C4 June 2013 Q2
2.
Give your answer in the form \(\dfrac{a}{b}\) where \(a\) and \(b\) are integers. (3)
| Scheme | Marks |
|---|---|
| \(\left\{\sqrt{\left(\dfrac{1 + x}{1 - x}\right)}\right\} = (1 + x)^{\frac{1}{2}}(1 - x)^{-\frac{1}{2}}\) \((1 + x)^{\frac{1}{2}}(1 - x)^{-\frac{1}{2}}\) | B1 |
| \(= \left(1 + \left(\dfrac{1}{2}\right)x + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}x^2 + \ldots\right) \times \left(1 + \left(-\dfrac{1}{2}\right)(-x) + \dfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-x)^2 + \ldots\right)\) See notes | M1 A1 A1 |
| \(= \left(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \ldots\right) \times \left(1 + \dfrac{1}{2}x + \dfrac{3}{8}x^2 + \ldots\right)\) | |
| \(= 1 + \dfrac{1}{2}x + \dfrac{3}{8}x^2 + \dfrac{1}{2}x + \dfrac{1}{4}x^2 - \dfrac{1}{8}x^2 + \ldots\) See notes | M1 |
| \(= 1 + x + \dfrac{1}{2}x^2\) Answer is given in the question. | A1 * |
| (6) |
Notes
B1: \((1 + x)^{\frac{1}{2}}(1 - x)^{-\frac{1}{2}}\) or \(\sqrt{(1 + x)}(1 - x)^{-\frac{1}{2}}\) seen or implied. (Also allow \(\left((1 + x)(1 - x)^{-1}\right)^{\frac{1}{2}}\)).
M1: Expands \((1 + x)^{\frac{1}{2}}\) to give any 2 out of 3 terms simplified or un-simplified,
Eg: \(1 + \dfrac{1}{2}x\) or \(+\left(\dfrac{1}{2}\right)x + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}x^2\) or \(1 + \ldots + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}x^2\)
or expands \((1 - x)^{-\frac{1}{2}}\) to give any 2 out of 3 terms simplified or un-simplified,
Eg: \(1 + \left(-\dfrac{1}{2}\right)(-x)\) or \(+\left(-\dfrac{1}{2}\right)(-x) + \dfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-x)^2\) or \(1 + \ldots + \dfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-x)^2\)
Also allow: \(1 + \ldots + \dfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(x)^2\) for M1.
A1: At least one binomial expansion correct (either un-simplified or simplified). (ignore \(x^3\) and \(x^4\) terms)
A1: Two binomial expansions are correct (either un-simplified or simplified). (ignore \(x^3\) and \(x^4\) terms)
Note: Candidates can give decimal equivalents when expanding out their binomial expansions.
M1: Multiplies out to give 1, exactly two terms in \(x\) and exactly three terms in \(x^2\).
A1: Candidate achieves the result on the exam paper. Make sure that their working is sound.
Special Case: Award SC FINAL M1A1 for a correct \(\left(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \ldots\right) \times \left(1 + \dfrac{1}{2}x + \dfrac{3}{8}x^2 + \ldots\right)\)
multiplied out with no errors to give either \(1 + x + \dfrac{3}{8}x^2 + \dfrac{1}{4}x^2 - \dfrac{1}{8}x^2\) or \(1 + \dfrac{1}{2}x + \dfrac{3}{8}x^2 + \dfrac{1}{2}x + \dfrac{1}{8}x^2\) or
\(1 + \dfrac{1}{2}x + \dfrac{1}{4}x^2 + \dfrac{1}{2}x + \dfrac{1}{4}x^2\) or \(1 + \dfrac{1}{2}x + \dfrac{5}{8}x^2 + \dfrac{1}{2}x - \dfrac{1}{8}x^2\) leading to the correct answer of \(1 + x + \dfrac{1}{2}x^2\).
Note: If a candidate writes down either \((1 + x)^{\frac{1}{2}} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \ldots\) or \((1 - x)^{-\frac{1}{2}} = 1 + \dfrac{1}{2}x + \dfrac{3}{8}x^2 + \ldots\)
with no working then you can award 1st M1, 1st A1.
