Past Exam Questions

(i) To express \(f(x) = x^2 - 4x + 7\) in the form \((x-a)^2 + b\), complete the square:

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

The range of \(f\) is \(f(x) > 3\) since \(x > 2\).

(ii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 3\).

Rearrange to solve for \(x\):

\(x - 2 = \pm \sqrt{y - 3}\).

Since \(x > 2\), we take the positive root:

\(x = 2 + \sqrt{y - 3}\).

Thus, \(f^{-1}(x) = 2 + \sqrt{x-3}\).

The domain of \(f^{-1}\) is \(x > 3\).

(iii) Given \(f = hg\) and \(g(x) = x - 2\), we have:

\(f(x) = h(x)g(x) = h(x)(x-2)\).

Since \(f(x) = x^2 - 4x + 7\), equate:

\(h(x)(x-2) = x^2 - 4x + 7\).

Assume \(h(x) = x^2 + 3\) to satisfy the equation.

(i) To express \(2x^2 - 12x + 11\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 - 12x + 11 = 2(x^2 - 6x) + 11\)

\(= 2((x - 3)^2 - 9) + 11\)

\(= 2(x - 3)^2 - 18 + 11\)

\(= 2(x - 3)^2 - 7\)

(ii) The function \(f(x) = 2x^2 - 12x + 11\) is a quadratic function that opens upwards. It is one-one on the interval where it is either increasing or decreasing. The vertex of the parabola is at \(x = 3\). Therefore, the largest value of \(k\) for which \(f\) is one-one is 3.

(iii) To find \(f^{-1}(x)\), start with \(y = 2(x - 3)^2 - 7\).

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

\(\frac{y + 7}{2} = (x - 3)^2\)

\(x - 3 = \pm \sqrt{\frac{y + 7}{2}}\)

Since \(x \leq 3\), choose the negative root:

\(x = 3 - \sqrt{\frac{y + 7}{2}}\)

Thus, \(f^{-1}(x) = 3 - \sqrt{\frac{1}{2}(x + 7)}\).

The domain of \(f^{-1}\) is \(x \geq -7\).

(iv) With \(k = 1\), \(x + 3 \leq 1\) implies \(x \leq -2\). Therefore, the largest \(p\) is -2.

The composite function \(fg(x) = f(g(x)) = f(x + 3)\).

\(f(x + 3) = 2(x + 3)^2 - 12(x + 3) + 11\)

\(= 2(x^2 + 6x + 9) - 12x - 36 + 11\)

\(= 2x^2 + 12x + 18 - 12x - 36 + 11\)

\(= 2x^2 - 7\)

(a)(i) The function \(f(x) = (x - 3)^2 - 1\) is defined for \(x < a\). The greatest possible value of \(a\) is 3, as the function is one-one up to this point.

(a)(ii) For \(a = 3\), the range of \(f\) is \(y > -1\). To find \(f^{-1}(x)\), set \(y = (x - 3)^2 - 1\), so \(y + 1 = (x - 3)^2\). Solving for \(x\), we get \(x = 3 \pm \sqrt{1 + y}\). Since \(x < 3\), we take \(x = 3 - \sqrt{1 + y}\), hence \(f^{-1}(x) = 3 - \sqrt{1 + x}\).

(b)(i) \(gg(2x) = [(2x - 3)^2 - 3]^2\). Expanding, \((2x - 3)^2 = 4x^2 - 12x + 9\). Then \(gg(2x) = (4x^2 - 12x + 6)^2\). This can be expressed as \((2x - 3)^4 - 6(2x - 3)^2 + 9\).

(b)(ii) Expanding \((2x - 3)^4 - 6(2x - 3)^2 + 9\), we get \(16x^4 - 96x^3 + 216x^2 - 216x + 81 - 24x^2 + 72x - 54 + 9\). Simplifying, \(16x^4 - 96x^3 + 192x^2 - 144x + 36\).

(i) The smallest value of \(c\) is 2. This is because the function \(f(x) = (x-2)^2 + 2\) is defined for \(x \geq c\), and the smallest value for which the function is one-to-one is when \(c = 2\).

