Standard +0.3 This is a structured Further Maths question requiring standard techniques: finding range by rearranging to a quadratic inequality, differentiation using quotient rule for turning points, and identifying horizontal asymptote by comparing degrees. While it involves multiple steps, each technique is routine for FP1 students and the question provides clear guidance through its parts.
9 The curve \(C\) has equation
$$y = \frac { 2 x ^ { 2 } + 2 x + 3 } { x ^ { 2 } + 2 }$$
Show that, for all \(x , 1 \leqslant y \leqslant \frac { 5 } { 2 }\).
Find the coordinates of the turning points on \(C\).
Find the equation of the asymptote of \(C\).
Sketch the graph of \(C\), stating the coordinates of any intersections with the \(y\)-axis and the asymptote.
From this \(\left(2, 2\dfrac{1}{2}\right)\) is the other turning point
B1, A1
7 marks
Part (i): Nature of turning points
Continuous function (implied by graph)
\(\Rightarrow (2,2.5)\) Max and \((-1,1)\) Min
Answer
Marks
Guidance
\(\Rightarrow 1\leq y\leq\dfrac{5}{2}\)
M1, M1A1, A1
(AG)
Question 9 (alternative approach):
Part: Forms quadratic equation in x
\(yx^2+2y=2x^2+2x+3\)
Answer
Marks
\(\Rightarrow (y-2)x^2-2x+(2y-3)=0\)
M1A1
Part: Uses discriminant
For real \(x\): \(4-4(y-2)(2y-3)\geq 0\)
\(\Rightarrow (2y-5)(y-1)\leq 0\)
Answer
Marks
Guidance
\(\Rightarrow 1\leq y\leq\dfrac{5}{2}\)
M1, A1
(AG)
Part: Differentiates and equates to zero
\(y'=0\)
\(\Rightarrow (x^2+2)(4x+2)-2x(2x^2+2x+3)=0\)
Answer
Marks
\(\Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1\) or \(x=2\)
M1
Part: States coordinates of turning points
Answer
Marks
Guidance
Turning points are \((-1,1)\) and \(\left(2, 2\dfrac{1}{2}\right)\)
A1A1
3 marks
Part: Asymptote
\(y=2+\dfrac{2x-1}{x^2+2}\)
Answer
Marks
Guidance
As \(x\rightarrow\pm\infty\), \(y\rightarrow 2\) \(\therefore y=2\)
M1A1
2 marks
Part: Intercepts and graph
Answer
Marks
Guidance
Shows \(\left(0,1\dfrac{1}{2}\right)\) and \(\left(\dfrac{1}{2},2\right)\)
B1
Completely correct graph
B1
2 marks
# Question 9(i) and (ii):
## Part: Possible approach - first two parts together
$y=\dfrac{2x^2+2x+3}{x^2+2}=1+\dfrac{(x+1)^2}{x^2+2}$ | B1 |
$\dfrac{(x+1)^2}{x^2+2}\geq 0 \Rightarrow y\geq 1$ | B1 |
From this $(-1,1)$ is a turning point | M1A1 |
$y=\dfrac{2x^2+2x+3}{x^2+2}=\dfrac{5}{2}-\dfrac{(x-2)^2}{2(x^2+2)}$ | B1 |
$\dfrac{(x-2)^2}{2(x^2+2)}\geq 0 \Rightarrow y\leq\dfrac{5}{2}$ | B1 |
From this $\left(2, 2\dfrac{1}{2}\right)$ is the other turning point | B1, A1 | 7 marks
## Part (i): Nature of turning points
Continuous function (implied by graph)
$\Rightarrow (2,2.5)$ Max and $(-1,1)$ Min
$\Rightarrow 1\leq y\leq\dfrac{5}{2}$ | M1, M1A1, A1 | (AG) | 4 marks
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# Question 9 (alternative approach):
## Part: Forms quadratic equation in x
$yx^2+2y=2x^2+2x+3$
$\Rightarrow (y-2)x^2-2x+(2y-3)=0$ | M1A1 |
## Part: Uses discriminant
For real $x$: $4-4(y-2)(2y-3)\geq 0$
$\Rightarrow (2y-5)(y-1)\leq 0$
$\Rightarrow 1\leq y\leq\dfrac{5}{2}$ | M1, A1 | (AG) | 4 marks
## Part: Differentiates and equates to zero
$y'=0$
$\Rightarrow (x^2+2)(4x+2)-2x(2x^2+2x+3)=0$
$\Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1$ or $x=2$ | M1 |
## Part: States coordinates of turning points
Turning points are $(-1,1)$ and $\left(2, 2\dfrac{1}{2}\right)$ | A1A1 | 3 marks
## Part: Asymptote
$y=2+\dfrac{2x-1}{x^2+2}$
As $x\rightarrow\pm\infty$, $y\rightarrow 2$ $\therefore y=2$ | M1A1 | 2 marks
## Part: Intercepts and graph
Shows $\left(0,1\dfrac{1}{2}\right)$ and $\left(\dfrac{1}{2},2\right)$ | B1 |
Completely correct graph | B1 | 2 marks
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9 The curve $C$ has equation
$$y = \frac { 2 x ^ { 2 } + 2 x + 3 } { x ^ { 2 } + 2 }$$
Show that, for all $x , 1 \leqslant y \leqslant \frac { 5 } { 2 }$.
Find the coordinates of the turning points on $C$.
Find the equation of the asymptote of $C$.
Sketch the graph of $C$, stating the coordinates of any intersections with the $y$-axis and the asymptote.
\hfill \mbox{\textit{CAIE FP1 2012 Q9 [11]}}