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1
Find the complete set of values of \(x\) for which \(2x^2 - 5x - 3 \geq 0\).
- A\(x \leq -\dfrac{1}{2}\) or \(x \geq 3\)
- B\(x \leq -\dfrac12\)
- C\(x \geq 3\)
- D\(x \leq -3\) or \(x \geq \dfrac{1}{2}\)
- E\(-3 \leq x \leq \dfrac12\)
- F\(-\dfrac{1}{2} \leq x \leq 3\)
Worked solution
\((2x+1)(x-3) \geq 0\): the roots are \(-\frac12\) and \(3\) and the parabola opens upwards, so \(x \leq -\frac12\) or \(x \geq 3\).
Full method — how and why
A quadratic inequality with a leading coefficient other than \(1\). Factorise, find the roots, then use the shape of the parabola.
- Factorise: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\).Two numbers multiplying to \(2 \times (-3) = -6\) and adding to \(-5\): \(-6\) and \(1\), so split \(-5x\) as \(-6x + x\).
- Roots: \(x = -\frac12\) and \(x = 3\).Set each bracket to zero.
- The coefficient of \(x^2\) is positive, so the parabola opens upwards and is \(\geq 0\) outside the roots.A quick sketch is the safest way to convert a factorisation into an inequality answer.
- Answer: \(x \leq -\frac12\) or \(x \geq 3\), with \(\leq\) because equality is allowed.The question has \(\geq 0\), so the roots themselves are included.
Where it goes wrong
- Giving \(-\frac12 \leq x \leq 3\), which is where the expression is negative.
- Mixing up the roots as \(-3\) and \(\frac12\) - the root of \((2x+1)\) is \(-\frac12\), not \(-2\).
Takeaway Factorise, mark the roots, sketch the parabola: 'outside' for \(>0\), 'between' for \(<0\) when \(a > 0\).
Answer A · Simultaneous equations and inequalities (MM1.5) · question SIM-02
2
Find the set of values of \(x\) satisfying both \(3x - 7 < 2\) and \(x^2 > 4\).
- A\(x < -2\) or \(2 < x < 3\)
- B\(2 < x < 3\)
- C\(-2 < x < 3\)
- D\(x < 3\)
- E\(x > 2\)
- F\(x < -2\)
Worked solution
\(3x < 9 \Rightarrow x < 3\). \(x^2 > 4 \Rightarrow x < -2\) or \(x > 2\). Combining: \(x < -2\) or \(2 < x < 3\).
Full method — how and why
Two inequalities joined by 'and'. Solve separately, then take the overlap - the second one produces two intervals, which is where the care is needed.
- Solve the linear one: \(3x - 7 < 2 \Rightarrow 3x < 9 \Rightarrow x < 3\).Dividing by a positive number leaves the inequality direction unchanged.
- Solve the quadratic one: \(x^2 > 4 \Rightarrow |x| > 2 \Rightarrow x < -2\) or \(x > 2\).\(x^2 > 4\) is satisfied by large positive and large negative \(x\) - taking only the positive branch is the classic error.
- Intersect the sets on a number line: from \(x < 3\), keep the parts of \(x < -2\) (all of it) and of \(x > 2\) (only up to \(3\)).'Both' means intersection; drawing both solution sets on a single line prevents mistakes.
- Answer: \(x < -2\) or \(2 < x < 3\).The answer is itself two intervals - a perfectly legitimate solution set.
Where it goes wrong
- Dropping the branch \(x < -2\) and answering \(2 < x < 3\).
- Solving \(x^2 > 4\) as \(x > \pm 2\), which is not a meaningful statement.
Takeaway \(x^2 > k\) gives two intervals; intersect solution sets on a number line rather than in your head.
Answer A · Simultaneous equations and inequalities (MM1.5) · question SIM-03
3
The line \(y = x + k\) meets the circle \(x^2 + y^2 = 8\) at exactly one point. Find the possible values of \(k\).
- A\(k = \pm 2\sqrt{2}\)
- B\(k = \pm 8\)
- C\(k = \pm 4\)
- D\(k = \pm \sqrt{8}\)
- E\(k = 4\) only
- F\(k = \pm 2\)
Worked solution
\(x^2 + (x+k)^2 = 8 \Rightarrow 2x^2 + 2kx + k^2 - 8 = 0\). Tangent when \(4k^2 - 8(k^2 - 8) = 0 \Rightarrow -4k^2 + 64 = 0 \Rightarrow k = \pm 4\).
