Question
This question asks you to investigate conditions for the existence of complex roots of polynomial equations of degree 3 and 4.
The cubic equation $x^3+p x^2+q x+r=0$, where $p, q, r \in \mathbb{R}$, has roots $\alpha, \beta$ and $\gamma$.
Consider the equation $x^3-7 x^2+q x+1=0$, where $q \in \mathbb{R}$.
Noah believes that if $p^2 \geq 3 q$ then $\alpha, \beta$ and $\gamma$ are all real.
Now consider polynomial equations of degree 4 .
The equation $x^4+p x^3+q x^2+r x+s=0$, where $p, q, r, s \in \mathbb{R}$, has roots $\alpha, \beta, \gamma$ and $\delta$.
In a similar way to the cubic equation, it can be shown that:
$$
\begin{aligned}
& p=-(\alpha+\beta+\gamma+\delta) \\
& q=\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta \\
& r=-(\alpha \beta \gamma+\alpha \beta \delta+\alpha \gamma \delta+\beta \gamma \delta) \\
& s=\alpha \beta \gamma \delta
\end{aligned}
$$
The equation $x^4-9 x^3+24 x^2+22 x-12=0$, has one integer root.
a. By expanding $(x-\alpha)(x-\beta)(x-\gamma)$ show that:
$$
\begin{aligned}
& p=-(\alpha+\beta+\gamma) \\
& q=\alpha \beta+\beta \gamma+\gamma \alpha \\
& r=-\alpha \beta \gamma
\end{aligned}
$$
b.i. Show that $p^2-2 q=\alpha^2+\beta^2+\gamma^2$.
b.iiHence show that $(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2=2 p^2-6 q$.
c. Given that $p^2<3 q$, deduce that $\alpha, \beta$ and $\gamma$ cannot all be real.
d. Using the result from part (c), show that when $q=17$, this equation has at least one complex root.
e.i. By varying the value of $q$ in the equation $x^3-7 x^2+q x+1=0$, determine the smallest positive integer value of $q$ required to show that Noah is incorrect.
e.ii.Explain why the equation will have at least one real root for all values of $q$.
f.i. Find an expression for $\alpha^2+\beta^2+\gamma^2+\delta^2$ in terms of $p$ and $q$.
f.ii. Hence state a condition in terms of $p$ and $q$ that would imply $x^4+p x^3+q x^2+r x+s=0$ has at least one complex root.
g. Use your result from part (f)(ii) to show that the equation $x^4-2 x^3+3 x^2-4 x+5=0$ has at least one complex root.
h.i. State what the result in part (f)(ii) tells us when considering this equation $x^4-9 x^3+24 x^2+22 x-12=0$.
h.ii.Write down the integer root of this equation.
h.iiiBy writing $x^4-9 x^3+24 x^2+22 x-12$ as a product of one linear and one cubic factor, prove that the equation has at least one complex root.
▶️Answer/Explanation
a. attempt to expand $(x-\alpha)(x-\beta)(x-\gamma) \quad$ M1
$$
\begin{aligned}
& =\left(x^2-(\alpha+\beta) x+\alpha \beta\right)(x-\gamma) \text { OR }=(x-\alpha)\left(x^2-(\beta+\gamma) x+\beta \gamma\right) \\
& \left(x^3+p x^2+q x+r\right)=x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma
\end{aligned}
$$
comparing coefficients:
$$
\begin{aligned}
& p=-(\alpha+\beta+\gamma) \quad \text { AG } \\
& q=(\alpha \beta+\beta \gamma+\gamma \alpha) \quad \text { AG } \\
& r=-\alpha \beta \gamma \quad \text { AG }
\end{aligned}
$$
Note: For candidates who do not include the $\boldsymbol{A} \boldsymbol{G}$ lines award full marks.
[3 marks]
b.i. $p^2-2 q=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)$
attempt to expand $(\alpha+\beta+\gamma)^2$
(M1)
$$
\begin{aligned}
& =\alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \gamma+\gamma \alpha)-2(\alpha \beta+\beta \gamma+\gamma \alpha) \text { or equivalent } \\
& =\alpha^2+\beta^2+\gamma^2 \quad \text { AG }
\end{aligned}
$$
A1
Note: Accept equivalent working from RHS to LHS.
