{3}\).

1. Start with the inequality:

\(x^2 - 4x + 8 < 9\)

2. Simplify the inequality:

\(x^2 - 4x + 8 - 9 < 0\)

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

3. Complete the square for \(x^2 - 4x\):

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

\((x-2)^2 < 5\)

4. Solve the inequality:

\(-\sqrt{5} < x - 2 < \sqrt{5}\)

5. Add 2 to each part:

\(2 - \sqrt{5} < x < 2 + \sqrt{5}\)

Start with the inequality:

\(2x^2 - 6x + 5 > 13\)

Simplify the inequality:

\(2x^2 - 6x + 5 - 13 > 0\)

\(2x^2 - 6x - 8 > 0\)

Factor the quadratic:

\(2(x^2 - 3x - 4) > 0\)

\(2(x - 4)(x + 1) > 0\)

Find the critical points by setting the factors to zero:

\(x - 4 = 0 \Rightarrow x = 4\)

\(x + 1 = 0 \Rightarrow x = -1\)

Test intervals around the critical points:

• For \(x < -1\), choose \(x = -2\): \(2(-2 - 4)(-2 + 1) > 0 \Rightarrow 2(-6)(-1) > 0 \Rightarrow 12 > 0\)
• For \(-1 < x < 4\), choose \(x = 0\): \(2(0 - 4)(0 + 1) > 0 \Rightarrow 2(-4)(1) > 0 \Rightarrow -8 > 0\) (false)
• For \(x > 4\), choose \(x = 5\): \(2(5 - 4)(5 + 1) > 0 \Rightarrow 2(1)(6) > 0 \Rightarrow 12 > 0\)

Thus, the solution is \(x > 4\) or \(x < -1\).

Start with the inequality:

\(x^2 + 6x + 2 > 9\)

Simplify it:

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

\(x^2 + 6x - 7 > 0\)

Factor the quadratic expression:

\((x + 7)(x - 1) > 0\)

The critical points are \(x = -7\) and \(x = 1\).

Test intervals around the critical points:

1. For \(x < -7\), choose \(x = -8\):

\((-8 + 7)(-8 - 1) > 0\)

\((-1)(-9) > 0\)

\(9 > 0\) (True)

2. For \(-7 < x < 1\), choose \(x = 0\):

\((0 + 7)(0 - 1) > 0\)

\(7(-1) > 0\)

\(-7 > 0\) (False)

3. For \(x > 1\), choose \(x = 2\):

\((2 + 7)(2 - 1) > 0\)

\(9(1) > 0\)

\(9 > 0\) (True)

Thus, the solution is \(x > 1\) or \(x < -7\).

Given the function \(f(x) = 6x - x^2 - 5\), we need to find the values of \(x\) such that \(f(x) \leq 3\).

Start by setting up the inequality:

\(6x - x^2 - 5 \leq 3\)

Rearrange the inequality:

\(-x^2 + 6x - 5 \leq 3\)

\(-x^2 + 6x - 8 \leq 0\)

Multiply through by \(-1\) (which reverses the inequality):

\(x^2 - 6x + 8 \geq 0\)

Factor the quadratic:

\((x - 2)(x - 4) \geq 0\)

The critical points are \(x = 2\) and \(x = 4\).

Test intervals around the critical points:

• For \(x < 2\), choose \(x = 0\): \((0 - 2)(0 - 4) = 8 \geq 0\)
• For \(2 < x < 4\), choose \(x = 3\): \((3 - 2)(3 - 4) = -1 \lt 0\)
• For \(x > 4\), choose \(x = 5\): \((5 - 2)(5 - 4) = 3 \geq 0\)

Thus, the solution is \(x < 2\) or \(x \geq 4\).

To solve the inequality \(4x^2 - 12x > 7\), follow these steps:

1. Rearrange the inequality:

\(4x^2 - 12x - 7 > 0\)

2. Factor the quadratic expression:

\(4x^2 - 12x - 7 = (2x - 7)(2x + 1)\)

3. Solve the inequality \((2x - 7)(2x + 1) > 0\):

- The critical points are \(x = \frac{7}{2}\) and \(x = -\frac{1}{2}\).