Full method — how and why
A line meeting a circle exactly once is a tangency condition: substitute and set the discriminant to zero.
- Substitute \(y = x + k\) into \(x^2 + y^2 = 8\): \(x^2 + (x + k)^2 = 8\).Eliminate \(y\) so the intersections are the roots of a quadratic in \(x\).
- Expand and collect: \(2x^2 + 2kx + (k^2 - 8) = 0\).\((x+k)^2 = x^2 + 2kx + k^2\), so the \(x^2\) terms combine to \(2x^2\).
- Exactly one point means one repeated root: discriminant \(= 0\), so \((2k)^2 - 4(2)(k^2 - 8) = 0\).Tangency is precisely the repeated-root case.
- Simplify: \(4k^2 - 8k^2 + 64 = 0 \Rightarrow -4k^2 = -64 \Rightarrow k^2 = 16 \Rightarrow k = \pm 4\).Two tangent lines exist - one on each side of the circle - so both signs are genuine.
Where it goes wrong
- Answering \(k = \pm 2\sqrt2\) by confusing the radius \(\sqrt8\) with the intercept.
- Giving only \(k = 4\); the geometry guarantees a matching tangent with \(k = -4\).
Takeaway Tangent to a circle: substitute and set the discriminant to zero - or use 'distance from centre \(=\) radius', which gives \(\frac{|k|}{\sqrt2} = 2\sqrt2\) just as fast.
Answer C · Simultaneous equations and inequalities (MM1.4) · question SIM-04
4
For how many integers \(n\) is \(n^2 - 10n + 16 < 0\)?
- A\(5\)
- B\(3\)
- C\(4\)
- D\(8\)
- E\(7\)
- F\(6\)
Worked solution
\((n-2)(n-8) < 0 \Rightarrow 2 < n < 8\), so \(n = 3, 4, 5, 6, 7\): five integers.
Full method — how and why
Counting integers in the solution set of a quadratic inequality. The counting at the end is where the marks are won or lost.
- Factorise: \(n^2 - 10n + 16 = (n - 2)(n - 8)\).Two numbers multiplying to \(16\), adding to \(-10\): \(-2\) and \(-8\).
- Negative between the roots: \(2 < n < 8\).Upward parabola, so it dips below the axis strictly between its roots.
- List the integers strictly inside: \(3, 4, 5, 6, 7\) - five values.Strict inequality excludes \(2\) and \(8\), where the expression equals zero.
Where it goes wrong
- Including the endpoints and answering \(7\).
- Counting \(8 - 2 = 6\) as the number of integers - the count between \(a\) and \(b\) exclusive is \(b - a - 1\).
Takeaway Write the integers out; the number strictly between \(a\) and \(b\) is \(b - a - 1\), not \(b - a\).
Answer A · Simultaneous equations and inequalities (MM1.5) · question SIM-05
5
Solve \(xy = 6\) and \(x - y = 1\). What is the sum of all possible values of \(x + y\)?
- A\(10\)
- B\(5\)
- C\(-5\)
- D\(7\)
- E\(1\)
- F\(0\)
Worked solution
\(x = y + 1\): \(y^2 + y - 6 = 0 \Rightarrow (y+3)(y-2) = 0\). Solutions \((3, 2)\) with \(x + y = 5\) and \((-2, -3)\) with \(x + y = -5\). The sum is \(0\).
Full method — how and why
A product and a difference. Substituting the linear relation gives a quadratic with two solution pairs, and the question wants a total over both.
- From \(x - y = 1\), write \(x = y + 1\).Substituting the linear equation keeps things to one variable.
- Substitute into \(xy = 6\): \((y + 1)y = 6 \Rightarrow y^2 + y - 6 = 0\).Expand before rearranging into standard form.
- Factorise: \((y + 3)(y - 2) = 0\), so \(y = -3\) or \(y = 2\).Both are valid - nothing in the question restricts the sign.
- Find the pairs: \(y = 2 \Rightarrow x = 3\), giving \(x + y = 5\); \(y = -3 \Rightarrow x = -2\), giving \(x + y = -5\). Sum of these: \(0\).The question asks for the sum of all possible values of \(x + y\), so compute \(x+y\) for each pair and add.
Where it goes wrong
- Finding only the positive pair and answering \(5\).