Note: If a candidate writes down both correct binomial expansions with no working, then you can award 1st M1, 1st A1, 2nd A1.
Aliter 2. (a) Way 2
| Scheme | Marks |
|---|---|
| \(\left\{\sqrt{\left(\dfrac{1 + x}{1 - x}\right)} = \sqrt{\dfrac{(1 + x)(1 - x)}{(1 + x)(1 - x)}} = \sqrt{\dfrac{(1 - x^2)}{(1 - x)^2}} = \right\} = (1 - x^2)^{\frac{1}{2}}(1 - x)^{-1}\) \((1 - x^2)^{\frac{1}{2}}(1 - x)^{-1}\) | B1 |
| \(= \left(1 + \left(\dfrac{1}{2}\right)(-x^2) + \ldots\right) \times \left(1 + (-1)(-x) + \dfrac{(-1)(-2)}{2!}(-x)^2 + \ldots\right)\) See notes | M1A1A1 |
| \(= \left(1 - \dfrac{1}{2}x^2 + \ldots\right) \times \left(1 + x + x^2 + \ldots\right)\) | |
| \(= 1 + x + x^2 - \dfrac{1}{2}x^2\) See notes | M1 |
| \(= 1 + x + \dfrac{1}{2}x^2\) Answer is given in the question. | A1 * |
| (6) |
B1: \((1 - x^2)^{\frac{1}{2}}(1 - x)^{-1}\) seen or implied.
M1: Expands \((1 - x^2)^{\frac{1}{2}}\) to give both terms simplified or un-simplified, \(1 + \left(\dfrac{1}{2}\right)(-x^2)\)
or expands \((1 - x)^{-1}\) to give any 2 out of 3 terms simplified or un-simplified,
Eg: \(1 + (-1)(-x)\) or \(\ldots + (-1)(-x) + \dfrac{(-1)(-2)}{2!}(-x)^2\) or \(1 + \ldots + \dfrac{(-1)(-2)}{2!}(-x)^2\)
A1: At least one binomial expansion correct (either un-simplified or simplified). (ignore \(x^3\) and \(x^4\) terms)
A1: Two binomial expansions are correct (either un-simplified or simplified). (ignore \(x^3\) and \(x^4\) terms)
M1: Multiplies out to give 1, exactly one term in \(x\) and exactly two terms in \(x^2\).
A1: Candidate achieves the result on the exam paper. Make sure that their working is sound.
Aliter 2. (a) Way 3
| Scheme | Marks |
|---|---|
| \(\left\{\sqrt{\left(\dfrac{1 + x}{1 - x}\right)} = \sqrt{\dfrac{(1 + x)(1 + x)}{(1 - x)(1 + x)}} = \right\} = (1 + x)(1 - x^2)^{-\frac{1}{2}}\) \((1 + x)(1 - x^2)^{-\frac{1}{2}}\) | B1 |
| \(= (1 + x)\left(1 + \dfrac{1}{2}x^2 + \ldots\right)\) Must follow on from above. | M1A1A1 |
| \(= 1 + x + \dfrac{1}{2}x^2\) | dM1A1 |
Note: The final M1 mark is dependent on the previous method mark for Way 3.
Aliter 2. (a) Way 4
Assuming the result on the Question Paper. (You need to be convinced that a candidate is applying this method before you apply the Mark Scheme for Way 4).
| Scheme | Marks |
|---|---|
| \(\left\{\sqrt{\left(\dfrac{1 + x}{1 - x}\right)} = \dfrac{\sqrt{(1 + x)}}{\sqrt{(1 - x)}} = 1 + x + \dfrac{1}{2}x^2\right\} \Rightarrow (1 + x)^{\frac{1}{2}} = \left(1 + x + \dfrac{1}{2}x^2\right)(1 - x)^{\frac{1}{2}}\) | B1 |
| \((1 + x)^{\frac{1}{2}} = 1 + \left(\dfrac{1}{2}\right)x + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}x^2 + \ldots\ \left\{= 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \ldots\right\}\), \((1 - x)^{\frac{1}{2}} = 1 + \left(\dfrac{1}{2}\right)(-x) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(-x)^2 + \ldots\ \left\{= 1 - \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \ldots\right\}\) | M1A1A1 |
| \(\text{RHS} = \left(1 + x + \dfrac{1}{2}x^2\right)(1 - x)^{\frac{1}{2}} = \left(1 + x + \dfrac{1}{2}x^2\right)\left(1 - \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \ldots\right)\) | |
| \(= 1 - \dfrac{1}{2}x - \dfrac{1}{8}x^2 + x - \dfrac{1}{2}x^2 + \dfrac{1}{2}x^2\) See notes | M1 |
| \(= 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2\) | |
| So, \(\text{LHS} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 = \text{RHS}\) | A1 * |
| (6) |
B1: \((1 + x)^{\frac{1}{2}} = \left(1 + x + \dfrac{1}{2}x^2\right)(1 - x)^{\frac{1}{2}}\) seen or implied.