4. Test intervals around the critical points:

- For \(x < -\frac{1}{2}\), choose \(x = -1\):

\((2(-1) - 7)(2(-1) + 1) = (-9)(-1) = 9 > 0\)

- For \(-\frac{1}{2} < x < \frac{7}{2}\), choose \(x = 0\):

\((2(0) - 7)(2(0) + 1) = (-7)(1) = -7 < 0\)

- For \(x > \frac{7}{2}\), choose \(x = 4\):

\((2(4) - 7)(2(4) + 1) = (1)(9) = 9 > 0\)

5. The solution is:

\(x > \frac{7}{2} \text{ or } x < -\frac{1}{2}\)

To find the set of \(x\) values for which \(y > 9\), start with the inequality:

\(2x^2 - 3x > 9\)

Subtract 9 from both sides:

\(2x^2 - 3x - 9 > 0\)

Factor the quadratic expression:

\((2x + 3)(x - 3) > 0\)

Find the roots by setting each factor to zero:

\(2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\)

\(x - 3 = 0 \Rightarrow x = 3\)

Test intervals around the roots to determine where the inequality holds:

• For \(x < -\frac{3}{2}\), choose \(x = -2\): \((2(-2) + 3)((-2) - 3) = (-1)(-5) = 5 > 0\)
• For \(-\frac{3}{2} < x < 3\), choose \(x = 0\): \((2(0) + 3)(0 - 3) = (3)(-3) = -9 < 0\)
• For \(x > 3\), choose \(x = 4\): \((2(4) + 3)(4 - 3) = (11)(1) = 11 > 0\)

The solution is \(x > 3\) or \(x < -\frac{3}{2}\).

To find the set of values of \(x\) for which \(f(x) > 4\), we start with the inequality:

\(x^2 - 3x > 4\)

Rearrange the inequality:

\(x^2 - 3x - 4 > 0\)

Factor the quadratic expression:

\((x - 4)(x + 1) > 0\)

Determine the critical points by setting each factor to zero:

\(x - 4 = 0 \Rightarrow x = 4\)

\(x + 1 = 0 \Rightarrow x = -1\)

Test intervals around the critical points:

• For \(x < -1\), choose \(x = -2\): \((x - 4)(x + 1) = (-2 - 4)(-2 + 1) = 6 > 0\)
• For \(-1 < x < 4\), choose \(x = 0\): \((x - 4)(x + 1) = (0 - 4)(0 + 1) = -4 < 0\)
• For \(x > 4\), choose \(x = 5\): \((x - 4)(x + 1) = (5 - 4)(5 + 1) = 6 > 0\)

Thus, the solution is \(x < -1\) or \(x > 4\).

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

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

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

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

Calculate the discriminant: \((3 - 2c)^2 - 4c(-c - 3)\).

Simplify: \((3 - 2c)^2 + 4c(c + 3) = 9 - 12c + 4c^2 + 4c^2 + 12c = 8c^2 + 9\).

The discriminant \(8c^2 + 9\) is always positive for all values of \(c\), indicating two real roots.

Thus, the line and curve intersect for all values of \(c\).

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

\((2k - 3)x^2 - kx - (k - 2) = 3x - 4\)

Rearrange to form a quadratic equation:

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

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

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

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

Calculate the discriminant:

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

Simplify:

\(k^2 + 6k + 9 + 8k^2 -60k + 72 = 0\)

\(9k^2 - 54k +81 = 0\)

Divide by 9:

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

\(k = 3\)

Since the line is tangent, \(k = 3\).

To find the points of intersection, set the equations equal: \(3x + k = x^2 + kx + 6\).

Rearrange to form a quadratic equation: \(x^2 + (k - 3)x + (6 - k) = 0\).

For two distinct points of intersection, the discriminant must be greater than zero: \(b^2 - 4ac > 0\).

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

Calculate the discriminant: \((k - 3)^2 - 4(6 - k) > 0\).

Simplify: \(k^2 - 2k - 15 > 0\).