[3 marks]
b.iiEITHER
attempt to expand $(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2$
(M1)
$$
\begin{aligned}
& =\left(\alpha^2+\beta^2-2 \alpha \beta\right)+\left(\beta^2+\gamma^2-2 \beta \gamma\right)+\left(\gamma^2+\alpha^2-2 \gamma \alpha\right) \\
& =2\left(\alpha^2+\beta^2+\gamma^2\right)-2(\alpha \beta+\beta \gamma+\gamma \alpha) \\
& =2\left(p^2-2 q\right)-2 q \text { or equivalent } \quad \text { A1 } \\
& =2 p^2-6 q \quad \text { AG }
\end{aligned}
$$
A1
OR
attempt to write $2 p^2-6 q$ in terms of $\alpha, \beta, \gamma$
(M1)
$$
\begin{aligned}
& =2\left(p^2-2 q\right)-2 q \\
& =2\left(\alpha^2+\beta^2+\gamma^2\right)-2(\alpha \beta+\beta \gamma+\gamma \alpha) \quad \text { A1 } \\
& =\left(\alpha^2+\beta^2-2 \alpha \beta\right)+\left(\beta^2+\gamma^2-2 \beta \gamma\right)+\left(\gamma^2+\alpha^2-2 \gamma \alpha\right) \\
& =(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2 \quad \text { AG }
\end{aligned}
$$
A1
AG
A1
Note: Accept equivalent working where LHS and RHS are expanded to identical expressions.
[3 marks]
c. $p^2<3 q \Rightarrow 2 p^2-6 q<0$
$\Rightarrow(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2<0$
A1
if all roots were real $(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2 \geq 0 \quad \boldsymbol{R 1}$
Note: Condone strict inequality in the $\boldsymbol{R} \mathbf{1}$ line.
Note: Do not award AOR1.
$\Rightarrow$ roots cannot all be real $\quad A G$
[2 marks]
d. $p^2=(-7)^2=49$ and $3 q=51$
A1
so $p^2<3 q \Rightarrow$ the equation has at least one complex root
R1
Note: Allow equivalent comparisons; e.g. checking $p^2<6 q$
[2 marks]
e.i. use of GDC (eg graphs or tables)
(M1)
$q=12 \quad \boldsymbol{A 1}$
[2 marks]
e.ii.complex roots appear in conjugate pairs (so if complex roots occur the other root will be real OR all 3 roots will be real).
OR
a cubic curve always crosses the $x$-axis at at least one point. $\quad \boldsymbol{R 1}$
[1 mark]
f.i. attempt to expand $(\alpha+\beta+\gamma+\delta)^2$
(M1)
$$
\begin{aligned}
& (\alpha+\beta+\gamma+\delta)^2=\alpha^2+\beta^2+\gamma^2+\delta^2+2(\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta) \\
& \Rightarrow \alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta) \\
& \left(\Rightarrow \alpha^2+\beta^2+\gamma^2+\delta^2=\right) p^2-2 q
\end{aligned}
$$
A1
[3 marks]
f.ii. $p^2<2 q$ OR $p^2-2 q<0$
A1
Note: Allow FT on their result from part (f)(i).
[1 mark]
g. $4<6$ OR $2^2-2 \times 3<0$
R1
hence there is at least one complex root.
AG
Note: Allow $\boldsymbol{F T}$ from part (f)(ii) for the $\boldsymbol{R}$ mark provided numerical reasoning is seen.
[1 mark]
h.i. $\left(p^2>2 q\right)(81>2 \times 24)$ (so) nothing can be deduced $\quad \boldsymbol{R 1}$
Note: Do not allow $\boldsymbol{F T}$ for the $\boldsymbol{R}$ mark.
