Past Exam Questions

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

1. **Translation**: The graph is translated vertically downwards by 2 units. This is represented by the vector \(\begin{pmatrix} 0 \\ -2 \end{pmatrix}\).

2. **Stretch**: The graph is stretched parallel to the x-axis with a scale factor of 2. This means that each x-coordinate is multiplied by 2, effectively widening the graph horizontally.

These transformations can be applied in either order to achieve the graph of \(y = g(x)\).

To transform \(y = f(x)\) to \(y = g(x)\), we perform the following transformations:

1. Stretch the graph of \(y = f(x)\) by a factor of 2 in the y-direction. This means that every y-coordinate of the graph is multiplied by 2.

2. Translate the resulting graph by \(\begin{pmatrix} -6 \\ 0 \end{pmatrix}\), which means shifting the graph 6 units to the left.

These transformations can be applied in either order to achieve the same result.

Start with the function \(f(x) = x^2 - 2x + 5\).

1. Stretch parallel to the x-axis with scale factor \(\frac{1}{2}\): Replace \(x\) with \(2x\). The function becomes \((2x)^2 - 2(2x) + 5 = 4x^2 - 4x + 5\).

2. Reflection in the y-axis: Replace \(x\) with \(-x\). The function becomes \((-2x)^2 - 2(-2x) + 5 = 4x^2 + 4x + 5\).

3. Stretch parallel to the y-axis with scale factor 3: Multiply the entire function by 3. The function becomes \(3(4x^2 + 4x + 5) = 12x^2 + 12x + 15\).

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

(a) The stretch in the y-direction changes the distance from the center to the maximum point from 2 to 6 (since the maximum point is at 12 and the minimum at 0, the center is at 6). Therefore, the scale factor of the stretch is \(\frac{6}{2} = 3\).

(b) The radius of the original circle is the distance from the center to the maximum point before the stretch, which is 2.

(c) After the translation \(\begin{pmatrix} 7 \\ -3 \end{pmatrix}\), the center of the circle moves from (1, 5) to \((1+7, 5-3) = (8, 2)\).

(d) The original center of the circle is found by reversing the translation from (8, 2): \((8-7, 2+3) = (1, 5)\).

To find \(g(x)\) in terms of \(f(x)\), we need to apply the given transformations to \(f(x)\).

1. The first transformation is a stretch in the \(x\)-direction by a factor of 0.5. This means we replace \(x\) with \(2x\) in the function \(f(x)\). So, the function becomes \(f(2x)\).

2. The second transformation is a translation by \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\), which means we add 1 to the entire function. Thus, the function becomes \(f(2x) + 1\).

Therefore, the expression for \(g(x)\) in terms of \(f(x)\) is \(g(x) = f(2x) + 1\).

(a) To express \(f(x) = x^2 - 4x + 9\) in the form \((x-a)^2 + b\), complete the square:

\(f(x) = (x-2)^2 + 5\)

(b) To express \(g(x) = 2x^2 + 4x + 12\) in the form \(2[(x+c)^2 + d]\), first factor out the 2:

\(g(x) = 2(x^2 + 2x + 6)\)

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

\(x^2 + 2x + 6 = (x+1)^2 + 5\)

Thus, \(g(x) = 2[(x+1)^2 + 5]\)

(c) Express \(g(x)\) in the form \(kf(x+h)\):

\(g(x) = 2(x^2 + 2x + 6) = 2[(x+3)^2 + 5]\)

So, \(g(x) = 2f(x+3)\) where \(k=2\) and \(h=3\)

(d) The transformations from \(y = f(x)\) to \(y = g(x)\) are:

- Translation by \(\begin{pmatrix} -3 \\ 0 \end{pmatrix}\)

- Stretch in the y-direction by a factor of 2

(a) To translate the curve \(y = x^2 + 2x - 5\) by \(\begin{pmatrix} -1 \\ 3 \end{pmatrix}\), we substitute \(x + 1\) for \(x\) in the equation and add 3 to the result:

\(y = (x+1)^2 + 2(x+1) - 5 + 3\)

Expanding, we get:

\(y = (x^2 + 2x + 1) + (2x + 2) - 5 + 3\)

\(y = x^2 + 4x + 1\)

(b) The transformation from \(y = x^2 + 2x - 5\) to \(y = 4x^2 + 4x - 5\) is a horizontal stretch. The coefficient of \(x^2\) changes from 1 to 4, indicating a stretch in the x-direction by a factor of \(\frac{1}{2}\), keeping the y-axis invariant.

