| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Solving Inequalities with Rational Functions |
| Difficulty | Standard +0.3 This is a standard FP1 rational function question covering routine techniques: showing numerator ≠ 0, identifying asymptotes from denominator zeros and end behavior, sketching, and solving a rational inequality. While it requires multiple steps and careful sign analysis for part (v), all techniques are textbook exercises without novel insight. Slightly above average difficulty due to the inequality requiring consideration of multiple intervals. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 0 \Rightarrow 3 + x^2 = 0\), no real solutions since \(3 + x^2 \geq 3 > 0\) | B1 | clear argument |
| Answer | Marks | Guidance |
|---|---|---|
| Vertical asymptotes: \(x = 2\), \(x = -2\) | B1 | |
| Horizontal asymptote: \(y = -1\) | B1 | (from \(\frac{x^2}{-x^2} \to -1\)) |
| All three stated | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| As \(x \to \pm\infty\), \(y \to -1\) from below (i.e. \(y < -1\)) | M1 A1 | with justification |
| Answer | Marks |
|---|---|
| Correct sketch showing asymptotes, two branches in \(x>2\) and \(x<-2\) regions approaching \(y=-1\), middle branch | B3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(3 + x^2 \leq -2(4-x^2)\) needs care with sign of \((4-x^2)\) | M1 | |
| \(3 + x^2 + 8 - 2x^2 \leq 0 \Rightarrow 11 - x^2 \leq 0\) (where \(4 - x^2 > 0\)) | M1 | |
| \(x^2 \geq 11\), combined with \(-2 < x < 2\): no solution in this region | A1 | |
| Where \(4 - x^2 < 0\): inequality reverses, \(11 - x^2 \geq 0\), so \(x^2 \leq 11\), combined with \( | x | >2\): |
| \(-\sqrt{11} \leq x < -2\) or \(2 < x \leq \sqrt{11}\) | A1 |
# Question 7:
**(i)**
$y = 0 \Rightarrow 3 + x^2 = 0$, no real solutions since $3 + x^2 \geq 3 > 0$ | B1 | clear argument
**(ii)**
Vertical asymptotes: $x = 2$, $x = -2$ | B1 |
Horizontal asymptote: $y = -1$ | B1 | (from $\frac{x^2}{-x^2} \to -1$)
All three stated | B1 |
**(iii)**
As $x \to \pm\infty$, $y \to -1$ from below (i.e. $y < -1$) | M1 A1 | with justification
**(iv)**
Correct sketch showing asymptotes, two branches in $x>2$ and $x<-2$ regions approaching $y=-1$, middle branch | B3 |
**(v)**
$\frac{3+x^2}{4-x^2} \leq -2$
$3 + x^2 \leq -2(4-x^2)$ needs care with sign of $(4-x^2)$ | M1 |
$3 + x^2 + 8 - 2x^2 \leq 0 \Rightarrow 11 - x^2 \leq 0$ (where $4 - x^2 > 0$) | M1 |
$x^2 \geq 11$, combined with $-2 < x < 2$: no solution in this region | A1 |
Where $4 - x^2 < 0$: inequality reverses, $11 - x^2 \geq 0$, so $x^2 \leq 11$, combined with $|x|>2$: | A1 |
$-\sqrt{11} \leq x < -2$ or $2 < x \leq \sqrt{11}$ | A1 |
---7 A curve has equation $y = \frac { 3 + x ^ { 2 } } { 4 - x ^ { 2 } }$.\\
(i) Show that $y$ can never be zero.\\
(ii) Write down the equations of the two vertical asymptotes and the one horizontal asymptote.\\
(iii) Describe the behaviour of the curve for large positive and large negative values of $x$, justifying your description.\\
(iv) Sketch the curve.\\
(v) Solve the inequality $\frac { 3 + x ^ { 2 } } { 4 - x ^ { 2 } } \leqslant - 2$.
\hfill \mbox{\textit{OCR MEI FP1 2006 Q7 [13]}}