- Adding all four numbers \(3 + 2 + (-2) + (-3)\) - which happens to be \(0\) too, but for the wrong reason; on another question that luck runs out.
Takeaway When a system has several solution pairs, compute the requested quantity for each pair and then combine.
Answer F · Simultaneous equations and inequalities (MM1.4) · question SIM-06
6
The inequality \(x^2 + bx + 9 > 0\) holds for all real \(x\). Find the complete set of values of \(b\).
- A\(-3 < b < 3\)
- B\(b < -6\) or \(b > 6\)
- C\(b > 6\)
- D\(-6 < b < 6\)
- E\(b \geq 6\)
- Fall real \(b\)
Worked solution
Need discriminant negative: \(b^2 - 36 < 0\), so \(-6 < b < 6\).
Full method — how and why
'True for all \(x\)' for an upward parabola means it never touches the axis: discriminant strictly negative.
- Note the leading coefficient is \(1 > 0\), so the parabola opens upwards.For \(>0\) everywhere you need both the right shape and no real roots; here the shape is automatic.
- Require no real roots: \(b^2 - 4(1)(9) < 0\).If there were real roots the curve would touch or cross the axis and the expression would fail to be positive there.
- Solve: \(b^2 < 36 \Rightarrow -6 < b < 6\).\(b^2 < 36\) means \(|b| < 6\), a single interval.
Where it goes wrong
- Solving \(b^2 < 36\) as two outer intervals - that is the solution of \(b^2 > 36\).
- Using \(\leq\): at \(b = \pm6\) the curve touches the axis, where the expression equals \(0\) and is not \(> 0\).
Takeaway 'Positive for all \(x\)' \(\Leftrightarrow\) \(a > 0\) and \(b^2 - 4ac < 0\); \(x^2 < k\) gives one interval, \(x^2 > k\) gives two.
Answer D · Simultaneous equations and inequalities (MM1.5) · question SIM-07
7
The simultaneous equations \(y = x^2 + 2x + c\) and \(y = 4x + 1\) have exactly one solution. Find the value of \(c\).
- A\(3\)
- B\(-1\)
- C\(2\)
- D\(1\)
- E\(-2\)
- F\(0\)
Worked solution
\(x^2 - 2x + (c - 1) = 0\) has one solution when \(4 - 4(c-1) = 0\), so \(c = 2\).
Full method — how and why
'Exactly one solution' for a line and a parabola is again the discriminant-zero (tangency) condition.
- Set the expressions equal: \(x^2 + 2x + c = 4x + 1\).A simultaneous solution is a point where both \(y\)-values agree.
- Collect: \(x^2 - 2x + (c - 1) = 0\).Move everything to one side; the \(2x - 4x\) gives \(-2x\).
- Discriminant zero: \((-2)^2 - 4(1)(c - 1) = 0\).One solution means a repeated root.
- Solve: \(4 - 4c + 4 = 0 \Rightarrow 4c = 8 \Rightarrow c = 2\).Careful with \(-4(c-1) = -4c + 4\).
Where it goes wrong
- Mishandling \(-4(c-1)\) and getting \(c = 0\).
- Setting the discriminant \(> 0\) ('has a solution') rather than \(= 0\) ('exactly one').
Takeaway Exactly one intersection between a line and a parabola always means discriminant \(= 0\).
Answer C · Simultaneous equations and inequalities (MM1.4) · question SIM-09
8
The curves \(y = x^2 - 4\) and \(y = -x^2 + 2x + k\) intersect at exactly two points. Find the complete set of values of \(k\).
- A\(k > \dfrac{9}{2}\)
- B\(k < -\dfrac{9}{2}\)
- C\(k > -\dfrac{9}{2}\)
- D\(k \neq -\dfrac92\)
- E\(k > -4\)
- Fall real \(k\)
Worked solution
\(2x^2 - 2x - 4 - k = 0\): discriminant \(4 + 8(4 + k) > 0 \Rightarrow 36 + 8k > 0 \Rightarrow k > -\frac92\).
Full method — how and why
Two parabolas meeting twice. Subtracting gives one quadratic, and 'two distinct points' is a positive discriminant.
- Set them equal: \(x^2 - 4 = -x^2 + 2x + k\).Intersections satisfy both equations, so their right-hand sides agree.
