Hard GCSE Exponential and trig graphs Questions

Challenging, exam-style GCSE Exponential and trig graphs questions with worked solutions. Stretch yourself on the hardest trig equation, sine graph, symmetry, inverse sine problems.

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GCSE Higher34 questionsStep-by-step solutions
Question 1
6 markschallenging
Solve 2sin⁡x∘=cos⁡60∘2\sin x^\circ = \cos 60^\circ in the interval 0≤x≤3600 \le x \le 360. Give your answers correct to 11 decimal place.
Show worked solution

Worked solution

  1. Recall the exact value

    cos⁡60∘=12\cos 60^\circ = \frac{1}{2}

    This is a standard exact value.

  2. Rewrite the equation

    2sin⁡x∘=122\sin x^\circ = \frac{1}{2}

    Replace cos 6060 with one half.

  3. Isolate sin x

    sin⁡x∘=14\sin x^\circ = \frac{1}{4}

    Divide both sides by 22.

  4. Write as a decimal

    sin⁡x∘=0.25\sin x^\circ = 0.25

    One quarter is 0.250.25.

  5. Use the inverse sine

    sin⁡−1(0.25)=14.477...∘\sin^{-1}(0.25) = 14.477...^\circ

    Make sure the calculator is in degree mode.

  6. Round the first solution

    x=14.5x = 14.5

    Correct to 11 decimal place.

  7. Sketch the sine curve

    y=sin⁡x∘ and y=0.25y = \sin x^\circ \text{ and } y = 0.25

    The line cuts the positive hump twice.

  8. Use the symmetry rule

    sin⁡(180−x)∘=sin⁡x∘\sin(180 - x)^\circ = \sin x^\circ

    The sine curve is symmetrical about x=90x = 90.

  9. Find the second solution

    180−14.477...=165.522...180 - 14.477... = 165.522...

    Subtract the unrounded value from 180180.

  10. Round the second solution

    x=165.5x = 165.5

    Correct to 11 decimal place.

  11. Check no other solutions

    sin⁡x∘<0 for 180<x<360\sin x^\circ < 0 \text{ for } 180 < x < 360

    The curve is below the x-axis there, so it cannot equal 0.250.25.

  12. Verify the first

    2sin⁡14.5∘=2×0.2504=0.5008≈0.52\sin 14.5^\circ = 2 \times 0.2504 = 0.5008 \approx 0.5

    Close to cos 6060, allowing for rounding.

  13. Verify the second

    2sin⁡165.5∘=2×0.2504=0.5008≈0.52\sin 165.5^\circ = 2 \times 0.2504 = 0.5008 \approx 0.5

    Also correct.

  14. Check the interval

    0≤14.5≤360, 0≤165.5≤3600 \le 14.5 \le 360,\ 0 \le 165.5 \le 360

    Both lie in the required interval.

  15. State the solutions

    x=14.5 or x=165.5x = 14.5 \text{ or } x = 165.5

    There are exactly two solutions.

Answer
x=14.5 or x=165.5x = 14.5 \text{ or } x = 165.5
Question 2
6 markschallenging
£8000\pounds 8000 is invested at r%r\% per year compound interest. After 33 years the investment is worth £9261\pounds 9261. Work out the value of the investment after 66 years, to the nearest penny.
Show worked solution

Worked solution

  1. Write the model

    8000×m3=92618000 \times m^{3} = 9261

    m is the yearly multiplier.

  2. Divide by 80008000

    m3=92618000m^{3} = \frac{9261}{8000}

    Isolate the cubed multiplier.

  3. Work out the fraction

    92618000=1.157625\frac{9261}{8000} = 1.157625

    Do the division carefully.

  4. Take the cube root

    m=1.1576253m = \sqrt[3]{1.157625}

    Undo the cube.

  5. Work out the multiplier

    m=1.05m = 1.05

    Because 1.05×1.05×1.05=1.1576251.05 \times 1.05 \times 1.05 = 1.157625.

  6. Interpret the multiplier

    1.05=100%+5%1.05 = 100\% + 5\%

    So the interest rate is 55% per year.

  7. Write the full model

    V=8000×1.05 tV = 8000 \times 1.05^{\,t}

    t is the number of complete years.

  8. Substitute t=6t = 6

    V=8000×1.056V = 8000 \times 1.05^{6}

    Six years means six lots of the multiplier.