(ii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 2\). Rearrange to solve for \(x\):

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

\(x - 2 = \pm \sqrt{y-2}\)

\(x = \pm \sqrt{y-2} + 2\)

Since \(x \geq 4\), we take the positive root: \(x = \sqrt{y-2} + 2\).

Thus, \(f^{-1}(x) = \sqrt{x-2} + 2\).

The domain of \(f^{-1}\) is \(x > 6\) because \(f(x)\) is defined for \(x \geq 4\) and \(f(4) = 6\).

(iii) Solve \(ff(x) = 51\):

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

Set \(((x-2)^2)^2 + 2 = 51\)

\(((x-2)^2)^2 = 49\)

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

\(x-2 = \pm \sqrt{7}\)

Since \(x \geq 4\), \(x = 2 + \sqrt{7}\).

(i) To find the points of intersection, set \(f(x) = g(x)\):

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

Rearrange to form a quadratic equation:

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

\(\frac{1}{2}x^2 - \frac{3}{2}x - 6 = 0\)

Multiply through by 2 to clear fractions:

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

Factorize or use the quadratic formula to find \(x = 4\) and \(x = -3\).

Substitute back to find \(y\) values: \((4, 0)\) and \((-3, -3.5)\).

(ii) Solve \(f(x) > g(x)\):

\(\frac{1}{2}x - 2 > 4 + x - \frac{1}{2}x^2\)

Rearrange to form a quadratic inequality:

\(\frac{1}{2}x^2 - \frac{3}{2}x - 6 < 0\)

Factorize or use the quadratic formula to find intervals: \(x > 4\) and \(x < -3\).

(iii) Find \(fg(x) = f(x) \cdot g(x)\):

\(fg(x) = \left(\frac{1}{2}x - 2\right)\left(4 + x - \frac{1}{2}x^2\right)\)

Expand and simplify:

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

Complete the square or use calculus to find the range:

\(y \leq \frac{1}{4}\)

(iv) For \(h(x)\) to have an inverse, it must be one-to-one. Find the vertex of the parabola:

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

Thus, \(k = 1\) (accept \(k \geq 1\)).

Answer: (i)(a) \(f(x)>2\). (i)(b) \(g(x)>6\). (i)(c) \(2<fg(x)<4\). (ii) \(gf\) cannot be formed because the range of \(f\) is partly outside the domain of \(g\).

(i)(a) For \(f(x)=\dfrac{8}{x-2}+2\), where \(x>2\), the term \(\dfrac{8}{x-2}\) is always positive. Therefore

\(f(x)>2.\)

(i)(b) For \(g(x)=\dfrac{8}{x-2}+2\), where \(2<x<4\), the value approaches infinity as \(x\to2^+\), and as \(x\to4^-\),

\(g(x)\to\frac{8}{2}+2=6.\)

Since \(x=4\) is not included, the range is

\(g(x)>6.\)

(i)(c) The composite \(fg\) means \(f(g(x))\). Since \(g(x)>6\), let \(u=g(x)\), so \(u>6\). Then

\(f(u)=\frac{8}{u-2}+2.\)

For \(u>6\), we have \(u-2>4\), so

\(0<\frac{8}{u-2}<2.\)

Therefore

\(2<fg(x)<4.\)

(ii) To form \(gf\), the range of \(f\) must lie inside the domain of \(g\). The range of \(f\) is \(f(x)>2\), but the domain of \(g\) is only \(2<x<4\). Since values of \(f\) greater than or equal to \(4\) are outside the domain of \(g\), the composite function \(gf\) cannot be formed.

(i) To express \(4x^2 + 12x + 10\) in the form \((ax + b)^2 + c\), we complete the square:

\(4x^2 + 12x + 10 = 4(x^2 + 3x) + 10\).

Complete the square for \(x^2 + 3x\):

\(x^2 + 3x = (x + \frac{3}{2})^2 - \frac{9}{4}\).