- Collect everything on one side: \(2x^2 - 2x - 4 - k = 0\).Note the \(-x^2\) becomes \(+x^2\) when moved across, giving \(2x^2\).
- Discriminant \(> 0\): \((-2)^2 - 4(2)(-4 - k) > 0\).Here \(c = -4 - k\); keeping the whole thing in brackets avoids sign errors.
- Simplify: \(4 + 32 + 8k > 0 \Rightarrow 8k > -36 \Rightarrow k > -\frac92\).\(-4 \times 2 \times (-4-k) = 32 + 8k\) - two negatives make a positive.
Where it goes wrong
- Losing a sign in \(c = -4 - k\) and getting \(k > \frac92\) or \(k > -4\), both of which are offered.
- Forgetting to combine \(x^2\) with \(-(-x^2)\) and working with \(x^2\) instead of \(2x^2\).
Takeaway Bring both curves to one side before using the discriminant, and bracket the constant term to keep signs straight.
Answer C · Simultaneous equations and inequalities (MM1.4) · question SIM-12
9
Solve \(2x + 3y = 12\) and \(x^2 - y^2 = 0\). How many solutions \((x, y)\) are there?
- A\(0\)
- B\(2\)
- C\(1\)
- D\(4\)
- E\(3\)
- Finfinitely many
Worked solution
\(x^2 = y^2\) means \(y = x\) or \(y = -x\). \(y = x\): \(5x = 12\), one solution. \(y = -x\): \(-x = 12\), one solution. Two in total.
Full method — how and why
\(x^2 - y^2 = 0\) factorises into two lines. The system is a line meeting a pair of lines, so count the intersections of each case.
- Factorise: \(x^2 - y^2 = (x - y)(x + y) = 0\), so \(y = x\) or \(y = -x\).A product is zero when either factor is zero - this splits the problem into two straight-line cases.
- Case \(y = x\): \(2x + 3x = 12 \Rightarrow 5x = 12\), one solution \(\left(\frac{12}{5}, \frac{12}{5}\right)\).Substitute into the linear equation; a linear equation in one unknown has exactly one root.
- Case \(y = -x\): \(2x - 3x = 12 \Rightarrow -x = 12\), one solution \((-12, 12)\).Same method; the coefficient of \(x\) is non-zero so again exactly one solution.
- Total: two solutions.The two cases give different points, so no double counting.
Where it goes wrong
- Taking the square root as \(y = x\) only, halving the answer.
- Concluding 'a line meets a quadratic in at most 2 points, so 2' without checking that neither case is parallel or degenerate.
Takeaway \(A^2 = B^2\) means \(A = B\) or \(A = -B\); both branches must be followed.
Answer B · Simultaneous equations and inequalities (MM1.4) · question SIM-14
10
Two numbers have sum \(10\) and the sum of their squares is \(58\). What is the positive difference between them?
- A\(3\)
- B\(4\)
- C\(6\)
- D\(1\)
- E\(2\)
- F\(8\)
Worked solution
\((x - y)^2 = 2(x^2 + y^2) - (x+y)^2 = 116 - 100 = 16\), so the difference is \(4\) (the numbers are \(3\) and \(7\)).
Full method — how and why
Symmetric information again: a sum and a sum of squares. The identity for \((x-y)^2\) gets the difference without finding the numbers.
- Use \((x - y)^2 = x^2 + y^2 - 2xy\) and \((x + y)^2 = x^2 + y^2 + 2xy\).Adding these gives \((x+y)^2 + (x-y)^2 = 2(x^2 + y^2)\) - the relation that links all three given quantities.
- Rearrange: \((x - y)^2 = 2(x^2 + y^2) - (x + y)^2\).This avoids having to find \(xy\) as an intermediate step.
- Substitute: \(2(58) - 10^2 = 116 - 100 = 16\).Careful: it is \(2(x^2+y^2)\), not \(x^2 + y^2\).
- Take the positive square root: the difference is \(4\) (the numbers are \(3\) and \(7\)).The question asks for the positive difference, so take \(+4\).
Where it goes wrong
- Forgetting the factor \(2\) and getting \((x-y)^2 = -42\), which is impossible.
- Giving \(16\) (the square of the difference) as the answer.
Takeaway \((x+y)^2 + (x-y)^2 = 2(x^2+y^2)\): any two of sum, difference and sum-of-squares give the third.
