Mathematics 9709/21 — May/June 2022

Cambridge AS Level · Pure Mathematics 2 · worked solutions for every part, with the mark scheme

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Topics Logarithmic and Exponential Functions · Trigonometry · Integration · Differentiation · Algebra · Numerical Solution of Equations

Q1Logarithmic and Exponential FunctionsFree sample

Fig. 1

The variables $x$ and $y$ satisfy the equation $y = 4^{2x-a}$ , where $a$ is an integer. As shown in the diagram, the graph of $\ln y$ against $x$ is a straight line passing through the point $(0, -20.8)$ , where the second coordinate is given correct to 3 significant figures.

(a)

Show that the gradient of the straight line is $\ln 16$ .

DifficultyMedium-Easy

Worked solution

Approach

Take the natural logarithm of both sides of the given equation to linearise it into the form $\ln y = mx + c$ . Then identify the gradient $m$ .

Working

Given $y = 4^{2x-a}$ .

Take $\ln$ of both sides:

\ln y = \ln(4^{2x-a})

Using the law of logarithms $\ln(b^k) = k \ln b$ :

\ln y = (2x - a) \ln 4

Expand the right side:

\ln y = 2x \ln 4 - a \ln 4

This is a linear equation in the form $\ln y = mx + c$ , where the gradient $m = 2 \ln 4$ .

Using the logarithm law $k \ln b = \ln(b^k)$ :

m = \ln(4^2) = \ln 16

Thus, the gradient of the straight line is $\ln 16$ .

Answer

The gradient is $\ln 16$ .

Final answer

The gradient is $\ln 16$ .

Detailed explanation

Walkthrough

We start with the exponential equation $y = 4^{2x-a}$ . To analyse the graph of $\ln y$ against $x$ , we take the natural logarithm of both sides. This allows us to bring the exponent down using the power rule for logarithms, $\ln(b^k) = k \ln b$ , giving $\ln y = (2x - a) \ln 4$ . Expanding this gives $\ln y = (2 \ln 4)x - a \ln 4$ , which is in the standard linear form $Y = mX + c$ where $Y = \ln y$ and $X = x$ . The gradient is the coefficient of $x$ , which is $2 \ln 4$ . We can rewrite this using the logarithm law $k \ln b = \ln(b^k)$ to get $\ln(4^2) = \ln 16$ , proving the required result.

Key Takeaways

Taking the logarithm of both sides of an exponential equation is a standard technique to linearise it. The power rule for logarithms $\ln(b^k) = k \ln b$ is essential for moving the exponent to the front. Comparing the result to $y = mx + c$ allows direct extraction of the gradient and y-intercept.

Common Mistakes

Failing to use brackets when expanding $(2x - a) \ln 4$ , leading to incorrect terms like $2x - a \ln 4$ instead of $2x \ln 4 - a \ln 4$ .
Not using the logarithm law $k \ln b = \ln(b^k)$ to simplify $2 \ln 4$ to $\ln 16$ as required by the question.

Things to Be Careful About

Ensure proper use of brackets when applying logarithm laws to expressions with multiple terms in the exponent.
The question asks to show the gradient is $\ln 16$ , so the final step must explicitly show $2 \ln 4 = \ln 16$ .

Techniques used

take natural logarithm of both sidesapply power rule for logarithmsidentify gradient from linear form

(b)

Determine the value of $a$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the y-intercept of the straight line from part (a) to find $a$ . The line passes through $(0, -20.8)$ , so when $x = 0$ , $\ln y = -20.8$ .

Working

From part (a), the equation of the line is:

\ln y = 2x \ln 4 - a \ln 4

The y-intercept occurs when $x = 0$ , giving:

\ln y = -a \ln 4

We are given that the line passes through $(0, -20.8)$ , so substitute $x = 0$ and $\ln y = -20.8$ :

-20.8 = -a \ln 4

Solve for $a$ :

a = \frac{20.8}{\ln 4}

Calculate the value:

a = \frac{20.8}{1.38629...} \approx 15.004...

Since $a$ is an integer, $a = 15$ .

Answer

a = 15

Final answer

a = 15

Detailed explanation

Walkthrough

From part (a), we established the linear equation $\ln y = 2x \ln 4 - a \ln 4$ . The y-intercept is the value of $\ln y$ when $x = 0$ . Setting $x = 0$ gives $\ln y = -a \ln 4$ . The problem states the graph passes through $(0, -20.8)$ , meaning the y-intercept is $-20.8$ . Equating the two expressions for the y-intercept gives $-20.8 = -a \ln 4$ . Solving this for $a$ yields $a = 20.8 / \ln 4 \approx 15.004$ . Since the question specifies that $a$ is an integer, we round to the nearest integer to get $a = 15$ .

Key Takeaways

Once an equation is linearised, substituting known coordinates allows you to solve for unknown parameters. Always check the question for constraints on the parameter, such as being an integer, which helps in rounding or selecting the correct value.

Common Mistakes

Forgetting that the point is $(0, -20.8)$ on the $\ln y$ vs $x$ graph, and mistakenly using $y = -20.8$ instead of $\ln y = -20.8$ .
Not rounding to the nearest integer when the question specifies $a$ is an integer, leading to a non-integer answer like $15.004$ .

Things to Be Careful About

The coordinate given is $(0, -20.8)$ on a graph of $\ln y$ against $x$ , so the second coordinate is the value of $\ln y$ , not $y$ itself.
The question states $a$ is an integer, so the final answer must be an exact integer. Condone $15.0$ but require an integer form.

Techniques used

substitute coordinates into linear equationsolve for unknown parameter

The rest of this paper

6 more questions

Q2Trigonometry6M
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Q6Differentiation · Logarithmic and Exponential Functions8M
Q7Algebra · Integration · Logarithmic and Exponential Functions9M

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