Hard A-Level Inequalities Questions

Challenging, exam-style A-Level Inequalities questions with worked solutions. Stretch yourself on the hardest discriminant, quadratic inequality, linear inequality, factorising problems.

discriminantquadratic inequalitylinear inequalityfactorisingsimultaneousintersection
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
The equation (k+1)x2−2kx+(k−2)=0(k + 1)x^2 - 2kx + (k - 2) = 0 has real roots. Find the set of values of kk.
Show worked solution

Worked solution

  1. Check the degenerate case k = -1

    k=−1:  0⋅x2+2x−3=0⇒x=32k = -1:\; 0\cdot x^2 + 2x - 3 = 0 \Rightarrow x = \tfrac{3}{2}

    If k = -1 the x^2 coefficient is zero and the equation becomes linear, which still has a real root.

  2. Now take k \ne -1 (a genuine quadratic)

    a=k+1,b=−2k,c=k−2a = k + 1, \quad b = -2k, \quad c = k - 2

    For a real quadratic we use the discriminant with these coefficients.

  3. Condition for real roots

    b2−4ac≥0b^2 - 4ac \ge 0

    Real roots (equal or distinct) require a non-negative discriminant.

  4. Substitute

    (−2k)2−4(k+1)(k−2)≥0(-2k)^2 - 4(k + 1)(k - 2) \ge 0

    Put the coefficients into b^2 - 4ac.

  5. Square the first term

    4k2−4(k+1)(k−2)≥04k^2 - 4(k + 1)(k - 2) \ge 0

    (-2k)^2 = 4k^2.

  6. Expand the product

    4k2−4(k2−k−2)≥04k^2 - 4(k^2 - k - 2) \ge 0

    Multiply out (k+1)(k-2) = k^2 - k - 2.

  7. Remove the bracket

    4k2−4k2+4k+8≥04k^2 - 4k^2 + 4k + 8 \ge 0

    Distribute the -4 across the bracket, changing every sign inside.

  8. Simplify

    4k+8≥04k + 8 \ge 0

    The 4k^2 terms cancel, leaving a linear inequality in k.

  9. Solve

    k≥−2k \ge -2

    Subtract 8 and divide by 4 (a positive divisor, so no flip).

  10. Check the degenerate case fits

    k=−1≥−2  ✓k = -1 \ge -2 \; \checkmark

    The linear case k = -1 already satisfies k \ge -2, so it does not need to be added separately.

  11. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  12. Interpret the final answer

    k≥−2k \ge -2

    The answer describes a set of numbers. Every value in this range makes the original statement true, and every value outside it does not.

  13. Check the boundary points

    closed endpoints included\text{closed endpoints included}

    Because the inequality allows 'equal to', the boundary values are part of the solution and are drawn as solid (filled) circles.

  14. Verify with a test value

    substitute a value from the range\text{substitute a value from the range}

    Pick a number from your solution and put it back into the original inequality. If it gives a true statement, you can be confident the region is correct.

  15. State the solution

    k≥−2k \ge -2

    For every k that is at least -2 the equation has real roots.

Answer
k≥−2k \ge -2
Question 2
8 markschallenging
A company's daily profit (in thousands of pounds) from making xx units is modelled by P=−x2+14x−40P = -x^2 + 14x - 40. Find the values of xx (with x>0x > 0) for which the company makes a profit (i.e. P>0P > 0).
Show worked solution

Worked solution

  1. Form the inequality

    −x2+14x−40>0-x^2 + 14x - 40 > 0

    'Makes a profit' means P is strictly greater than zero.

  2. Multiply by -1 and flip the sign

    x2−14x+40<0x^2 - 14x + 40 < 0

    Multiplying by -1 makes the x^2 term positive but reverses the inequality from '>' to '<'.

  3. Factorise

    (x−4)(x−10)<0(x - 4)(x - 10) < 0

    Numbers multiplying to 40 and adding to -14 are -4 and -10.

  4. Critical values

    x=4  or  x=10x = 4 \;\text{or}\; x = 10

    These are the production levels giving exactly zero profit (break-even).

  5. Sketch the parabola

    y=x2−14x+40y = x^2 - 14x + 40

    Upward parabola crossing at 4 and 10; below the axis is between the roots.

  6. Solve the inequality

    4<x<104 < x < 10

    Between the roots, endpoints excluded because the sign is strict.

  7. Check the context

    x>0 and 4<x<10⇒4<x<10x > 0 \text{ and } 4 < x < 10 \Rightarrow 4 < x < 10

    The range 4 < x < 10 is already positive, so the restriction x > 0 changes nothing.

  8. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  9. Interpret the final answer

    4<x<104 < x < 10

    The answer describes a set of numbers. Every value in this range makes the original statement true, and every value outside it does not.

