C2 January 2011 Q6
6. \[y = \frac{5}{3x^2 - 2}\]
| \(x\) | 2 | 2.25 | 2.5 | 2.75 | 3 |
|---|---|---|---|---|---|
| \(y\) | 0.5 | 0.38 | 0.2 |
Figure 2 shows a sketch of part of the curve with equation \(y = \dfrac{5}{3x^2 - 2}\), \(x \gt 1\).
At the points \(A\) and \(B\) on the curve, \(x = 2\) and \(x = 3\) respectively.
The region \(S\) is bounded by the curve, the straight line through \(B\) and \((2, 0)\), and the line through \(A\) parallel to the \(y\)-axis. The region \(S\) is shown shaded in Figure 2.
| \(x\) | 2 | 2.25 | 2.5 | 2.75 | 3 |
|---|---|---|---|---|---|
| \(y\) | 0.5 | 0.38 | 0.298507… | 0.241691… | 0.2 |
| Scheme | Marks |
|---|---|
| At \(\{x = 2.5,\}\ y = 0.30\) (only) At least one \(y\)-ordinate correct. | B1 |
| At \(\{x = 2.75,\}\ y = 0.24\) (only) Both \(y\)-ordinates correct. | B1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2} \times 0.25\) ; Outside brackets \(\tfrac{1}{2} \times 0.25\) or \(\tfrac{1}{8}\) | B1 aef |
| \(\times\underline{\left\{0.5 + 0.2 + 2\left(0.38 + \text{their } 0.30 + \text{their } 0.24\right)\right\}}\) For structure of \(\{\ldots\ldots\ldots\}\) ; | M1 |
| Correct expression inside brackets which all must be multiplied by their “outside constant”. | A1 √ |
| \(\left\{= \tfrac{1}{8}(2.54)\right\} = \text{awrt } 0.32\) | A1 |
| (4) |
Notes
B1 for using \(\tfrac{1}{2} \times 0.25\) or \(\tfrac{1}{8}\) or equivalent.
M1 requires the correct \(\{\ldots\ldots\}\) bracket structure. This is for the first bracket to contain first \(y\)-ordinate plus last \(y\)-ordinate and the second bracket to be the summation of the remaining \(y\)-ordinates in the table.
No errors (eg. an omission of a \(y\)-ordinate or an extra \(y\)-ordinate or a repeated \(y\)-ordinate) are allowed in the second bracket and the second bracket must be multiplied by 2. Only one copying error is allowed here in the \(2\left(0.38 + \text{their } 0.30 + \text{their } 0.24\right)\) bracket.
A1ft for the correct bracket \(\{\ldots\ldots\}\) following through candidate’s \(y\)-ordinates found in part (a).
A1 for answer of awrt 0.32.
Bracketing mistake: Unless the final answer implies that the calculation has been done correctly then award M1A0A0 for either \(\dfrac{1}{2} \times 0.25 \times 0.5 + 2\left(0.38 + \text{their } 0.30 + \text{their } 0.24\right) + 0.2\) (nb: yielding final answer of 2.1025) so that the 0.5 is only multiplied by \(\dfrac{1}{2} \times 0.25\)
or \(\dfrac{1}{2} \times 0.25 \times (0.5 + 0.2) + 2\left(0.38 + \text{their } 0.30 + \text{their } 0.24\right)\) (nb: yielding final answer of 1.9275) so that the \((0.5 + 0.2)\) is multiplied by \(\dfrac{1}{2} \times 0.25\).
Need to see trapezium rule – answer only (with no working) gains no marks.
Alternative: Separate trapezia may be used, and this can be marked equivalently. (See appendix, shown below as Way 2.)
Alternative (b) Way 2
| Scheme | Marks |
|---|---|
| \(0.25 \times \left\{\dfrac{0.5 + 0.38}{2} + \dfrac{0.38 + 0.30}{2} + \dfrac{0.30 + 0.24}{2} + \dfrac{0.24 + 0.2}{2}\right\}\) 0.25 and a divisor of 2 on all terms inside brackets. | B1 |
| which is equivalent to: One of first and last ordinates, two of the middle ordinates inside brackets ignoring the denominator of 2. | M1 |
| \(\dfrac{1}{2} \times 0.25\) ;\(\times\underline{\left\{(0.5 + 0.2) + 2\left(0.38 + \text{their } 0.30 + \text{their } 0.24\right)\right\}}\) Correct expression inside brackets if \(\tfrac{1}{2}\) was to be factorised out. | A1 √ |
| \(\left\{= \tfrac{1}{8}(2.54)\right\} = \text{awrt } 0.32\) awrt 0.32 | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Area of triangle \(= \dfrac{1}{2} \times 1 \times 0.2 = 0.1\) | B1 |
| \(\text{Area}(S) = \text{“}0.3175\text{”} - 0.1\) | M1 |
| \(= 0.2175\) | A1 ft |
| (3) | |
| [9] |
Notes
B1 for the area of the triangle identified as either \(\dfrac{1}{2} \times 1 \times 0.2\) or 0.1. May be identified on the diagram.
M1 for “part (b) answer” – “0.1 only” or “part (b) answer – their attempt at 0.1 only”. (Strict attempt!)
A1ft for correctly following through “part (b) answer” – 0.1. This is also dependent on the answer to (b) being greater than 0.1. Note: candidates may round answers here, so allow A1ft if they round their answer correct to 2 dp.