[1 mark]
h.ii.-1
A1
[1 mark]
h.iiiattempt to express as a product of a linear and cubic factor $\quad$ M1
$$
(x+1)\left(x^3-10 x^2+34 x-12\right)
$$
A1A1
Note: Award A1 for each factor. Award at most A1AO if not written as a product.
since for the cubic, $p^2<3 q(100<102)$
R1
there is at least one complex root
AG
[4 marks]
Question
This question asks you to explore cubic polynomials of the form $(x-r)\left(x^2-2 a x+a^2+b^2\right)$ for $x \in \mathbb{R}$ and corresponding cubic equations with one real root and two complex roots of the form $(z-r)\left(z^2-2 a z+a^2+b^2\right)=0$ for $z \in \mathbb{C}$.
In parts (a), (b) and (c), let $r=1, a=4$ and $b=1$.
Consider the equation $(z-1)\left(z^2-8 z+17\right)=0$ for $z \in \mathbb{C}$.
Consider the function $f(x)=(x-1)\left(x^2-8 x+17\right)$ for $x \in \mathbb{R}$.
Consider the function $g(x)=(x-r)\left(x^2-2 a x+a^2+b^2\right)$ for $x \in \mathbb{R}$ where $r, a \in \mathbb{R}$ and $b \in \mathbb{R}, b>0$.
The equation $(z-r)\left(z^2-2 a z+a^2+b^2\right)=0$ for $z \in \mathbb{C}$ has roots $r$ and $a \pm b$ i where $r, a \in \mathbb{R}$ and $b \in \mathbb{R}, b>0$.
On the Cartesian plane, the points $\mathrm{C}_1\left(a, \sqrt{g^{\prime}(a)}\right)$ and $\mathrm{C}_2\left(a,-\sqrt{g^{\prime}(a)}\right)$ represent the real and imaginary parts of the complex roots of the equation $(z-r)\left(z^2-2 a z+a^2+b^2\right)=0$.
The following diagram shows a particular curve of the form $y=(x-r)\left(x^2-2 a x+a^2+16\right)$ and the tangent to the curve at the point $\mathrm{A}(a, 80)$. The curve and the tangent both intersect the $x$-axis at the point $\mathrm{R}(-2,0)$. The points $\mathrm{C}_1$ and $\mathrm{C}_2$ are also shown.
Consider the curve $y=(x-r)\left(x^2-2 a x+a^2+b^2\right)$ for $a \neq r, b>0$. The points $\mathrm{A}(a, g(a))$ and $\mathrm{R}(r, 0)$ are as defined in part (d)(ii). The curve has a point of inflexion at point $\mathrm{P}$.
Consider the special case where $a=r$ and $b>0$.
a.i. Given that 1 and $4+\mathrm{i}$ are roots of the equation, write down the third root.
a.ii.Verify that the mean of the two complex roots is 4 .
b. Show that the line $y=x-1$ is tangent to the curve $y=f(x)$ at the point $\mathrm{A}(4,3)$.
c. Sketch the curve $y=f(x)$ and the tangent to the curve at point A, clearly showing where the tangent crosses the $x$-axis.
d.i.Show that $g^{\prime}(x)=2(x-r)(x-a)+x^2-2 a x+a^2+b^2$.
d.iiHence, or otherwise, prove that the tangent to the curve $y=g(x)$ at the point $\mathrm{A}(a, g(a))$ intersects the $x$-axis at the point $\mathrm{R}(r, 0)$.
e. Deduce from part (d)(i) that the complex roots of the equation $(z-r)\left(z^2-2 a z+a^2+b^2\right)=0$ can be expressed as $a \pm \mathrm{i} \sqrt{g^{\prime}(a)}$.
f.i. Use this diagram to determine the roots of the corresponding equation of the form $(z-r)\left(z^2-2 a z+a^2+16\right)=0$ for $z \in \mathbb{C}$.
f.ii. State the coordinates of $\mathrm{C}_2$.
g.i. Show that the $x$-coordinate of $\mathrm{P}$ is $\frac{1}{3}(2 a+r)$.
You are not required to demonstrate a change in concavity.
g.ii.Hence describe numerically the horizontal position of point $\mathrm{P}$ relative to the horizontal positions of the points $\mathrm{R}$ and $\mathrm{A}$.
h.i. Sketch the curve $y=(x-r)\left(x^2-2 a x+a^2+b^2\right)$ for $a=r=1$ and $b=2$.
h.ii.For $a=r$ and $b>0$, state in terms of $r$, the coordinates of points $\mathrm{P}$ and $\mathrm{A}$.
