Test of Mathematics for University Admission

TMUA Algebra questions

12 original TMUA algebra questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Split a modulus into casesLocate quadratic roots: discriminant, sum, productSolve a quadratic inequality, then count integersCoefficients in a product of expansionsUse the discriminant to find a range of valuesRationalise the denominator with the conjugateWrite everything as powers of one baseSolve an inequality with a sign table, then count integers

Question 1TMUA Paper 1Warm-up

How many real solutions does the equation

\lvert x^2 - 4 \rvert = 3

have?

A $0$
B $2$
C $3$
D $4$
E $6$

Show the answer and worked solution

Answer: D · Method: Split a modulus into cases

Remove the modulus: either $x^2 - 4 = 3$ or $x^2 - 4 = -3$ .
So $x^2 = 7$ or $x^2 = 1$ , giving $x = \pm\sqrt{7}$ or $x = \pm 1$ : four distinct real values.

Why the tempting wrong answers are wrong

B: You only solved $x^2 - 4 = 3$ . A modulus equation splits into two cases.

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Question 2TMUA Paper 1Warm-up

For how many integers

n

n^2 - 10n + 16 < 0

A $4$
B $5$
C $6$
D $7$
E $8$

Show the answer and worked solution

Answer: B · Method: Solve a quadratic inequality, then count integers

Factorise: $(n - 2)(n - 8) < 0$ , so $2 < n < 8$ .
The integers are $3, 4, 5, 6, 7$ : five of them.

Why the tempting wrong answers are wrong

D: The inequality is strict, so $n = 2$ and $n = 8$ don't count.

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Question 3TMUA Paper 1Warm-up

What is the coefficient of

x^{2}

in the expansion of

(2 - 3x)^{6}

A $144$
B $2159$
C $2160$
D $2161$
E $4860$

Show the answer and worked solution

Answer: C · Method: Coefficients in a product of expansions

The $x^{2}$ term is $\binom{6}{2} \times 2^{4} \times (-3x)^{2}$ .
That's $15 \times 16 \times 9 = 2160$ .

Why the tempting wrong answers are wrong

A: Don't forget the binomial coefficient $\binom{6}{2} = 15$ .
E: The power of $x$ comes from the second term, so it's $2^{4}(-3x)^{2}$ , not the other way round.

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Question 4TMUA Paper 1Warm-up

Given that

\dfrac{2 + \sqrt{7}}{4 - \sqrt{7}} = p + q\sqrt{7}

, where

p

and

q

are rational, what is

q

A $-\dfrac{2}{3}$
B $-\dfrac{1}{3}$
C $\dfrac{6}{23}$
D $\dfrac{2}{3}$
E $\dfrac{5}{3}$

Show the answer and worked solution

Answer: D · Method: Rationalise the denominator with the conjugate

Multiply top and bottom by the conjugate $4 + \sqrt{7}$ . The bottom becomes $16 - 7 = 9$ .
The top becomes $(2 + \sqrt{7})(4 + \sqrt{7}) = 15 + 6\sqrt{7}$ .
So $p = \dfrac{5}{3}$ and $q = \dfrac{2}{3}$ .

Why the tempting wrong answers are wrong

C: $(4 - \sqrt{7})(4 + \sqrt{7}) = 16 - 7$ , not $16 + 7$ .
E: That's $p$ , the rational part.

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Question 5TMUA Paper 1Warm-up

Given that

\dfrac{1 + \sqrt{7}}{2 - \sqrt{7}} = p + q\sqrt{7}

, where

p

and

q

are rational, what is

p + q

A $-5$
B $-4$
C $-3$
D $\dfrac{12}{11}$
E $4$

Show the answer and worked solution

Answer: B · Method: Rationalise the denominator with the conjugate

Multiply top and bottom by the conjugate $2 + \sqrt{7}$ . The bottom becomes $4 - 7 = -3$ .
The top becomes $(1 + \sqrt{7})(2 + \sqrt{7}) = 9 + 3\sqrt{7}$ .
So $p = -3$ and $q = -1$ , giving $p + q = -4$ .

Why the tempting wrong answers are wrong

C: That's $p$ on its own.
D: $(2 - \sqrt{7})(2 + \sqrt{7}) = 4 - 7$ , not $4 + 7$ .

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Question 6TMUA Paper 1Warm-up

Solve

2^{x + 1} \times 4^{x} = 8^{2x - 1}

A $-\dfrac{4}{3}$
B $\dfrac{1}{3}$
C $\dfrac{4}{3}$
D $\dfrac{7}{3}$
E $\dfrac{10}{3}$

Show the answer and worked solution

Answer: C · Method: Write everything as powers of one base

Write everything as a power of $2$ : $2^{x + 1} \times 2^{2x} = 2^{6x - 3}$ .
Add the powers on the left and compare: $3x + 1 = 6x - 3$ .
So $-3x = -4$ and $x = \dfrac{4}{3}$ .