Thus, \(4(x^2 + 3x) = 4((x + \frac{3}{2})^2 - \frac{9}{4}) = 4(x + \frac{3}{2})^2 - 9\).

Therefore, \(4x^2 + 12x + 10 = 4(x + \frac{3}{2})^2 - 9 + 10 = 4(x + \frac{3}{2})^2 + 1\).

So, \((ax + b)^2 + c = (2x + 3)^2 + 1\) with \(a = 2, b = 3, c = 1\).

(ii) Given \(fg(x) = 4x^2 + 12x + 10\) and \(f(x) = x^2 + 1\), we have:

\(g(x) = \frac{4x^2 + 12x + 10}{x^2 + 1}\).

From part (i), \(4x^2 + 12x + 10 = (2x + 3)^2 + 1\), so:

\(g(x) = \frac{(2x + 3)^2 + 1}{x^2 + 1} = 2x + 3\).

(iii) To find \((fg)^{-1}(x)\), solve \(y = (2x + 3)^2 + 1\) for \(x\):

\(y - 1 = (2x + 3)^2\).

\(2x + 3 = \pm \sqrt{y - 1}\).

\(x = \frac{\pm \sqrt{y - 1} - 3}{2}\).

Choose the positive root for \(x > 0\):

\((fg)^{-1}(x) = \frac{1}{2}\sqrt{x-1} - \frac{3}{2}\).

The domain of \((fg)^{-1}\) is \(x > 10\) since \(y = (2x + 3)^2 + 1 > 10\).

(i) Given \(fg(x) = f(x) \cdot g(x) = 6x^2 - 21\), substitute \(f(x) = 2x + 3\) and \(g(x) = ax^2 + b\):

\(2(ax^2 + b) + 3 = 6x^2 - 21\)

\(2ax^2 + 2b + 3 = 6x^2 - 21\)

Equating coefficients, \(2a = 6\) gives \(a = 3\), and \(2b + 3 = -21\) gives \(b = -12\).

(ii) For \(fg(x) = 6x^2 - 21\) to be defined, \(x \leq q\). Solve \(6x^2 - 21 \geq 3\):

\(6x^2 \geq 24\)

\(x^2 \geq 4\)

\(x \leq -2\) or \(x \geq 2\). Since \(x \leq q\), the greatest possible \(q = -2\).

(iii) Given \(q = -3\), find the range of \(fg\):

\(y = 6(-3)^2 - 21 = 33\)

Thus, the range is \(y \geq 33\).

(iv) Solve \(y = 6x^2 - 21\) for \(x\):

\(y + 21 = 6x^2\)

\(x^2 = \frac{y + 21}{6}\)

\(x = \pm \sqrt{\frac{y + 21}{6}}\)

Since \(x \leq q\), choose the negative root: \(x = -\sqrt{\frac{y + 21}{6}}\)

Thus, \((fg)^{-1}(x) = -\sqrt{\frac{x + 21}{6}}\).

The domain of \((fg)^{-1}\) is \(x \geq 33\).

(a) To find the minimum point, complete the square for \(y = x^2 - 8x + 5\):

\(y = (x-4)^2 - 16 + 5 = (x-4)^2 - 11\).

The minimum point is at \(x = 4\), giving \(y = -11\). Thus, the coordinates are \((4, -11)\).

(b) The curve is stretched by a factor of 2 parallel to the y-axis, so the y-coordinate of the minimum point becomes \(-11 \times 2 = -22\). Then, the curve is translated by \(\begin{pmatrix} 4 \\ 1 \end{pmatrix}\), so the x-coordinate becomes \(4 + 4 = 8\) and the y-coordinate becomes \(-22 + 1 = -21\). Thus, the coordinates of the minimum point of the transformed curve are \((8, -21)\).

(c) The equation of the transformed curve is found by applying the transformations to the original equation:

First, stretch by a factor of 2: \(y = 2(x^2 - 8x + 5) = 2x^2 - 16x + 10\).

Then, translate by \(\begin{pmatrix} 4 \\ 1 \end{pmatrix}\):

Replace \(x\) with \(x - 4\): \(y = 2((x-4)^2 - 11) + 1\).