  10. Check the boundary points

    open endpoints excluded\text{open endpoints excluded}

    Because the inequality is strict, the boundary values are NOT part of the solution and are drawn as open (hollow) circles.

  11. Verify with a test value

    substitute a value from the range\text{substitute a value from the range}

    Pick a number from your solution and put it back into the original inequality. If it gives a true statement, you can be confident the region is correct.

  12. Check a value outside the range

    substitute a value not in the range\text{substitute a value not in the range}

    Also try a value outside your solution: it should make the inequality false. This double-check catches a reversed inequality sign.

  13. Link to prior learning

    forming an inequality\text{forming an inequality}

    This question builds on forming an inequality from earlier work. Recognising the familiar skill inside a new problem is a key exam habit.

  14. Show the solution on a number line

    mark critical values, then shade\text{mark critical values, then shade}

    A number line makes the answer clear: mark the boundary values, decide open or closed circles, then shade the part that satisfies the inequality.

  15. Interpret the answer

    4<x<104 < x < 10

    The company makes a profit when it produces between 4 and 10 units per day.

Answer
4<x<104 < x < 10
Question 3
8 markschallenging
Find the set of values of the constant aa for which ax2−2ax+3>0ax^2 - 2ax + 3 > 0 for all real values of xx.
Show worked solution

Worked solution

  1. First requirement: opens upward

    a>0a > 0

    For an upward parabola that can stay above the axis, the x^2 coefficient a must be positive (if a < 0 it eventually goes negative, and a = 0 is not a quadratic).

  2. Second requirement: no real roots

    b2−4ac<0b^2 - 4ac < 0

    To stay strictly above the axis the quadratic must have no real roots, so the discriminant is negative.

  3. Identify a, b and c

    coeff a,  b=−2a,  c=3\text{coeff } a,\; b = -2a,\; c = 3

    Here the leading coefficient is the parameter a itself; b = -2a and c = 3.

  4. Substitute into the discriminant

    (−2a)2−4(a)(3)<0(-2a)^2 - 4(a)(3) < 0

    Put the coefficients into b^2 - 4ac.

  5. Square and simplify

    4a2−12a<04a^2 - 12a < 0

    (-2a)^2 = 4a^2 and -4(a)(3) = -12a.

  6. Factorise

    4a(a−3)<04a(a - 3) < 0

    Take out the common factor 4a; the critical values are 0 and 3.

  7. Solve the discriminant inequality

    0<a<30 < a < 3

    Below the axis for an upward parabola (in a) is between the roots 0 and 3.

  8. Combine with a > 0

    (a>0)  and  (0<a<3)⇒0<a<3(a > 0) \;\text{and}\; (0 < a < 3) \Rightarrow 0 < a < 3

    The condition 0 < a < 3 already guarantees a > 0, so both requirements are met together.

  9. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  10. Interpret the final answer

    0<a<30 < a < 3

    The answer describes a set of numbers. Every value in this range makes the original statement true, and every value outside it does not.

  11. Check the boundary points

    open endpoints excluded\text{open endpoints excluded}

    Because the inequality is strict, the boundary values are NOT part of the solution and are drawn as open (hollow) circles.

  12. Verify with a test value

    substitute a value from the range\text{substitute a value from the range}

    Pick a number from your solution and put it back into the original inequality. If it gives a true statement, you can be confident the region is correct.

  13. Check a value outside the range

    substitute a value not in the range\text{substitute a value not in the range}

    Also try a value outside your solution: it should make the inequality false. This double-check catches a reversed inequality sign.

  14. Link to prior learning

    condition a>0\text{condition a>0}

    This question builds on condition a>0 from earlier work. Recognising the familiar skill inside a new problem is a key exam habit.

  15. State the solution

    0<a<30 < a < 3

    For these values of a the expression is positive for every real x.

Answer
0<a<30 < a < 3
Question 4
8 markschallenging
Solve the inequality (x2−1)(x2−9)>0(x^2 - 1)(x^2 - 9) > 0.
Show worked solution

Worked solution

  1. Factorise each bracket

    (x−1)(x+1)(x−3)(x+3)>0(x - 1)(x + 1)(x - 3)(x + 3) > 0

    Each bracket is a difference of two squares, giving four linear factors.

  2. List the critical values

    x=−3,−1,1,3x = -3, -1, 1, 3

    The product is zero at these four points, which split the number line into five regions.

  3. Test a value in x > 3

    x=4:(+)(+)(+)(+)>0x = 4: (+)(+)(+)(+) > 0

    Pick x = 4: all four factors are positive, so the product is positive. This region is part of the solution.

  4. Test a value in 1 < x < 3

    x=2:(+)(+)(−)(+)<0x = 2: (+)(+)(-)(+) < 0

    Pick x = 2: the factor (x - 3) is negative, so the product is negative. Not in the solution.