  9. Work out the power

    1.056=1.3400956406251.05^{6} = 1.340095640625

    Keep the full accuracy.

  10. Multiply

    V=8000×1.340095640625V = 8000 \times 1.340095640625

    Apply the six-year multiplier.

  11. Work out the value

    V=10720.765125V = 10720.765125

    This is the value in pounds.

  12. Round to the nearest penny

    V=£10 720.77V = \pounds 10\,720.77

    The third decimal place is 55, so round the pence up.

  13. Check with a shortcut

    9261×1.053=9261×1.157625=10720.765...9261 \times 1.05^{3} = 9261 \times 1.157625 = 10720.765...

    Applying three more years to the 33-year value gives the same result.

  14. Check the growth is exponential

    10720.779261=1.157625\frac{10720.77}{9261} = 1.157625

    The same 33-year multiplier applies again, as expected.

  15. State the answer

    £10 720.77\pounds 10\,720.77

    This is the value after 66 years.

Answer
£10 720.77\pounds 10\,720.77
Question 3
6 markschallenging
The area covered by a water lily is modelled by A=2×1.15 tA = 2 \times 1.15^{\,t} square metres, where tt is the number of days. The pond has an area of 5050 square metres. Work out the smallest whole number of days after which the lily covers more than half of the pond.
Show worked solution

Worked solution

  1. Find half the pond area

    502=25 m2\frac{50}{2} = 25 \text{ m}^2

    We need the lily to cover more than 2525 square metres.

  2. Write the inequality

    2×1.15 t>252 \times 1.15^{\,t} > 25

    Use the model with the target area.

  3. Divide by 22

    1.15 t>12.51.15^{\,t} > 12.5

    Isolate the power.

  4. Try t=15t = 15

    1.1515=8.137...1.15^{15} = 8.137...

    This is less than 12.512.5, so not yet.

  5. Check t=15t = 15 directly

    2×8.137=16.27 m22 \times 8.137 = 16.27 \text{ m}^2

    Still below 2525 square metres.

  6. Try t=18t = 18

    1.1518=12.375...1.15^{18} = 12.375...

    Just below 12.512.5, so still not enough.

  7. Check t=18t = 18 directly

    2×12.375=24.75 m22 \times 12.375 = 24.75 \text{ m}^2

    This is 24.7524.75 square metres, just short of 2525.

  8. Try t=19t = 19

    1.1519=14.232...1.15^{19} = 14.232...

    This is greater than 12.512.5.

  9. Check t=19t = 19 directly

    2×14.232=28.46 m22 \times 14.232 = 28.46 \text{ m}^2

    This is more than 2525 square metres.

  10. Identify the first day

    t=19t = 19

    Day 1818 fails and day 1919 works.

  11. State the answer

    19 days19 \text{ days}

    The lily first covers more than half the pond after 1919 days.

  12. Interpret the growth rate

    1.15=100%+15%1.15 = 100\% + 15\%

    The area grows by 1515% each day.

  13. Comment on the shape

    exponential growth curve\text{exponential growth curve}

    The graph rises slowly at first and then very steeply.

  14. Note the limitation of the model

    the lily cannot exceed the pond\text{the lily cannot exceed the pond}

    Once the pond is full the model must break down.

  15. Check the y-intercept

    A=2×1.150=2A = 2 \times 1.15^{0} = 2

    On day 00 the lily covered 22 square metres, as stated.

Answer
19 days19 \text{ days}
Question 4
6 markschallenging
At 1010:0000 there are 30003000 bacteria in a dish. The number of bacteria doubles every 2020 minutes. Work out the time at which the number of bacteria first exceeds one million.
Show worked solution

Worked solution

  1. Set up the model

    N=3000×2 dN = 3000 \times 2^{\,d}

    d is the number of 20-minute doubling periods after 10:00.

  2. Write the inequality

    3000×2 d>1 000 0003000 \times 2^{\,d} > 1\,000\,000

    We want the count to exceed one million.

  3. Divide by 30003000

    2 d>10000003000=333.33...2^{\,d} > \frac{1000000}{3000} = 333.33...

    Isolate the power of 22.

  4. Try d=8d = 8

    28=2562^{8} = 256

    256256 is less than 333.33333.33, so this is not enough.