Expand: \(y = 2(x^2 - 8x + 16 - 11) + 1 = 2x^2 - 16x + 10 + 1 = 2x^2 - 16x + 11\).

Thus, the equation of the transformed curve is \(y = 2x^2 - 32x + 107\).

(a) To express \(2x^2 - 8x + 14\) in the form \(2[(x-a)^2 + b]\), we complete the square for the quadratic expression:

Start with \(x^2 - 4x\). Complete the square:

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

Thus, \(2x^2 - 8x = 2((x-2)^2 - 4) = 2(x-2)^2 - 8\).

Now, add 14: \(2(x-2)^2 - 8 + 14 = 2(x-2)^2 + 6\).

Therefore, \(2x^2 - 8x + 14 = 2[(x-2)^2 + 3]\).

(b) To map \(y = f(x) = x^2\) to \(y = g(x) = 2x^2 - 8x + 14\), we perform the following transformations:

1. Translate by \(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\): This shifts the graph right by 2 units and up by 3 units.

OR

1. Stretch in the \(y\) direction by a factor of 2: This scales the graph vertically.

The transformation from \(y = f(x)\) to \(y = 3 - f(x)\) involves two steps:

1. Reflection in the x-axis: This changes \(y = f(x)\) to \(y = -f(x)\).

2. Translation 3 units in the positive y-direction: This changes \(y = -f(x)\) to \(y = 3 - f(x)\).

Alternatively, the transformations can be performed in reverse order:

1. Translation 3 units in the negative y-direction: This changes \(y = f(x)\) to \(y = f(x) - 3\).

2. Reflection in the x-axis: This changes \(y = f(x) - 3\) to \(y = 3 - f(x)\).

(a) The transformation from \(y = f(x)\) to \(y = f(2x)\) involves a horizontal compression by a factor of \(\frac{1}{2}\). This is because the function \(f(2x)\) compresses the x-values by a factor of 2, effectively scaling the x-direction by \(\frac{1}{2}\).

The transformation from \(y = f(2x)\) to \(y = f(2x) - 3\) involves a vertical translation downwards by 3 units. This is because subtracting 3 from the function shifts the graph down by 3 units.

(b) To find the corresponding point on the original curve \(y = f(x)\), we reverse the transformations. The point \(P(5, 6)\) on \(y = f(2x) - 3\) corresponds to \(y = f(2x)\) at \((5, 9)\) after reversing the vertical translation.

Next, reverse the horizontal compression by multiplying the x-coordinate by 2, giving \((10, 9)\) on \(y = f(x)\).

(a) To express \(f(x) = -3x^2 + 12x + 2\) in the form \(-3(x-a)^2 + b\), we complete the square:

Start with \(-3(x^2 - 4x) + 2\).

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

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

Substitute back:

\(-3((x-2)^2 - 4) + 2 = -3(x-2)^2 + 12 + 2 = -3(x-2)^2 + 14\).

Thus, \(a = 2\) and \(b = 14\).

(e) Given the translation vector \(\begin{pmatrix} -3 \\ 1 \end{pmatrix}\), the function \(g(x)\) is obtained by substituting \(x+3\) into \(f(x)\) and adding 1:

\(g(x) = -3((x+3)-2)^2 + 14 + 1\).

Simplify:

\(g(x) = -3(x+1)^2 + 15\).

Expand \(-3(x+1)^2\):

\(-3(x^2 + 2x + 1) = -3x^2 - 6x - 3\).

Thus, \(g(x) = -3x^2 - 6x - 3 + 15 = -3x^2 - 6x + 12\).

(a) To express \(f(x) = x^2 - 2x + 5\) in completed square form, we complete the square:

\(f(x) = (x-1)^2 + 4\).

For \(g(x) = x^2 + 4x + 13\), complete the square:

\(g(x) = (x+2)^2 + 9\).

Now express \(g(x)\) in the form \(f(x+p) + q\):

\(g(x) = (x+2)^2 + 9 = ((x+3)-1)^2 + 4 + 5 = f(x+3) + 5\).

