June 2022 Paper 2 Q3
3
| Amaya: | \(\displaystyle\int (1 + x)^2\,\mathrm{d}x = \frac{(1 + x)^3}{3} + c\) | \(= \tfrac{1}{3} + x + x^2 + \tfrac{1}{3}x^3 + c\) |
| Ben: | \(\displaystyle\int (1 + x)^2\,\mathrm{d}x = \int (1 + 2x + x^2)\,\mathrm{d}x\) | \(= x + x^2 + \tfrac{1}{3}x^3 + c\) |
Explain whether you agree with Charlie’s statement. [1]
Determine the value of \(a\). [2]
Find the exact value of \(\displaystyle\int_0^{\frac{1}{12}\pi} \frac{\cos 2x}{\sin 2x + 2}\,\mathrm{d}x\), giving your answer in its simplest form. [4]
| Scheme | Marks | AO |
|---|---|---|
| (No because) they differ only by a constant or eg \(c_2 = c_1 + \frac{1}{3}\), or \(\frac{1}{3}\) is part of Ben’s \(c\) If definite integral found, answers are same If differentiate, answers same | B1 | 1.2 |
| [1] |
Notes
B1: oe, eg They may have different constants of integration
Only the “\(c\)”s are different
Not “Both are correct” or “just different correct methods”
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\left[\dfrac{(1 + x)^{-1}}{-1}\right]_1^a\) or \(\left[-\dfrac{1}{u}\right]_2^{a+1}\) or \(\left[-\dfrac{1}{\sqrt{u}}\right]_4^{(a+1)^2}\) oe | M1 | 1.1 |
| \(= \dfrac{(1 + a)^{-1}}{-1} + \dfrac{1}{2}\) oe | M1 | 1.1 |
| \(\left(= \dfrac{1}{2} - \dfrac{1}{1 + a}\right) = \dfrac{a - 1}{2(a + 1)}\) | A1 | 1.1 |
| [3] | ||
| (ii) \(\dfrac{a - 1}{2(a + 1)} = \dfrac{1}{3}\) or their (b)(i) (limits subst’d) \(= \dfrac{1}{3}\) | M1 | 1.1 |
| \(a = 5\) | A1 | 1.1 |
| [2] |
Notes
(b)(i)
M1: Attempt integral, must be of form \(k(1 + x)^{-1}\) or \(ku^{-1}\) or \(ku^{-0.5}\) (if from substitution \(u = (1 + x)^2\))
Ignore limits
M1: Attempt substitute appropriate limits into their integral
A1: cao oe single fraction
(b)(ii)
M1: or their new attempt at \(\displaystyle\int_1^a \frac{1}{(1 + x)^2}\,\mathrm{d}x = \frac{1}{3}\)
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| DR \(\frac{1}{2}\left[\ln|\sin 2x + 2|\right]_0^{\frac{1}{12}\pi}\) | M1 | 1.1 |
| \(= \frac{1}{2}\left[\ln\left(\sin\frac{1}{6}\pi + 2\right) - \ln(0 + 2)\right]\) | M1 | 1.1 |
| \(= \frac{1}{2}\left(\ln\left(\frac{5}{2}\right) - \ln 2\right)\) | A1 | 1.1 |
| \(= \frac{1}{2}\ln\frac{5}{4}\) oe, eg \(\ln\dfrac{\sqrt{5}}{2}\) | A1 | 2.1 |
| [4] |
Notes
Allow incorrect use of brackets throughout
Allow \(\ln(\ldots)\) instead of \(\ln|\ldots|\)
M1: Allow \(k\ln(\sin 2x + 2)\), \(k\) any constant. Ignore limits
M1: Attempt substitute both correct limits into their log integral.
Allow numerical errors
A1: Allow \(\times\) any \(k\), otherwise any correct form without trig.
A1: Correct one-term exact result. ISW, eg ignore decimal
NB No working, no marks.
Alternative methods
| Scheme | Marks |
|---|---|
| \(u = \sin 2x + 2\), or \(u = \sin 2x\) \(\frac{1}{2}\displaystyle\int_2^{5/2} \frac{1}{u}\,\mathrm{d}u\) or \(\frac{1}{2}\displaystyle\int_0^{1/2} \frac{1}{u + 2}\,\mathrm{d}u\) \(\frac{1}{2}\left[\ln u\right]_2^{5/2}\) or \(\frac{1}{2}\left[\ln(u + 2)\right]_0^{1/2}\) | M1 |
| \(\left(= \frac{1}{2}\left[\ln(\sin 2x + 2)\right]_0^{\frac{1}{12}\pi}\right)\) | M1 |
| \(\frac{1}{2}\left(\ln\left(\frac{5}{2}\right) - \ln 2\right)\) | A1 |
| \(= \frac{1}{2}\ln\frac{5}{4}\) oe, eg \(\ln\dfrac{\sqrt{5}}{2}\) | A1 |
M1: Attempt substitute and integrate and obtain \(k\ln u\) or \(k\ln(u + 2)\), \(k\) any constant; ignore limits
(May not see this step)
M1: Attempt substitute their limits into their log integral.
but not limits for wrong variable, eg not \(\ln\frac{\pi}{12} - \ln 0\)
Allow numerical errors
A1: Correct exact result, any form without trig. Allow \(\times\) any \(k\)
A1: cao, ISW, eg ignore decimal answer