(a) Start with the equation \(\log_3(2x + 1) = 1 + 2\log_3(x - 1)\).

Using the law of logarithms, \(2\log_3(x - 1) = \log_3((x - 1)^2)\).

Thus, the equation becomes \(\log_3(2x + 1) = \log_3(3) + \log_3((x - 1)^2)\).

This simplifies to \(\log_3(2x + 1) = \log_3(3(x - 1)^2)\).

Equating the arguments gives \(2x + 1 = 3(x - 1)^2\).

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

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

Rearranging gives \(3x^2 - 8x + 2 = 0\).

(b) Use the quadratic equation from part (a) to solve for \(y\):

\(3(2y)^2 - 8(2y) + 2 = 0\) simplifies to \(12y^2 - 16y + 2 = 0\).

Divide through by 2: \(6y^2 - 8y + 1 = 0\).

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6, b = -8, c = 1\).

\(y = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 6 \cdot 1}}{12}\).

\(y = \frac{8 \pm \sqrt{64 - 24}}{12}\).

\(y = \frac{8 \pm \sqrt{40}}{12}\).

\(y = \frac{8 \pm 2\sqrt{10}}{12}\).

\(y = \frac{4 \pm \sqrt{10}}{6}\).

Calculating gives \(y \approx 1.19\) (to 2 decimal places).

Start by using the property \(\log_{10} 10 = 1\) and \(\log_{10} 10^{-1} = -1\).

Rearrange the equation:

\(\log_{10}(2x + 1) = 2\log_{10}(x + 1) - 1\)

\(\log_{10}(2x + 1) + 1 = 2\log_{10}(x + 1)\)

\(\log_{10}(10(2x + 1)) = \log_{10}((x + 1)^2)\)

Equate the arguments:

\(10(2x + 1) = (x + 1)^2\)

Expand and simplify:

\(20x + 10 = x^2 + 2x + 1\)

\(x^2 - 18x - 9 = 0\)

Solve the quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1, b = -18, c = -9\).

\(x = \frac{18 \pm \sqrt{324 + 36}}{2}\)

\(x = \frac{18 \pm \sqrt{360}}{2}\)

\(x = \frac{18 \pm 18.9737}{2}\)

\(x = 18.487\) and \(x = -0.487\)

(i) Start with the equation \(\log_{10}(x-4) = 2 - \log_{10} x\).

Using the law of logarithms for subtraction, \(\log_{10}(x-4) + \log_{10} x = 2\).

This can be rewritten as \(\log_{10}((x-4)x) = 2\).

Since \(\log_{10} 100 = 2\), we have \((x-4)x = 100\).

Expanding gives \(x^2 - 4x = 100\).

Rearranging, we obtain the quadratic equation \(x^2 - 4x - 100 = 0\).

(ii) Solve the quadratic equation \(x^2 - 4x - 100 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = -100\).

Calculate the discriminant: \(b^2 - 4ac = (-4)^2 - 4 \times 1 \times (-100) = 16 + 400 = 416\).

Thus, \(x = \frac{4 \pm \sqrt{416}}{2}\).

Calculate \(\sqrt{416} \approx 20.396\).

So, \(x = \frac{4 \pm 20.396}{2}\).

This gives solutions \(x = 12.198\) and \(x = -8.198\).

Since \(x\) must be positive, the solution is \(x = 12.2\) (to 3 significant figures).

Start with the equation: \(2\log_2 x = 3 + \log_2(x + 1)\).

Apply the power rule of logarithms: \(\log_2 x^2 = 3 + \log_2(x + 1)\).

Use the property \(\log_a b - \log_a c = \log_a \left(\frac{b}{c}\right)\):

\(\log_2 \left(\frac{x^2}{x+1}\right) = 3\).