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Question 7TMUA Paper 1Warm-up

Solve

3^{x + 1} \times 9^{2x} = 27^{2x + 3}

A $-9$
B $-8$
C $-7$
D $2$
E $8$

Show the answer and worked solution

Answer: B · Method: Write everything as powers of one base

Write everything as a power of $3$ : $3^{x + 1} \times 3^{4x} = 3^{6x + 9}$ .
Add the powers on the left and compare: $5x + 1 = 6x + 9$ .
So $-x = 8$ and $x = -8$ .

Why the tempting wrong answers are wrong

D: Write $9 = 3^2$ and $27 = 3^3$ before comparing powers.

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Question 8TMUA Paper 1Warm-up

How many integers

x

satisfy

x^2 - 2x - 3 < 0

A $1$
B $2$
C $3$
D $4$
E $5$

Show the answer and worked solution

Answer: C · Method: Solve an inequality with a sign table, then count integers

Factorise: $(x + 1)(x - 3) < 0$ , so $-1 < x < 3$ .
That's $3$ integers.

Why the tempting wrong answers are wrong

E: The inequality is strict, so the roots themselves don't count.

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Question 9TMUA Paper 1Standard

For which values of

k

x^2 + 2kx + (-3k)

positive for all real values of

x

A $-3 < k < 0$
B $0 < k < 3$
C $k < -3 \text{ or } k > 0$
D $-3 \le k \le 0$
E $k > -3$

Show the answer and worked solution

Answer: A · Method: Use the discriminant to find a range of values

A quadratic with a positive $x^2$ coefficient is positive for every $x$ exactly when it has no real roots: its discriminant is negative.
The discriminant is $(2k)^2 - 4(-3k) = 4(k^2 + 3k)$ .
That's $4(k + 3)(k)$ , which is negative exactly when $-3 < k < 0$ .

Why the tempting wrong answers are wrong

C: That's where the discriminant is positive, so the quadratic crosses the $x$ -axis.
D: At $k = -3$ or $k = 0$ the discriminant is $0$ : the quadratic touches zero, so it isn't positive for every $x$ .

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Question 10TMUA Paper 1Standard

How many real solutions does the equation

\lvert x^2 - 5x \rvert = 1

have?

A $1$
B $2$
C $4$
D $5$
E $6$

Show the answer and worked solution

Answer: C · Method: Split a modulus into cases

Split the modulus: $x^2 - 5x - 1 = 0$ or $x^2 - 5x + 1 = 0$ .
The first has discriminant $25 + 4 > 0$ : two solutions.
The second has discriminant $25 - 4 = 21 > 0$ : two more.
That's $4$ solutions in total.

Why the tempting wrong answers are wrong

B: Don't forget the case where the inside of the modulus is negative.

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Question 11TMUA Paper 1Standard

What is the coefficient of

x^{2}

in the expansion of

(1 + 2x)^{6}(1 + 3x)^{2}

A $-3$
B $60$
C $69$
D $141$
E $142$

Show the answer and worked solution

Answer: D · Method: Coefficients in a product of expansions

Expand each bracket up to $x^{2}$ : the coefficients are $1, 12, 60$ and $1, 6, 9$ .
Add up every pair whose powers total 2: $(1)(9) + (12)(6) + (60)(1) = 141$ .

Why the tempting wrong answers are wrong

A: Check the signs in the second bracket.
C: You also need the cross terms, where the powers from each bracket add to the power you want.

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Question 12TMUA Paper 1Challenging

The equation

x^2 + kx + 9 = 0

has two distinct real roots, both greater than

1

. Which of the following describes all possible values of

k

A $k < -6$
B $-10 < k < -6$
C $k > 6$
D $-10 < k < -2$
E $k < -10$

Show the answer and worked solution

Answer: B · Method: Locate quadratic roots: discriminant, sum, product

Distinct real roots need $k^2 - 36 > 0$ , so $k < -6$ or $k > 6$ .
Both roots greater than $1$ needs the sum of the roots, $-k$ , to exceed $2$ (so $k < -2$ ), and the curve to be positive at $x = 1$ : $1 + k + 9 > 0$ , so $k > -10$ .
Combining all three conditions gives $-10 < k < -6$ .

Why the tempting wrong answers are wrong

A: This ignores the condition at $x = 1$ . For example, $k = -11$ gives one root below $1$ .
C: For $k > 6$ the sum of the roots, $-k$ , is negative, so both roots are negative.
D: This forgets that the roots must be real and distinct: you also need $k < -6$ .

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