Mathematics 9709/31 — May/June 2016

Cambridge A-Level · Pure Mathematics 3 · worked solutions for every part, with the mark scheme

questions

marks

105

minutes

Topics Algebra · Trigonometry · Differentiation · Logarithmic and Exponential Functions · Integration · Differential Equations · +3 more

Q1AlgebraLogarithmic and Exponential FunctionsFree sample

(i)

Solve the equation $2|x - 1| = 3|x|$ .

DifficultyMedium-Easy

Worked solution

Approach

The equation involves absolute values. Since both sides are non-negative, squaring both sides removes the modulus signs without changing the solutions. Then solve the resulting quadratic.

Working

Squaring both sides:

4(x-1)^2 = 9x^2

Expand and rearrange:

4(x^2 - 2x + 1) = 9x^2

4x^2 - 8x + 4 = 9x^2

0 = 5x^2 + 8x - 4

Using the quadratic formula:

x = \frac{-8 \pm \sqrt{8^2 - 4(5)(-4)}}{2(5)}

x = \frac{-8 \pm \sqrt{64 + 80}}{10} = \frac{-8 \pm 12}{10}

Therefore:

x = \frac{-8 + 12}{10} = \frac{2}{5} \quad\text{or}\quad x = \frac{-8 - 12}{10} = -2

Answer

x = -2 \quad\text{or}\quad x = \frac{2}{5}

Final answer

x = -2 or x = 2/5

Detailed explanation

Walkthrough

The equation $2|x - 1| = 3|x|$ has two absolute-value terms, so the sign of each expression is unknown. A reliable method is to square both sides, because both sides are non-negative. This produces a quadratic equation.

Expanding $4(x-1)^2$ gives $4x^2 - 8x + 4$ , and rearranging gives $5x^2 + 8x - 4 = 0$ . Applying the quadratic formula gives two solutions: $x = \frac{2}{5}$ and $x = -2$ . Both values satisfy the original equation, so no solution is lost.

Key Takeaways

This question tests solving modulus equations by squaring, expanding brackets carefully, and solving a quadratic. It also shows that squaring is valid when both sides are guaranteed non-negative.

Common Mistakes

Expanding $(x-1)^2$ incorrectly as $x^2 - 1$ instead of $x^2 - 2x + 1$ .
Losing the negative solution when taking square roots.
Making sign errors when substituting into the quadratic formula.
Forgetting that squaring is only safe here because both sides are non-negative.

Things to Be Careful About

The mark scheme also accepts solving the linear equations $2(x-1) = \pm 3x$ . If using that method, make sure both cases are solved. With the squaring method, no extraneous roots are introduced because both sides of the original equation are non-negative.

Techniques used

square both sides to remove modulus signsexpand and rearrange into a quadratic equationsolve a quadratic equation using the quadratic formula

(ii)

Hence solve the equation $2|5^x - 1| = 3|5^x|$ , giving your answer correct to 3 significant figures.

DifficultyMedium

Worked solution

Approach

Let $y = 5^x$ . The equation then has exactly the same form as part (i). Use the solutions from part (i), discard the impossible negative value because $5^x > 0$ , and solve the remaining exponential equation using logarithms.

Working

Let $y = 5^x$ . From part (i):

2|y - 1| = 3|y| \Rightarrow y = -2 \text{ or } y = \frac{2}{5}

Since $y = 5^x > 0$ for all real $x$ , the solution $y = -2$ is impossible. Hence:

5^x = \frac{2}{5}

Take natural logarithms of both sides:

x \ln 5 = \ln\left(\frac{2}{5}\right)

x = \frac{\ln\left(\frac{2}{5}\right)}{\ln 5}

Evaluating:

x = \frac{-0.91629\ldots}{1.60944\ldots} = -0.56932\ldots

Correct to 3 significant figures:

x = -0.569

Answer

x = -0.569

Final answer

x = -0.569

Detailed explanation

Walkthrough

The key is to notice that the equation $2|5^x - 1| = 3|5^x|$ has the same shape as $2|x - 1| = 3|x|$ if we replace $x$ by $5^x$ . Writing $y = 5^x$ lets us reuse the answers from part (i): $y = -2$ or $y = \frac{2}{5}$ .

However, $y = 5^x$ is always positive, so $y = -2$ cannot occur. This is why the mark scheme asks for $-0.569$ only. The remaining equation is $5^x = \frac{2}{5}$ . Taking natural logarithms gives $x \ln 5 = \ln(2/5)$ , so $x = \frac{\ln(2/5)}{\ln 5} = -0.569$ to 3 significant figures.

Key Takeaways

This question shows how a substitution can turn a new-looking equation into one already solved. It also reinforces that exponentials are always positive, and that equations of the form $a^x = b$ are solved by taking logarithms.

Common Mistakes

Trying to solve the modulus equation from scratch instead of using the substitution.
Forgetting that $5^x > 0$ and attempting to solve $5^x = -2$ .
Rounding $-0.56932\ldots$ incorrectly; to 3 significant figures it is $-0.569$ .
Using a logarithm with the wrong base, or dividing the logarithms in the wrong order.

Things to Be Careful About

The mark scheme awards the final mark for $-0.569$ only, because the negative solution from part (i) is impossible. Make sure the final answer is given to 3 significant figures and that the negative sign is kept.

Techniques used

substitute 5^x as a new variableapply the previous modulus solutionsdiscard the negative solution using positivity of exponentialssolve an exponential equation by taking logarithms

The rest of this paper

9 more questions

Q2Integration5M
Q3Trigonometry5M
Q4Differential Equations6M
Q5Differentiation · Trigonometry6M
Q6Numerical Solution of Equations7M
Q7Differentiation9M
Q8Algebra10M
Q9Vectors11M
Q10Complex Numbers11M

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