▶️Answer/Explanation
a.i. 4 – i
A1
[1 mark]a.i.mean $=\frac{1}{2}(4+\mathrm{i}+4-\mathrm{i}) \quad \boldsymbol{A 1}$
$=4 \quad \boldsymbol{A G}$
[1 mark]
b. METHOD 1
attempts product rule differentiation
(M1)
Note: Award (M1) for attempting to express $f(x)$ as $f(x)=x^3-9 x^2+25 x-17$
$$
\begin{array}{ll}
f^{\prime}(x)=(x-1)(2 x-8)+x^2-8 x+17 & \left(f^{\prime}(x)=3 x^2-18 x+25\right) \quad \text { A1 } \\
f^{\prime}(4)=1 \quad \text { A1 }
\end{array}
$$
Note: Where $f^{\prime}(x)$ is correct, award $\mathbf{A} 1$ for solving $f^{\prime}(x)=1$ and obtaining $x=4$.
EITHER
$$
y-3=1(x-4) \quad \text { A1 }
$$
OR
$$
\begin{aligned}
& y=x+c \\
& 3=4+c \Rightarrow c=-1
\end{aligned}
$$
A1
OR
states the gradient of $y=x-1$ is also 1 and verifies that $(4,3)$ lies on the line $y=x-1$
A1
THEN
so $y=x-1$ is the tangent to the curve at $\mathrm{A}(4,3) \quad$ AG
Note: Award a maximum of (MO)AOA1A1 to a candidate who does not attempt to find $f^{\prime}(x)$.
METHOD 2
sets $f(x)=x-1$ to form $x-1=(x-1)\left(x^2-8 x+17\right)$
(M1)
EITHER
$$
(x-1)\left(x^2-8 x+16\right)=0\left(x^3-9 x^2+24 x-16=0\right)
$$
attempts to solve a correct cubic equation
(M1)
$$
(x-1)(x-4)^2=0 \Rightarrow x=1,4
$$
OR
recognises that $x \neq 1$ and forms $x^2-8 x+17=1\left(x^2-8 x+16=0\right)$
A1
attempts to solve a correct quadratic equation
(M1)
$$
(x-4)^2=0 \Rightarrow x=4
$$
THEN
$x=4$ is a double root
R1
so $y=x-1$ is the tangent to the curve at $\mathrm{A}(4,3)$
AG
Note: Candidates using this method are not required to verify that $y=3$.
[4 marks]
c.
a positive cubic with an $x$-intercept $(x=1)$, and a local maximum and local minimum in the first quadrant both positioned to the left of $\mathrm{A}$
Note: As the local minimum and point $A$ are very close to each other, condone graphs that seem to show these points coinciding. For the point of tangency, accept labels such as $\mathrm{A},(4,3)$ or the point labelled from both axes. Coordinates are not required.
a correct sketch of the tangent passing through $\mathrm{A}$ and crossing the $x$-axis at the same point $(x=1)$ as the curve
A1
Note: Award A1AO if both graphs cross the $x$-axis at distinctly different points.