Convert the logarithmic equation to an exponential equation:

\(\frac{x^2}{x+1} = 2^3 = 8\).

Multiply both sides by \(x+1\):

\(x^2 = 8(x + 1)\).

Expand and rearrange to form a quadratic equation:

\(x^2 - 8x - 8 = 0\).

Solve the quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -8\), \(c = -8\).

Calculate the discriminant: \(b^2 - 4ac = (-8)^2 - 4(1)(-8) = 64 + 32 = 96\).

\(x = \frac{8 \pm \sqrt{96}}{2}\).

\(x = \frac{8 \pm 4\sqrt{6}}{2}\).

\(x = 4 \pm 2\sqrt{6}\).

Calculate the positive root: \(x = 4 + 2\sqrt{6} \approx 8.90\).

Thus, the solution is \(x = 8.90\).

Start with the equation \(\log_{10}(x+9) = 2 + \log_{10} x\).

Use the logarithm property: \(\log_{10}(x+9) = \log_{10}(10^2) + \log_{10} x\).

This simplifies to \(\log_{10}(x+9) = \log_{10}(100x)\).

Since the logarithms are equal, set the arguments equal: \(x+9 = 100x\).

Rearrange to solve for \(x\):

\(x + 9 = 100x\)

\(9 = 99x\)

\(x = \frac{9}{99} = \frac{1}{11}\).

(i) Start with the equation \(\log_2(x+5) = 5 - \log_2 x\).

Using the law of logarithms, \(\log_2(x+5) + \log_2 x = 5\).

This can be rewritten as \(\log_2((x+5)x) = 5\).

Therefore, \((x+5)x = 2^5\).

So, \(x^2 + 5x = 32\).

Rearranging gives the quadratic equation \(x^2 + 5x - 32 = 0\).

(ii) Solve the quadratic equation \(x^2 + 5x - 32 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 5\), \(c = -32\).

Calculate the discriminant: \(b^2 - 4ac = 5^2 - 4 \times 1 \times (-32) = 25 + 128 = 153\).

Thus, \(x = \frac{-5 \pm \sqrt{153}}{2}\).

The positive solution is \(x = \frac{\sqrt{153} - 5}{2}\), which is approximately \(x = 3.68\).

(i) Start with the equation \(\log_{10}(x+5) = 2 - \log_{10} x\).

Use the property of logarithms: \(\log_{10} a - \log_{10} b = \log_{10} \left( \frac{a}{b} \right)\).

Rearrange the equation: \(\log_{10}(x+5) + \log_{10} x = 2\).

Combine the logarithms: \(\log_{10}((x+5)x) = 2\).

Convert the logarithmic equation to an exponential equation: \((x+5)x = 10^2\).

Simplify: \(x^2 + 5x = 100\).

Thus, the quadratic equation is \(x^2 + 5x - 100 = 0\).

(ii) Solve the quadratic equation \(x^2 + 5x - 100 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 5\), and \(c = -100\).

Calculate the discriminant: \(b^2 - 4ac = 5^2 - 4 \times 1 \times (-100) = 25 + 400 = 425\).

Find the roots: \(x = \frac{-5 \pm \sqrt{425}}{2}\).

Since \(x\) must be positive, choose the positive root: \(x = \frac{\sqrt{425} - 5}{2}\).

Approximate the value: \(x \approx 7.81\).

Start with the equation \(4^{x-2} = 4^x - 4^2\).

Rewrite \(4^{x-2}\) as \(\frac{4^x}{4^2}\), so the equation becomes:

\(\frac{4^x}{16} = 4^x - 16\).

Multiply through by 16 to clear the fraction:

\(4^x = 16(4^x - 16)\).

Expand the right side:

\(4^x = 16 \cdot 4^x - 256\).

Rearrange to isolate terms involving \(4^x\):

\(4^x - 16 \cdot 4^x = -256\).

Factor out \(4^x\):

\(4^x(1 - 16) = -256\).

Simplify:

\(-15 \cdot 4^x = -256\).

Divide both sides by -15:

\(4^x = \frac{256}{15}\).