  5. Test a value in -1 < x < 1

    x=0:(−)(+)(−)(+)>0x = 0: (-)(+)(-)(+) > 0

    Pick x = 0: two factors are negative, and a negative times a negative is positive. In the solution.

  6. Test a value in -3 < x < -1

    x=−2:(−)(−)(−)(+)<0x = -2: (-)(-)(-)(+) < 0

    Pick x = -2: three factors are negative, giving an overall negative. Not in the solution.

  7. Test a value in x < -3

    x=−4:(−)(−)(−)(−)>0x = -4: (-)(-)(-)(-) > 0

    Pick x = -4: all four factors are negative, and an even number of negatives multiplies to positive. In the solution.

  8. Collect the positive regions

    x<−3,−1<x<1,x>3x < -3, \quad -1 < x < 1, \quad x > 3

    The product is greater than zero in these three regions.

  9. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  10. Interpret the final answer

    x<−3or−1<x<1or\quadx>3x < -3\quad\text{or}\quad-1 < x < 1\quad\text{or}\quadx > 3

    The answer describes a set of numbers. Every value in this range makes the original statement true, and every value outside it does not.

  11. Check the boundary points

    open endpoints excluded\text{open endpoints excluded}

    Because the inequality is strict, the boundary values are NOT part of the solution and are drawn as open (hollow) circles.

  12. Verify with a test value

    substitute a value from the range\text{substitute a value from the range}

    Pick a number from your solution and put it back into the original inequality. If it gives a true statement, you can be confident the region is correct.

  13. Check a value outside the range

    substitute a value not in the range\text{substitute a value not in the range}

    Also try a value outside your solution: it should make the inequality false. This double-check catches a reversed inequality sign.

  14. Link to prior learning

    factorising fully\text{factorising fully}

    This question builds on factorising fully from earlier work. Recognising the familiar skill inside a new problem is a key exam habit.

  15. State the solution

    x<−3  or  −1<x<1  or  x>3x < -3 \;\text{or}\; -1 < x < 1 \;\text{or}\; x > 3

    All endpoints are excluded because the inequality is strict.

Answer
x<−3or−1<x<1or\quadx>3x < -3\quad\text{or}\quad-1 < x < 1\quad\text{or}\quadx > 3
Question 5
8 markschallenging
Find the set of values of kk for which x2+2kx+(3k+4)>0x^2 + 2kx + (3k + 4) > 0 for all real values of xx.
Show worked solution

Worked solution

  1. Interpret 'positive for all x'

    upward parabola never touches the axis\text{upward parabola never touches the axis}

    The x^2 coefficient is positive, so the parabola opens upward. It stays above the axis only if it has no real roots.

  2. No-real-roots condition

    b2−4ac<0b^2 - 4ac < 0

    No real roots means a negative discriminant.

  3. Identify a, b and c

    a=1,b=2k,c=3k+4a = 1, \quad b = 2k, \quad c = 3k + 4

    Read the coefficients; note k appears in both b and c.

  4. Substitute into the discriminant

    (2k)2−4(1)(3k+4)<0(2k)^2 - 4(1)(3k + 4) < 0

    Put the coefficients into b^2 - 4ac.

  5. Square and expand

    4k2−12k−16<04k^2 - 12k - 16 < 0

    (2k)^2 = 4k^2 and -4(3k+4) = -12k - 16.

  6. Divide by 4

    k2−3k−4<0k^2 - 3k - 4 < 0

    Dividing by the positive number 4 keeps the direction of the inequality.

  7. Factorise

    (k−4)(k+1)<0(k - 4)(k + 1) < 0

    Numbers multiplying to -4 and adding to -3 are -4 and +1.

  8. Critical values

    k=4  or  k=−1k = 4 \;\text{or}\; k = -1

    Set each bracket to zero.

  9. Sketch the parabola in k

    y=k2−3k−4y = k^2 - 3k - 4

    Upward parabola crossing at -1 and 4; below the axis is between the roots.

  10. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  11. Interpret the final answer

    −1<k<4-1 < k < 4

    The answer describes a set of numbers. Every value in this range makes the original statement true, and every value outside it does not.

  12. Check the boundary points

    open endpoints excluded\text{open endpoints excluded}

    Because the inequality is strict, the boundary values are NOT part of the solution and are drawn as open (hollow) circles.

  13. Verify with a test value

    substitute a value from the range\text{substitute a value from the range}

    Pick a number from your solution and put it back into the original inequality. If it gives a true statement, you can be confident the region is correct.

  14. Check a value outside the range

    substitute a value not in the range\text{substitute a value not in the range}

    Also try a value outside your solution: it should make the inequality false. This double-check catches a reversed inequality sign.

  15. State the solution

    −1<k<4-1 < k < 4

    For these values of k the discriminant is negative, so the quadratic is positive for all x.

Answer
−1<k<4-1 < k < 4

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