[2 marks]
d.i.EITHER
$$
g^{\prime}(x)=(x-r)(2 x-2 a)+x^2-2 a x+a^2+b^2 \quad \text { (M1)A1 }
$$
OR
$$
g(x)=x^3-(2 a+r) x^2+\left(a^2+b^2+2 a r\right) x-\left(a^2+b^2\right) r
$$
attempts to find $g^{\prime}(x) \quad$ M1
$$
\begin{aligned}
& g^{\prime}(x)=3 x^2-2(2 a+r) x+a^2+b^2+2 a r \\
& =2 x^2-2(a+r) x+2 a r+x^2-2 a x+a^2+b^2 \\
& \left(=2\left(x^2-a x-r x+a r\right)+x^2-2 a x+a^2+b^2\right)
\end{aligned}
$$
THEN
$$
g^{\prime}(x)=2(x-r)(x-a)+x^2-2 a x+a^2+b^2
$$
$A G$
[2 marks]
d.ii.METHOD 1
$$
\begin{aligned}
& g(a)=b^2(a-r) \\
& g^{\prime}(a)=b^2
\end{aligned}
$$
(A1)
attempts to substitute their $g(a)$ and $g^{\prime}(a)$ into $y-g(a)=g^{\prime}(a)(x-a) \quad$ M1
$$
y-b^2(a-r)=b^2(x-a)
$$
EITHER
$$
\begin{aligned}
& y=b^2(x-r)\left(y=b^2 x-b^2 r\right) \quad \text { A1 } \\
& \text { sets } y=0 \text { so } b^2(x-r)=0 \quad \text { M1 } \\
& b>0 \Rightarrow x=r \text { OR } b \neq 0 \Rightarrow x=r \quad \text { R1 }
\end{aligned}
$$
A1
R1
OR
$$
\begin{aligned}
& \text { sets } y=0 \text { so }-b^2(a-r)=b^2(x-a) \quad \text { M1 } \\
& b>0 \text { OR } b \neq 0 \Rightarrow-(a-r)=x-a \quad \text { R1 } \\
& x=r \quad \text { A1 }
\end{aligned}
$$
THEN
so the tangent intersects the $x$-axis at the point $\mathrm{R}(r, 0) \quad$ AG
METHOD 2
$$
\begin{aligned}
& g^{\prime}(a)=b^2 \\
& g(a)=b^2(a-r)
\end{aligned}
$$
attempts to substitute their $g(a)$ and $g^{\prime}(a)$ into $y=g^{\prime}(a) x+c$ and attempts to find $c \quad$ M1
$$
c=-b^2 r
$$
EITHER
$$
y=b^2(x-r)\left(y=b^2 x-b^2 r\right)
$$
sets $y=0$ so $b^2(x-r)=0 \quad$ M1
$$
b>0 \Rightarrow x=r \text { OR } b \neq 0 \Rightarrow x=r
$$
R1
OR
$$
\begin{array}{ll}
\text { sets } y=0 \text { so }-b^2(a-r)=b^2(x-a) & \text { M1 } \\
b>0 \text { OR } b \neq 0 \Rightarrow-(a-r)=x-a & \text { R1 } \\
x=r \quad \text { A1 }
\end{array}
$$
A1
THEN
so the tangent intersects the $x$-axis at the point $\mathrm{R}(r, 0) \quad$ AG
METHOD 2
$$
\begin{aligned}
& g^{\prime}(a)=b^2 \quad \text { (A1) } \\
& g(a)=b^2(a-r)
\end{aligned}
$$
(A1)
attempts to substitute their $g(a)$ and $g^{\prime}(a)$ into $y=g^{\prime}(a) x+c$ and attempts to find $c \quad$ M1 $c=-b^2 r$
EITHER
$$
\begin{aligned}
& y=b^2(x-r)\left(y=b^2 x-b^2 r\right) \quad \text { A1 } \\
& \text { sets } y=0 \text { so } b^2(x-r)=0 \quad \text { M1 } \\
& b>0 \Rightarrow x=r \text { OR } b \neq 0 \Rightarrow x=r \quad \text { R1 }
\end{aligned}
$$
OR
$$
\begin{aligned}
& \text { sets } y=0 \text { so } b^2(x-r)=0 \quad \text { M1 } \\
& b>0 \text { OR } b \neq 0 \Rightarrow x-r=0 \quad \text { R1 } \\
& x=r \quad \text { A1 }
\end{aligned}
$$
THEN
so the tangent intersects the $x$-axis at the point $\mathrm{R}(r, 0) \quad \mathbf{A G}$
METHOD 2
$$
\begin{aligned}
& g^{\prime}(a)=b^2 \\
& g(a)=b^2(a-r)
\end{aligned}
$$
attempts to substitute their $g(a)$ and $g^{\prime}(a)$ into $y=g^{\prime}(a) x+c$ and attempts to find $c \quad$ M1
$$
c=-b^2 r
$$
EITHER
$$