M1: For Way 4, this M1 mark is dependent on the first B1 mark.
Expands \((1 + x)^{\frac{1}{2}}\) to give any 2 out of 3 terms simplified or un-simplified,
Eg: \(1 + \dfrac{1}{2}x\) or \(+\left(\dfrac{1}{2}\right)x + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}x^2\) or \(1 + \ldots + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}x^2\)
or expands \((1 - x)^{\frac{1}{2}}\) to give any 2 out of 3 terms simplified or un-simplified,
Eg: \(1 + \left(\dfrac{1}{2}\right)(-x)\) or \(+\left(\dfrac{1}{2}\right)(-x) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(-x)^2\) or \(1 + \ldots + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(-x)^2\)
A1: At least one binomial expansion correct (either un-simplified or simplified). (ignore \(x^3\) and \(x^4\) terms)
A1: Two binomial expansions are correct (either un-simplified or simplified). (ignore \(x^3\) and \(x^4\) terms)
M1: For Way 4, this M1 mark is dependent on the first B1 mark.
Multiplies out RHS to give 1, exactly two terms in \(x\) and exactly three terms in \(x^2\).
A1: Candidate achieves the result on the exam paper. Candidate needs to have correctly processed both the LHS and RHS of \((1 + x)^{\frac{1}{2}} = \left(1 + x + \dfrac{1}{2}x^2\right)(1 - x)^{\frac{1}{2}}\).
| Scheme | Marks |
|---|---|
| \(\sqrt{\left(\dfrac{1 + \left(\frac{1}{26}\right)}{1 - \left(\frac{1}{26}\right)}\right)} = 1 + \left(\dfrac{1}{26}\right) + \dfrac{1}{2}\left(\dfrac{1}{26}\right)^2\) | M1 |
| ie: \(\dfrac{3\sqrt{3}}{5} = \dfrac{1405}{1352}\) | B1 |
| so, \(\sqrt{3} = \dfrac{7025}{4056}\) \(\dfrac{7025}{4056}\) | A1 cao |
| (3) | |
| (9 marks) |
Notes
M1: Substitutes \(x = \dfrac{1}{26}\) into both sides of \(\sqrt{\left(\dfrac{1 + x}{1 - x}\right)}\) and \(1 + x + \dfrac{1}{2}x^2\)
B1: For sight of \(\sqrt{\dfrac{27}{25}}\) (or better) and \(\dfrac{1405}{1352}\) or equivalent fraction
Eg: \(\dfrac{3\sqrt{3}}{5}\) and \(\dfrac{1405}{1352}\) or \(0.6\sqrt{3}\) and \(\dfrac{1405}{1352}\) or \(\dfrac{3\sqrt{3}}{5}\) and \(1\dfrac{53}{1352}\) or \(\sqrt{3}\) and \(\dfrac{5}{3}\left(\dfrac{1405}{1352}\right)\)
are fine for B1.
A1: \(\dfrac{7025}{4056}\) or any equivalent fraction, eg: \(\dfrac{14050}{8112}\) or \(\dfrac{182650}{105456}\) etc.
Special Case: Award SC: M1B1A0 for \(\sqrt{3} \approx 1.732001972..\) or truncated 1.732001 or awrt 1.732002.
Note that \(\dfrac{7025}{4056} = 1.732001972...\) and \(\sqrt{3} = 1.732050808...\)