Thus, \(p = 3\) and \(q = 5\).

(b) The transformation from \(y = f(x)\) to \(y = g(x)\) is a translation. The completed square forms show a horizontal shift by \(-3\) and a vertical shift by \(5\). Therefore, the transformation is a translation by \(\begin{pmatrix} -3 \\ 5 \end{pmatrix}\).

The transformation from \(y = f(x)\) to \(y = 2f(x - 1)\) involves two steps:

1. Translation: The expression \(f(x - 1)\) indicates a translation of the graph by 1 unit in the positive x-direction.

2. Stretch: The factor of 2 in \(2f(x - 1)\) indicates a vertical stretch by a factor of 2.

Therefore, the graph is translated 1 unit to the right and then stretched vertically by a factor of 2.

(a) The transformation involves two steps:

1. A stretch parallel to the \(y\)-axis by a factor of 3. This means every \(y\)-coordinate of the graph \(y = f(x)\) is multiplied by 3.

2. A translation 4 units in the positive \(x\)-direction. This shifts the graph horizontally to the right by 4 units.

(b) The equation of the transformed graph can be written as \(y = 3f(x-4)\). This accounts for the stretch by a factor of 3 and the horizontal translation by 4 units.

(a) To express \(x^2 + 6x + 5\) in the form \((x + a)^2 + b\), we complete the square:

Start with \(x^2 + 6x\). Take half of the coefficient of \(x\), which is 3, and square it to get 9.

Add and subtract 9 inside the expression: \(x^2 + 6x = (x^2 + 6x + 9) - 9 = (x + 3)^2 - 9\).

Now, include the constant term 5: \((x + 3)^2 - 9 + 5 = (x + 3)^2 - 4\).

Thus, \(x^2 + 6x + 5 = (x + 3)^2 - 4\).

(b) The transformation from \(y = x^2\) to \(y = x^2 + 6x + 5\) involves completing the square as shown in part (a): \(y = (x + 3)^2 - 4\).

This represents a translation of the graph of \(y = x^2\) by \(\begin{pmatrix} -3 \\ -4 \end{pmatrix}\), which means 3 units to the left and 4 units down.

(a) The transformation shown is a reflection in the y-axis. The equation of the transformed graph is \(y = f(-x)\).

(b) The transformation shown is a vertical stretch with a scale factor of 2. The equation of the transformed graph is \(y = 2f(x)\).

(c) The transformation shown is a translation 4 units to the left and 3 units down. The equation of the transformed graph is \(y = f(x+4) - 3\).

The transformation from \(y = f(x)\) to \(y = f\left(\frac{1}{2}x\right)\) involves a horizontal stretch. Specifically, it is a stretch by a factor of 2 in the x-direction, as the x-values are effectively halved, making the graph wider.

The transformation from \(y = f(x)\) to \(y = 1 + f(x)\) involves a vertical translation. Specifically, it is a translation of 1 unit upwards in the y-direction, as 1 is added to the function value.

Therefore, the two transformations are: a stretch by a factor of 2 in the x-direction and a translation of 1 unit in the y-direction.

(a) Start with the equation \(y = x^2\). First, apply the reflection in the x-axis: \(y = -x^2\). Then apply the translation \(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\), which shifts the graph 3 units to the right and 1 unit down. The equation becomes \(y = -(x-3)^2 - 1\).

(b) Start with the equation \(y = x^2\). First, apply the translation \(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\), which shifts the graph 3 units to the right and 1 unit down: \(y = (x-3)^2 - 1\). Then apply the reflection in the x-axis: \(y = -(x-3)^2 + 1\).

(c) To map \(y = g(x)\) onto \(y = h(x)\), observe that \(y = g(x)\) is \(-(x-3)^2 - 1\) and \(y = h(x)\) is \(-(x-3)^2 + 1\). The transformation is a vertical translation of 2 units up, represented by \(\begin{pmatrix} 0 \\ 2 \end{pmatrix}\).