\begin{aligned}
& y=b^2(x-r)\left(y=b^2 x-b^2 r\right) \quad \text { A1 } \\
& \text { sets } y=0 \text { so } b^2(x-r)=0 \quad \text { M1 } \\
& b>0 \Rightarrow x=r \text { OR } b \neq 0 \Rightarrow x=r \quad \text { R1 }
\end{aligned}
$$
OR
$$
\begin{aligned}
& \text { sets } y=0 \text { so } b^2(x-r)=0 \quad \text { M1 } \\
& b>0 \text { OR } b \neq 0 \Rightarrow x-r=0 \quad \text { R1 } \\
& x=r \\
& \text { A1 } \\
&
\end{aligned}
$$
METHOD 3
$$
g^{\prime}(a)=b^2
$$
the line through $R(r, 0)$ parallel to the tangent at $\mathrm{A}$ has equation
$$
y=b^2(x-r)
$$
A1
sets $g(x)=b^2(x-r)$ to form $b^2(x-r)=(x-r)\left(x^2-2 a x+a^2+b^2\right)$
M1
$$
\begin{aligned}
& b^2=x^2-2 a x+a^2+b^2,(x \neq r) \\
& (x-a)^2=0
\end{aligned}
$$
A1
A1
since there is a double root $(x=a)$, this parallel line through $R(r, 0)$ is the required tangent at $\mathrm{A}$
R1
e. EITHER
$g^{\prime}(a)=b^2 \Rightarrow b=\sqrt{g^{\prime}(a)}($ since $b>0) \quad R \mathbf{R 1}$
Note: Accept $b= \pm \sqrt{g^{\prime}(a)}$.
OR
$(a \pm b \mathrm{i}=) a \pm \mathrm{i} \sqrt{b^2}$ and $g^{\prime}(a)=b^2 \quad \boldsymbol{R 1}$
THEN
hence the complex roots can be expressed as $a \pm \mathrm{i} \sqrt{g^{\prime}(a)}$
AG
[1 mark]
f.i. $b=4$ (seen anywhere)
A1
EITHER
attempts to find the gradient of the tangent in terms of $a$ and equates to 16
(M1)
OR
substitutes $r=-2, x=a$ and $y=80$ to form $80=(a-(-2))\left(a^2-2 a^2+a^2+16\right)$
(M1)
OR
substitutes $r=-2, x=a$ and $y=80$ into $y=16(x-r)$
(M1)
THEN
$\frac{80}{a+2}=16 \Rightarrow a=3$
roots are -2 (seen anywhere) and $3 \pm 4 \mathrm{i}$
A1A1
Note: Award $\boldsymbol{A 1}$ for -2 and $\boldsymbol{A 1}$ for $3 \pm 4$ i. Do not accept coordinates.
[4 marks]
f.ii. $(3,-4)$
A1
Note: Accept ” $x=3$ and $y=-4$ “.
Do not award A1FT for $(a,-4)$.
[1 mark]
$$
\text { g.i. } g^{\prime}(x)=2(x-r)(x-a)+x^2-2 a x+a^2+b^2
$$
attempts to find $g “(x) \quad$ M1
$$
g^{\prime \prime}(x)=2(x-a)+2(x-r)+2 x-2 a(=6 x-2 r-4 a)
$$
sets $g “(x)=0$ and correctly solves for $x$
A1
for example, obtaining $x-r+2(x-a)=0$ leading to $3 x=2 a+r$
so $x=\frac{1}{3}(2 a+r) \quad$ AG
Note: Do not award $\boldsymbol{A 1}$ if the answer does not lead to the $\boldsymbol{A G}$.
[2 marks]
g.iipoint $\mathrm{P}$ is $\frac{2}{3}$ of the horizontal distance (way) from point $\mathrm{R}$ to point $\mathrm{A}$
A1
Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship. Award $\mathbf{A O}$ for non-numerical statements such as “P is between $\mathrm{R}$ and $\mathrm{A}$, closer to $\mathrm{A}$ “.
[1 mark]
h.i. $y=(x-1)\left(x^2-2 x+5\right)$
a positive cubic with no stationary points and a non-stationary point of inflexion at $x=1$
A1
Note: Graphs may appear approximately linear. Award this $\boldsymbol{A 1}$ if a change of concavity either side of $x=1$ is apparent. Coordinates are not required and the $y$-intercept need not be indicated.
[2 marks]
$$
\text { h.ii. }(r, 0)
$$
A1
[1 mark]