Answer B · Simultaneous equations and inequalities (MM1.4) · question SIM-17
11
\(x\) is a real number. \(P\): '\(x^2 < 4\)'. \(Q\): '\(x < 2\)'. Which one of the following is true?
- AP is necessary and sufficient for Q.
- BP is necessary but not sufficient for Q.
- CP is sufficient but not necessary for Q.
- DP is neither necessary nor sufficient for Q.
Worked solution
\(x^2 < 4\) means \(-2 < x < 2\), which implies \(x < 2\) (sufficient); but \(x = -5\) satisfies \(Q\) and not \(P\) (not necessary).
Full method — how and why
Solve the quadratic inequality properly, then compare the two solution sets as intervals.
- \(x^2<4\) means \(|x|<2\), i.e. \(-2A quadratic inequality of this shape gives an interval between the roots.Every \(x\) in \((-2,2)\) satisfies \(x<2\), so \(P \Rightarrow Q\): sufficient.\((-2,2)\) is contained in \((-\infty,2)\).Is it necessary? Take \(x=-5\): \(x<2\) holds but \(x^2 = 25 \not<4\). So \(Q \not\Rightarrow P\): not necessary.The larger set contains points outside the smaller one.So \(P\) is sufficient but not necessary.Whenever one solution set sits strictly inside another, the inner condition is sufficient but not necessary.Where it goes wrong
- Solving \(x^2<4\) as \(x<2\) and concluding the conditions are equivalent - that drops the lower bound.
- Solving it as \(x<-2\) or \(x>2\), which is the solution of \(x^2>4\).
Takeaway Compare conditions as sets: a subset is sufficient for the superset, the superset necessary for the subset.Answer C · Simultaneous equations and inequalities (MM1.5) · question X2-14
12
Solve the simultaneous equations \(x + 2y = 7\) and \(x^2 + y^2 = 10\). Find the sum of the possible values of \(x\).
- A\(7\)
- B\(2\)
- C\(\dfrac{7}{5}\)
- D\(\dfrac{14}{5}\)
- E\(\dfrac{28}{5}\)
- F\(4\)
Worked solution
\(x = 7 - 2y\): \((7-2y)^2 + y^2 = 10 \Rightarrow 5y^2 - 28y + 39 = 0 \Rightarrow (5y - 13)(y - 3) = 0\). \(y = 3 \Rightarrow x = 1\); \(y = \frac{13}{5} \Rightarrow x = \frac{9}{5}\). Sum of \(x\) values \(= 1 + \frac95 = \frac{14}{5}\).
Full method — how and why
One linear and one quadratic equation: substitution is the only method on the specification. The question then asks for the sum of the \(x\)-values, so both solutions matter.
- Make \(x\) the subject of the linear equation: \(x = 7 - 2y\).Substitute the linear one into the quadratic, never the other way round - it keeps the algebra to one variable and one squaring.
- Substitute: \((7 - 2y)^2 + y^2 = 10\).Now a single quadratic in \(y\).
- Expand and tidy: \(49 - 28y + 4y^2 + y^2 = 10 \Rightarrow 5y^2 - 28y + 39 = 0\).\((7-2y)^2 = 49 - 28y + 4y^2\) - the cross term is \(2 \times 7 \times (-2y)\).
- Factorise: \((5y - 13)(y - 3) = 0\), so \(y = 3\) or \(y = \frac{13}{5}\).Check by expanding mentally: \(5y^2 - 15y - 13y + 39\). If factorising is awkward, use the formula.
- Back-substitute for \(x\): \(y = 3 \Rightarrow x = 1\); \(y = \frac{13}{5} \Rightarrow x = 7 - \frac{26}{5} = \frac95\). Sum \(= 1 + \frac95 = \frac{14}{5}\).Always return to the linear equation for the second variable - it cannot introduce extra solutions.
Where it goes wrong
- Finding only one solution and answering \(1\) or \(\frac95\).
- Summing the \(y\)-values (\(3 + \frac{13}{5} = \frac{28}{5}\)) instead of the \(x\)-values - that wrong answer is offered.
Takeaway Substitute the linear equation into the quadratic, solve, then back-substitute into the linear equation; read carefully which variable's values are wanted.
Answer D · Simultaneous equations and inequalities (MM1.4) · question SIM-01
13
Which of the following inequalities has exactly the same solution set as \(\dfrac{1}{x} > 2\)?