Take the logarithm of both sides to solve for \(x\):

\(x \log(4) = \log\left(\frac{256}{15}\right)\).

Solve for \(x\):

\(x = \frac{\log\left(\frac{256}{15}\right)}{\log(4)}\).

Calculate the value:

\(x \approx 2.047\).

Start with the equation:

\(\frac{2^x + 1}{2^x - 1} = 5\)

Multiply both sides by \(2^x - 1\):

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

Expand the right side:

\(2^x + 1 = 5 \cdot 2^x - 5\)

Rearrange the terms:

\(2^x + 1 = 5 \cdot 2^x - 5\)

\(1 + 5 = 5 \cdot 2^x - 2^x\)

\(6 = 4 \cdot 2^x\)

Divide both sides by 4:

\(2^x = \frac{6}{4} = 1.5\)

Take the logarithm base 2 of both sides:

\(x = \log_2(1.5)\)

Calculate \(x\) using a calculator:

\(x \approx 0.585\)

Thus, the solution is \(x = 0.585\) to 3 significant figures.

Start with the equation \(3^{x+2} = 3^x + 3^2\).

Rewrite \(3^{x+2}\) as \(3^x imes 3^2\), giving \(3^x imes 9 = 3^x + 9\).

Factor out \(3^x\) from the left side: \(3^x (9 - 1) = 9\).

This simplifies to \(3^x imes 8 = 9\).

Divide both sides by 8: \(3^x = \frac{9}{8}\).

Take the logarithm of both sides: \(x \ln 3 = \ln \left(\frac{9}{8}\right)\).

Solve for \(x\): \(x = \frac{\ln \left(\frac{9}{8}\right)}{\ln 3}\).

Calculate \(x\) to 3 significant figures: \(x \approx 0.107\).

Start by substituting \(u = 3^x\). Then \(3^{-x} = \frac{1}{3^x} = \frac{1}{u}\).

The equation becomes \(u = 2 + \frac{1}{u}\).

Multiply through by \(u\) to clear the fraction: \(u^2 = 2u + 1\).

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

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

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4(1)(-1) = 4 + 4 = 8\).

Thus, \(u = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\).

Since \(u = 3^x\) must be positive, choose \(u = 1 + \sqrt{2}\).

Then \(3^x = 1 + \sqrt{2}\).

Take the logarithm of both sides: \(x \log 3 = \log(1 + \sqrt{2})\).

Thus, \(x = \frac{\log(1 + \sqrt{2})}{\log 3}\).

Calculate \(x \approx 0.802\) to 3 significant figures.

(i) Given \(y = 2^x\), we have \(2^{-x} = \frac{1}{y}\). Substitute into the equation:

\(2^x - 2^{-x} = y - \frac{1}{y} = 1\).

Multiply through by \(y\) to clear the fraction:

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

Rearrange to form a quadratic equation:

\(y^2 - y - 1 = 0\).

(ii) Solve the quadratic equation \(y^2 - y - 1 = 0\) using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -1, c = -1\).

\(y = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\).

Since \(y = 2^x\) and \(y > 0\), choose the positive root:

\(y = \frac{1 + \sqrt{5}}{2}\).

Now solve \(2^x = \frac{1 + \sqrt{5}}{2}\) by taking logarithms:

\(x = \log_2 \left( \frac{1 + \sqrt{5}}{2} \right)\).

Calculate \(x \approx 0.694\).

Let \(u = 4^x\). Then \(4^{-x} = \frac{1}{u}\).

The equation becomes \(u = 3 + \frac{1}{u}\).

Multiply through by \(u\) to clear the fraction: \(u^2 = 3u + 1\).

Rearrange to form a quadratic equation: \(u^2 - 3u - 1 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -3\), \(c = -1\).

\(u = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}\).

Since \(u = 4^x\) must be positive, take the positive root: \(u = \frac{3 + \sqrt{13}}{2}\).

Calculate \(u \approx 3.30277563773\).

Since \(u = 4^x\), solve for \(x\): \(x = \log_4(u)\).

\(x = \frac{\log(u)}{\log(4)}\).

\(x \approx \frac{\log(3.30277563773)}{\log(4)} \approx 0.862\).