Question 10 - A-Level Maths

CAIE FP1 2011 November — Question 10 13 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2011
Session	November
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Curve Sketching
Type	Range restriction with discriminant (quadratic denominator)
Difficulty	Challenging +1.2 This is a multi-part Further Maths question requiring factorization, axis intersections, range restriction via discriminant analysis (rearranging to quadratic in x and requiring Δ≥0), and curve sketching with turning points. While it involves several techniques, each step follows standard FP1 procedures without requiring novel insight—the discriminant method for range is a textbook technique for this topic.
Spec	1.02k Simplify rational expressions: factorising, cancelling, algebraic division 1.02n Sketch curves: simple equations including polynomials 1.07n Stationary points: find maxima, minima using derivatives

10 A curve $C$ has equation $$y = \frac { 5 \left( x ^ { 2 } - x - 2 \right) } { x ^ { 2 } + 5 x + 10 }$$ Find the coordinates of the points of intersection of $C$ with the axes. Show that, for all real values of $x , - 1 \leqslant y \leqslant 15$. Sketch $C$, stating the coordinates of any turning points and the equation of the horizontal asymptote.
[0pt] [Question 11 is printed on the next page.]

Show mark scheme Show mark scheme source

Question 10:

Curve Sketching

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$(-1,0),\ (2,0)$ and $(0,-1)$	B1, B1	2 marks, intersections with axes
$(y-5)x^2+(5y+5)x+10(y+1)=0$		Rearranges as quadratic
For real $x:\ b^2-4ac\geq 0$		Uses discriminant
$(5y+5)^2-40(y-5)(y+1)\geq 0\ldots$	M1A1
$(y-15)(y+1)\leq 0 \Rightarrow -1\leq y\leq 15$	M1A1 (AG)	4 marks, solves inequality
$y=-1\Rightarrow x=0\quad y=15\Rightarrow x=-4$	M1A1	Finds turning points
Turning points are $(-4,15)$ and $(0,-1)$
$y=5$	B1	States asymptote
Axes and asymptote correct	B1
Graph correct	B1B1	7 marks, total [13]

# Question 10:

## Curve Sketching

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(-1,0),\ (2,0)$ and $(0,-1)$ | B1, B1 | 2 marks, intersections with axes |
| $(y-5)x^2+(5y+5)x+10(y+1)=0$ | | Rearranges as quadratic |
| For real $x:\ b^2-4ac\geq 0$ | | Uses discriminant |
| $(5y+5)^2-40(y-5)(y+1)\geq 0\ldots$ | M1A1 | |
| $(y-15)(y+1)\leq 0 \Rightarrow -1\leq y\leq 15$ | M1A1 (AG) | 4 marks, solves inequality |
| $y=-1\Rightarrow x=0\quad y=15\Rightarrow x=-4$ | M1A1 | Finds turning points |
| Turning points are $(-4,15)$ and $(0,-1)$ | | |
| $y=5$ | B1 | States asymptote |
| Axes and asymptote correct | B1 | |
| Graph correct | B1B1 | 7 marks, total [13] |

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Show LaTeX source

10 A curve $C$ has equation

$$y = \frac { 5 \left( x ^ { 2 } - x - 2 \right) } { x ^ { 2 } + 5 x + 10 }$$

Find the coordinates of the points of intersection of $C$ with the axes.

Show that, for all real values of $x , - 1 \leqslant y \leqslant 15$.

Sketch $C$, stating the coordinates of any turning points and the equation of the horizontal asymptote.\\[0pt]
[Question 11 is printed on the next page.]

\hfill \mbox{\textit{CAIE FP1 2011 Q10 [13]}}

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Q1 6 Q2 6 Q3 7 Q4 7 Q5 7 Q6 8 Q7 9 Q8 10 Q9 13 Q10 13 Q11 EITHER Q11 OR