Detailed solution

(a) The quadratic function is of the form \( f(x) = ax^2 + bx + c \), where each coefficient \( a \), \( b \), and \( c \) is determined by rolling a six-sided die. A standard six-sided die has faces numbered 1 through 6, so each coefficient can take any integer value from 1 to 6. Since the rolls are independent for \( a \), \( b \), and \( c \):

Number of choices for \( a \): 6
Number of choices for \( b \): 6
Number of choices for \( c \): 6

The total number of possible combinations is the product of these choices:

\[ 6 \times 6 \times 6 = 6^3 = 216 \]

Thus, there are 216 possible quadratic functions that can be generated.

(b) To find the x-intercepts of \( f(x) = x^2 + 4x + 4 \), set the function equal to zero and solve for \( x \):

\[ x^2 + 4x + 4 = 0 \]

Notice that this is a perfect square trinomial:

\[ x^2 + 4x + 4 = (x + 2)^2 \]

So the equation becomes:

\[ (x + 2)^2 = 0 \]

Taking the square root of both sides:

\[ x + 2 = 0 \]
\[ x = -2 \]

The equation has exactly one solution, \( x = -2 \), because the quadratic is a perfect square and touches the x-axis at a single point (the vertex). Alternatively, we can check the discriminant (\( \Delta = b^2 – 4ac \)):

– \( a = 1 \), \( b = 4 \), \( c = 4 \)
– \( \Delta = 4^2 – 4 \cdot 1 \cdot 4 = 16 – 16 = 0 \)

A discriminant of zero confirms that there is exactly one real root. Thus, \( f(x) = x^2 + 4x + 4 \) has only one x-intercept.

(c) The number of x-intercepts depends on the discriminant \( \Delta = b^2 – 4ac \). For a quadratic to have exactly one x-intercept, \( \Delta = 0 \), meaning:

\[ b^2 – 4ac = 0 \]
OR \[ b^2 = 4ac \]

Since \( a \), \( b \), and \( c \) are integers from 1 to 6, we need to find all triples \( (a, b, c) \) that satisfy this equation. Also, \( b^2 \) is a perfect square (since \( b \) is an integer), and \( 4ac \) must be positive and a perfect square. Possible values of \( b \) are 1 to 6, so \( b^2 \) can be 1, 4, 9, 16, 25, or 36. We need \( 4ac = b^2 \), and since \( a \) and \( c \) are between 1 and 6, \( ac \) ranges from 1 to 36, and \( 4ac \) from 4 to 144. Let’s find the valid cases:

\( b = 1 \), \( b^2 = 1 \):
\[ 4ac = 1 \]
\[ ac = \frac{1}{4} \]
Not possible since \( ac \) must be an integer and at least 1.

\( b = 2 \), \( b^2 = 4 \):
\[ 4ac = 4 \]
\[ ac = 1 \]
Pairs: \( (a, c) = (1, 1) \)
Number of triples: 1

\( b = 3 \), \( b^2 = 9 \):
\[ 4ac = 9 \]
\[ ac = \frac{9}{4} \]
Not an integer, so no solutions.

\( b = 4 \), \( b^2 = 16 \):
\[ 4ac = 16 \]
\[ ac = 4 \]
Pairs: \( (1, 4) \), \( (2, 2) \), \( (4, 1) \)
Number of triples: 3

\( b = 5 \), \( b^2 = 25 \):
\[ 4ac = 25 \]
\[ ac = \frac{25}{4} \]
Not an integer, so no solutions.

