AS June 2020 Paper 1 Q14
14
(a) Given\[\frac{x + 7}{x + 1} \leqslant x + 1\]
show that
\[\frac{(x + a)(x + b)}{x + c} \geqslant 0\]where \(a\), \(b\), and \(c\) are integers to be found. [4 marks]
(b) Briefly explain why this statement is incorrect.\[\frac{(x + p)(x + q)}{x + r} \geqslant 0 \Leftrightarrow (x + p)(x + q)(x + r) \geqslant 0\]
[1 mark]
(c) Solve\[\frac{x + 7}{x + 1} \leqslant x + 1\]
[2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes both terms on one side of the inequality, with a common denominator. | M1 | 1.1a |
| Correctly combines their two fractions into one fraction. | A1F | 1.1b |
| Obtains a single fraction in which the numerator is a three-term quadratic. | M1 | 1.1a |
| Completes a rigorous argument to show the correct inequality in the required form. Accept \(0 \leqslant \frac{(x + -2)(x + 3)}{x + 1}\). | R1 | 2.1 |
Typical solution
\[0 \leqslant \frac{(x + 1)^2}{x + 1} - \frac{x + 7}{x + 1}\]\[0 \leqslant \frac{x^2 + 2x + 1 - x - 7}{x + 1}\]\[0 \leqslant \frac{x^2 + x - 6}{x + 1}\]\[\frac{(x + 3)(x - 2)}{x + 1} \geqslant 0\]| Scheme | Marks | AO |
|---|---|---|
| Explains that \(x = -r\) is a solution of the inequality on the RHS, but not the one of the LHS. | B1 | 2.4 |
Typical solution
\(x = -r\) is a solution of the inequality on the right, but not the one on the left.
| Scheme | Marks | AO |
|---|---|---|
| Obtains one correct region, FT their three critical values. Condone \(-3 \leqslant x \leqslant -1\). | M1 | 1.1a |
| Obtains both correct regions. \(-3 \leqslant x \lt -1\), \(x \geqslant 2\) | A1 | 1.1b |
| (7 marks) |
Typical solution