  5. Check d=8d = 8 directly

    3000×256=768 0003000 \times 256 = 768\,000

    Still below one million.

  6. Try d=9d = 9

    29=5122^{9} = 512

    512512 is greater than 333.33333.33.

  7. Check d=9d = 9 directly

    3000×512=1 536 0003000 \times 512 = 1\,536\,000

    This is above one million.

  8. Choose the smallest whole number of doublings

    d=9d = 9

    The count first exceeds one million after 99 doublings.

  9. Convert doublings to minutes

    9×20=180 minutes9 \times 20 = 180 \text{ minutes}

    Each doubling takes 2020 minutes.

  10. Convert minutes to hours

    180 minutes=3 hours180 \text{ minutes} = 3 \text{ hours}

    180180 divided by 6060 is 33.

  11. Add the time on

    10:00+3 hours=13:0010{:}00 + 3 \text{ hours} = 13{:}00

    Work forwards from the starting time.

  12. Check the previous doubling time

    10:00+8×20 min=12:4010{:}00 + 8 \times 20\text{ min} = 12{:}40

    At 1212:4040 there are only 768768 000000 bacteria.

  13. Confirm the crossing

    768 000<1 000 000<1 536 000768\,000 < 1\,000\,000 < 1\,536\,000

    The million mark is passed between 1212:4040 and 1313:0000.

  14. Explain the whole-doubling answer

    the model only gives whole doublings\text{the model only gives whole doublings}

    The count is only defined at 2020-minute intervals in this model.

  15. State the answer

    13:0013{:}00

    The number of bacteria first exceeds one million at 1313:0000.

Answer
13:0013{:}00
Question 5
5 markschallenging
Which of the following is a correct proof that tan⁡30∘×sin⁡60∘=cos⁡60∘\tan 30^\circ \times \sin 60^\circ = \cos 60^\circ?
Show worked solution

Worked solution

  1. Write down what must be shown

    tan⁡30∘×sin⁡60∘=cos⁡60∘\tan 30^\circ \times \sin 60^\circ = \cos 60^\circ

    Start from the left-hand side and aim for the right-hand side.

  2. Recall tan 3030

    tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}

    From the half-equilateral triangle, opposite 11 and adjacent root 33.

  3. Recall sin 6060

    sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}

    From the same triangle, opposite root 33 and hypotenuse 22.

  4. Recall cos 6060

    cos⁡60∘=12\cos 60^\circ = \frac{1}{2}

    The adjacent side is 11 and the hypotenuse is 22.

  5. Substitute into the left-hand side

    LHS=13×32\text{LHS} = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2}

    Replace both trig values with their exact forms.

  6. Multiply the numerators

    1×3=31 \times \sqrt{3} = \sqrt{3}

    The top of the fraction is root 33.

  7. Multiply the denominators

    3×2=23\sqrt{3} \times 2 = 2\sqrt{3}

    The bottom of the fraction is 22 root 33.

  8. Write the single fraction

    LHS=323\text{LHS} = \frac{\sqrt{3}}{2\sqrt{3}}

    Combine the two fractions.

  9. Cancel the surds

    323=12\frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}

    Root 33 is a common factor of the numerator and the denominator.

  10. Compare with the right-hand side

    RHS=cos⁡60∘=12\text{RHS} = \cos 60^\circ = \frac{1}{2}

    The right-hand side is also one half.

  11. Conclude

    LHS=RHS\text{LHS} = \text{RHS}

    The identity is proved.

  12. Check numerically

    0.5774×0.8660=0.50000.5774 \times 0.8660 = 0.5000

    The decimal check agrees to 44 decimal places.

  13. Note the common error

    tan⁡30∘≠3\tan 30^\circ \ne \sqrt{3}

    Root 33 is tan 6060, not tan 30−30 - mixing these up is a frequent mistake.

  14. Note a second common error

    323≠32\frac{\sqrt{3}}{2\sqrt{3}} \ne \frac{\sqrt{3}}{2}

    The surds must be cancelled properly.

  15. Select the correct proof

    13×32=12\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{1}{2}

    Only one option carries out every step correctly.

Answer
tan⁡30∘×sin⁡60∘=13×32=12=cos⁡60∘\tan 30^\circ \times \sin 60^\circ = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{1}{2} = \cos 60^\circ

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