- A\(x > \dfrac{1}{2}\)
- B\(0 < x < \dfrac{1}{2}\)
- C\(x < \dfrac{1}{2}\)
- D\(x < 0\) or \(x > \dfrac12\)
- E\(x > 2\)
- F\(0 < x < 2\)
Worked solution
If \(x > 0\): \(1 > 2x\), so \(0 < x < \frac12\). If \(x < 0\): \(\frac1x\) is negative and cannot exceed \(2\). So \(0 < x < \frac12\). (Multiplying by \(x\) without considering its sign is the classic error.)
Full method — how and why
An inequality with the variable in the denominator. You must not multiply by \(x\) without knowing its sign - split into cases instead.
- Case \(x > 0\): multiplying by \(x\) keeps the direction, giving \(1 > 2x\), i.e. \(x < \frac12\). Combined with \(x > 0\): \(0 < x < \frac12\).Multiplying an inequality by a positive quantity is safe; the case assumption must be carried into the answer.
- Case \(x < 0\): then \(\frac1x < 0 < 2\), so the inequality can never hold.Rather than multiplying by a negative (and flipping), just observe the sign - a negative number is never greater than \(2\).
- Case \(x = 0\): \(\frac1x\) is undefined, so \(x = 0\) is excluded.Always check the value that breaks the expression.
- Combine: the solution set is exactly \(0 < x < \frac12\).Only the first case contributes.
Where it goes wrong
- Multiplying through by \(x\) blindly to get \(x < \frac12\), which wrongly includes every negative number.
- Answering \(x > 2\) by 'inverting' both sides without justification.
Takeaway Never multiply an inequality by an expression of unknown sign: split into cases, or bring everything to one side and analyse the sign of the fraction.
Answer B · Simultaneous equations and inequalities (MM1.5) · question SIM-08
14
Find the complete set of values of \(x\) for which \((x-1)(x-3)(x-5) > 0\).
- A\(1 < x < 3\) or \(x > 5\)
- B\(x > 5\)
- C\(1 < x < 5\)
- D\(x < 1\) or \(3 < x < 5\)
- E\(x > 1\)
- F\(x < 1\) or \(x > 5\)
Worked solution
Sign chart: the product is positive for \(x > 5\), negative for \(3 < x < 5\), positive for \(1 < x < 3\), negative for \(x < 1\).
Full method — how and why
A cubic inequality. With three distinct linear factors, use a sign chart - the sign alternates as you cross each root.
- The roots are \(x = 1, 3, 5\); mark them on a number line, splitting it into four regions.A product of linear factors can only change sign at a root.
- Test the far-right region, e.g. \(x = 6\): \((5)(3)(1) > 0\).One test value fixes the whole chart; the far right is easiest because every factor is positive.
- Alternate leftwards: positive for \(x > 5\), negative on \((3,5)\), positive on \((1,3)\), negative for \(x < 1\).Each simple (unrepeated) root flips the sign exactly once.
- Collect the positive regions: \(1 < x < 3\) or \(x > 5\).The answer is a union of two intervals.
Where it goes wrong
- Answering \(x > 5\) only, forgetting the middle positive region.
- Assuming the expression is positive whenever \(x\) is bigger than all the roots - true here, but the alternation is what you must use.
Takeaway Sign chart: find the roots, test one region, then alternate; repeated factors are the only ones that do not flip the sign.
Answer A · Simultaneous equations and inequalities (MM1.5) · question SIM-10
15
Given that \(|2x - 3| < 5\), which of the following statements is/are true?
- I\(-1 < x < 4\)
- II\(x^2 < 16\)
- III\(x > -1\)
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
\(-5 < 2x - 3 < 5 \Rightarrow -1 < x < 4\), so I is true and III follows from I. For II: every \(x\) in \((-1, 4)\) has \(x^2 < 16\) (the largest possible \(|x|\) is just under \(4\)), so II is also true.
Full method — how and why
A modulus inequality plus three consequences. Solve the modulus first, then test each statement against that interval.
- Unpack the modulus: \(|2x - 3| < 5\) means \(-5 < 2x - 3 < 5\).\(|A| < k\) means \(A\) lies within \(k\) of zero - a double inequality, not two separate cases.
- Solve the double inequality: add \(3\) throughout to get \(-2 < 2x < 8\), then halve: \(-1 < x < 4\). So I is true.Whatever you do to one part of a double inequality you do to all three parts.