Factorize: \((k + 3)(k - 5) > 0\).

The solution to this inequality is \(k < -3\) or \(k > 5\).

To find the points of intersection, set the equations equal: \(3x^2 - 4x + 4 = mx + m - 1\).

Rearrange to form a quadratic equation: \(3x^2 - (4 + m)x + (5 - m) = 0\).

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

Here, \(a = 3\), \(b = -(4 + m)\), and \(c = 5 - m\).

Calculate the discriminant: \((4 + m)^2 - 4 imes 3 imes (5 - m)\).

Simplify: \((4 + m)^2 - 12(5 - m) = m^2 + 20m - 44\).

Set the discriminant greater than zero: \(m^2 + 20m - 44 > 0\).

Factorize: \((m + 22)(m - 2) > 0\).

Using the sign chart method, the solution is \(m > 2\) and \(m < -22\).

To find the points of intersection, set the equations equal:

\(2x^2 + m(2x + 1) = 6x + 4\).

Rearrange to form a quadratic equation:

\(2x^2 + 2mx + m - 6x - 4 = 0\).

Simplify to:

\(2x^2 + (2m - 6)x + (m - 4) = 0\).

Use the discriminant \(b^2 - 4ac\) to determine the nature of the roots. Here, \(a = 2\), \(b = 2m - 6\), \(c = m - 4\).

The discriminant is:

\((2m - 6)^2 - 4(2)(m - 4)\).

Calculate:

\(4m^2 - 24m + 36 - 8m + 32\).

Simplify to:

\(4m^2 - 32m + 68\).

Factor or complete the square:

\((2m - 8)^2 + 1\).

This expression is always positive, indicating two distinct real roots for all values of \(m\).

Thus, the line intersects the curve at two distinct points for all values of \(m\).

To find the values of \(m\) for which the line and the curve do not meet, we set the equations equal to each other:

\(mx - 3 = 2x^2 + 5\)

Rearrange to form a quadratic equation:

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

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

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

Calculate the discriminant:

\((-m)^2 - 4(2)(8) = m^2 - 64\)

Set the discriminant less than zero:

\(m^2 - 64 < 0\)

\(m^2 < 64\)

Taking the square root of both sides gives:

\(-8 < m < 8\)

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

\(3x^2 + 2x + 4 = mx + 1\)

Rearranging gives:

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

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

\((2 - m)^2 - 4 imes 3 imes 3 > 0\)

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

\((2 - m)^2 > 36\)

Taking the square root of both sides:

\(|2 - m| > 6\)

This gives two inequalities:

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

Solving these inequalities:

1. \(2 - m > 6\) gives \(m < -4\)

2. \(2 - m < -6\) gives \(m > 8\)

Thus, the set of values of m for which the line and the curve intersect at two distinct points is:

\(m < -4\) or \(m > 8\)

To find the value of \(k\), we set the curve equation equal to the line equation since the line is tangent to the curve:

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

Rearrange to form a quadratic equation:

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

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

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

Here, \(a = 2\), \(b = k-2\), \(c = k-4\). Substitute these into the discriminant formula:

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

Simplify:

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

Expanding both sides:

\(k^2 - 4k + 4 = 8k - 32\)

Rearrange to form a quadratic equation:

\(k^2 - 12k + 36 = 0\)

Factorize:

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

Thus, \(k = 6\).

(a) To find when the line is tangent to the curve, substitute \(y = mx + c\) into \(xy = 16\):

\(x(mx + c) = 16\)

\(mx^2 + cx - 16 = 0\)

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

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

Here, \(a = m\), \(b = c\), \(c = -16\).

\(c^2 - 4(m)(-16) = 0\)

\(c^2 + 64m = 0\)

\(m = -\frac{c^2}{64}\)

(b) Substitute \(m = -4\) into the equation:

\(x(-4x + c) = 16\)

\(-4x^2 + cx - 16 = 0\)

For two distinct points, the discriminant must be positive:

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

\(c^2 - 4(-4)(-16) > 0\)

\(c^2 - 256 > 0\)

\(c^2 > 256\)

\(c > 16\) or \(c < -16\)

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