- III says \(x > -1\), which is half of I. True.A weaker statement implied by a true one is automatically true.
- II: for \(-1 < x < 4\), the largest \(|x|\) can get is just under \(4\), so \(x^2 < 16\) always. True.For squares, what matters is the largest absolute value in the interval, not the largest value.
- All three hold.Match to the option list.
Where it goes wrong
- Treating \(|2x-3| < 5\) as \(2x - 3 < 5\) only, losing the lower bound.
- Rejecting II by testing a value like \(x = -3\) that is not in the interval at all.
Takeaway \(|A| < k \Leftrightarrow -k < A < k\); when squaring an interval, check the end with the larger absolute value.
Answer H · Simultaneous equations and inequalities (MM1.5) · question SIM-11
16
How many integers \(x\) satisfy \(x^2 < 5x + 24\) and \(x^2 > 3x\)?
- A\(4\)
- B\(8\)
- C\(5\)
- D\(7\)
- E\(6\)
- F\(9\)
Worked solution
\(x^2 - 5x - 24 < 0 \Rightarrow (x-8)(x+3) < 0 \Rightarrow -3 < x < 8\). \(x(x-3) > 0 \Rightarrow x < 0\) or \(x > 3\). Integers: \(-2, -1\) and \(4, 5, 6, 7\): six.
Full method — how and why
Two quadratic inequalities, then an integer count. The second one factorises without a constant term, which changes its shape.
- First: \(x^2 - 5x - 24 < 0 \Rightarrow (x - 8)(x + 3) < 0 \Rightarrow -3 < x < 8\).Between the roots for \(< 0\).
- Second: \(x^2 - 3x > 0 \Rightarrow x(x - 3) > 0 \Rightarrow x < 0\) or \(x > 3\).Roots \(0\) and \(3\); outside the roots for \(> 0\). Do not divide by \(x\) - that would lose the \(x<0\) branch.
- Intersect: from \((-3, 8)\) keep \((-3, 0)\) and \((3, 8)\).Overlap both solution sets on one number line.
- Count integers: \(-2, -1\) from the first piece and \(4, 5, 6, 7\) from the second: six.Endpoints \(-3, 0, 3, 8\) are all excluded by strict inequalities.
Where it goes wrong
- Dividing \(x^2 > 3x\) by \(x\) to get \(x > 3\), losing all the negative solutions.
- Including \(0\) or \(3\) in the count.
Takeaway Never divide an inequality by the variable; factorise instead, then count integers explicitly.
Answer E · Simultaneous equations and inequalities (MM1.5) · question SIM-13
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The solution set of \(ax^2 + bx + c < 0\) (with \(a > 0\)) is \(-2 < x < 5\). What is the solution set of \(ax^2 - bx + c < 0\)?
- A\(-2 < x < 5\)
- B\(x < -5\) or \(x > 2\)
- C\(2 < x < 5\)
- D\(x < -2\) or \(x > 5\)
- E\(-5 < x < -2\)
- F\(-5 < x < 2\)
Worked solution
Replacing \(x\) by \(-x\) reflects the graph in the \(y\)-axis, so the roots \(-2\) and \(5\) become \(2\) and \(-5\): \(-5 < x < 2\).
Full method — how and why
A transformation question hidden in an inequality. Replacing \(x\) by \(-x\) reflects the graph in the \(y\)-axis, and the solution set reflects with it.
- Note that \(a(-x)^2 + b(-x) + c = ax^2 - bx + c\): the second expression is the first with \(x\) replaced by \(-x\).Spotting this makes the question a one-liner; the alternative is to reconstruct \(a\), \(b\), \(c\) explicitly.
- So the graph of the second is the reflection of the first in the \(y\)-axis.\(f(-x)\) is always the \(y\)-axis reflection of \(f(x)\).
- The original is negative on \(-2 < x < 5\); reflecting sends \(x\) to \(-x\), so the roots \(-2\) and \(5\) move to \(2\) and \(-5\).Every point of the solution set flips sign, and the interval flips with it.
- The new solution set is \(-5 < x < 2\).Write the interval with the smaller number first.
Where it goes wrong
- Leaving the interval as \(2 < x < -5\), which is empty as written.
- Assuming the solution set is unchanged because 'only the sign of \(b\) changed'.
Takeaway Changing \(b\) to \(-b\) reflects a quadratic in the \(y\)-axis; everything about the graph, including solution sets, reflects too.
Answer F · Simultaneous equations and inequalities (MM1.5) · question SIM-15
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Find the complete set of values of \(x\) for which \(\dfrac{x-2}{x+1} \leq 0\).
- A\(-1 \leq x \leq 2\)
- B\(x < -1\) or \(x \geq 2\)
- C\(-1 < x \leq 2\)
- D\(x \leq 2\)
- E\(x \leq -1\) or \(x \geq 2\)
- F\(-2 \leq x < 1\)
Worked solution
A fraction is \(\leq 0\) when numerator and denominator have opposite signs (or the numerator is 0): \(x - 2 \leq 0\) and \(x + 1 > 0\) gives \(-1 < x \leq 2\); \(x - 2 \geq 0\) and \(x + 1 < 0\) is impossible. \(x = -1\) is excluded as the fraction is undefined.
Full method — how and why
A rational inequality. Never multiply by \(x + 1\) (its sign is unknown) - analyse the sign of the fraction instead.
- Find the critical values: numerator zero at \(x = 2\), denominator zero at \(x = -1\).The fraction can only change sign where the numerator or denominator vanishes.
- Test the three regions. For \(x < -1\) (say \(x = -2\)): \(\frac{-4}{-1} = 4 > 0\). For \(-1 < x < 2\) (say \(x = 0\)): \(\frac{-2}{1} = -2 < 0\). For \(x > 2\) (say \(x = 3\)): \(\frac{1}{4} > 0\).One test value per region settles the sign throughout that region.
- Include \(x = 2\) (the fraction is \(0\), and \(\leq 0\) allows it) but exclude \(x = -1\) (undefined).The two endpoints behave differently: a zero numerator is allowed by \(\leq\); a zero denominator never is.
- Answer: \(-1 < x \leq 2\).Note the mixed brackets - strict at \(-1\), inclusive at \(2\).
Where it goes wrong
- Multiplying by \(x + 1\) to get \(x - 2 \leq 0\), i.e. \(x \leq 2\), which wrongly includes everything below \(-1\).
- Writing \(-1 \leq x \leq 2\) and including a value where the expression is undefined.
Takeaway For \(\frac{A}{B} \leq 0\), use a sign chart on \(A\) and \(B\); a zero denominator is always excluded, a zero numerator is included when the inequality is not strict.
Answer C · Simultaneous equations and inequalities (MM1.5) · question SIM-16
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Consider the statement: for all real \(x\), if \(x > 2\) then \(x^2 - 3x + 2 > 0\). Which of the following is true?
- AThe statement is true, and so is its converse.
- BThe statement is true, but its converse is false.
- CThe statement is false, but its converse is true.
- DThe statement and its converse are both false.
Worked solution
\(x^2 - 3x + 2 = (x-1)(x-2) > 0\) exactly when \(x < 1\) or \(x > 2\). So \(x > 2\) does imply it (true), but the converse fails, e.g. \(x = 0\) gives \(2 > 0\) without \(x > 2\).
Full method — how and why
A Paper 2-style wrapper on a quadratic inequality: decide the statement and its converse separately.
- Factorise the quadratic: \(x^2 - 3x + 2 = (x-1)(x-2)\), which is positive exactly when \(x < 1\) or \(x > 2\).Knowing precisely where the expression is positive lets you test both directions quickly.
- The statement: does \(x > 2\) guarantee positivity? Yes - \(x > 2\) lies inside the region \(x > 2\). True.An implication is true when the hypothesis region sits inside the conclusion region.
- The converse: does positivity guarantee \(x > 2\)? No - \(x = 0\) gives \((0-1)(0-2) = 2 > 0\) but \(0 \not> 2\). False.The positivity region is strictly larger than \(x > 2\), so a counterexample must come from the extra part \(x < 1\).
- So the statement is true but its converse is false.Report both verdicts in the order the options use them.
Where it goes wrong
- Assuming the converse holds because the implication does - converses are logically independent.
- Only testing values greater than \(2\), which can never refute the converse.
Takeaway An implication is true when one region is contained in the other; its converse needs the reverse containment, so look for a counterexample in the leftover region.
Answer B · Simultaneous equations and inequalities (MM1.5) · question SIM-18
How this topic has been examined
Real TMUA questions on Simultaneous equations and inequalities, newest first. Links open the official paper at the question and the online marking page.