\( b = 6 \), \( b^2 = 36 \):
\[ 4ac = 36 \]
\[ ac = 9 \]
Pairs: \( (3, 3) \)
Number of triples: 1

Total number of triples: \( 1 + 3 + 1 = 5 \). The triples are:

1. \( (1, 2, 1) \): \( x^2 + 2x + 1 = (x + 1)^2 \)
2. \( (1, 4, 4) \): \( x^2 + 4x + 4 = (x + 2)^2 \)
3. \( (2, 4, 2) \): \( 2x^2 + 4x + 2 = 2(x + 1)^2 \)
4. \( (4, 4, 1) \): \( 4x^2 + 4x + 1 = (2x + 1)^2 \)
5. \( (3, 6, 3) \): \( 3x^2 + 6x + 3 = 3(x + 1)^2 \)

Each has \( \Delta = 0 \), confirming one x-intercept. Total possible functions = 216, so:

\[ \text{Probability} = \frac{5}{216} \]

(d) For two distinct x-intercepts, \( \Delta > 0 \):

\[ b^2 – 4ac > 0 \]
OR \[ b^2 > 4ac \]
OR \[ 4ac < b^2 \]

Since \( b \) ranges from 1 to 6, \( b^2 \) is 1, 4, 9, 16, 25, 36, and \( a \) and \( c \) from 1 to 6, so \( ac \) ranges from 1 to 36, and \( 4ac \) from 4 to 144. We need \( 4ac < b^2 \), and \( ac \) must be an integer. Possible \( ac \) values depend on \( b \), but overall, \( 4ac < 36 \) (since \( b^2 \leq 36 \)):

\[ 4ac < 36 \]
\[ ac < 9 \]
\[ ac \leq 8 \]

Possible \( ac \): 1, 2, 3, 4, 5, 6, 8 (since \( a, c \geq 1 \), \( ac = 7 \) is possible with \( (1, 7) \) etc., but we test up to 8 based on pairs).

(e) (i) For  \( ac = 1 \), so \( (a, c) = (1, 1) \). \( b \) from 1 to 6:

\( \Delta = b^2 – 4 \cdot 1 \cdot 1 = b^2 – 4 \)
\( b = 1 \): \( 1 – 4 < 0 \)
\( b = 2 \): \( 4 – 4 = 0 \)
\( b = 3 \): \( 9 – 4 = 5 > 0 \)
\( b = 4 \): \( 16 – 4 = 12 > 0 \)
\( b = 5 \): \( 25 – 4 = 21 > 0 \)
\( b = 6 \): \( 36 – 4 = 32 > 0 \)

\( b = 3, 4, 5, 6 \) work. Number of functions = 4.

(e) (ii) For \( ac = 2 \): \( (1, 2) \), \( (2, 1) \). For each:

\( a = 1, c = 2 \): \( \Delta = b^2 – 4 \cdot 1 \cdot 2 = b^2 – 8 \)
\( b = 1, 2 \): \( < 0 \)
\( b = 3 \): \( 9 – 8 = 1 > 0 \)
\( b = 4 \): \( 16 – 8 = 8 > 0 \)
 \( b = 5, 6 \): \( > 0 \)
4 cases
\( a = 2, c = 1 \): Same \( \Delta \), 4 cases

Total = 8.

(f) the probability of two distinct x-intercepts.

Count all cases for \( ac = 1 \) to 8, \( \Delta > 0 \):

\( ac = 1 \): 4
\( ac = 2 \): 8
\( ac = 3 \): \( (1, 3) \), \( (3, 1) \), \( b = 4, 5, 6 \) (3 each), 6
\( ac = 4 \): \( (1, 4) \), \( (2, 2) \), \( (4, 1) \), \( b = 5, 6 \) (2 each), 6
\( ac = 5 \): \( (1, 5) \), \( (5, 1) \), \( b = 5, 6 \) (2 each), 4
\( ac = 6 \): \( (1, 6) \), \( (2, 3) \), \( (3, 2) \), \( (6, 1) \), \( b = 5, 6 \) (except \( 2, 3 \) needs \( b \geq 5 \)), 8
\( ac = 8 \): \( (2, 4) \), \( (4, 2) \), \( b = 6 \), 2

Total = 38

Therefore, probability \( p = \frac{38}{216} = \frac{19}{108} \).

(g) Probability that \( f(x) = x^2 + 2Zx + 1 \) has two x-intercepts, \( Z \sim N(0, 1) \).

\( \Delta = (2Z)^2 – 4 \cdot 1 \cdot 1 = 4Z^2 – 4 \):

\[ 4Z^2 – 4 > 0 \]
\[ Z^2 > 1 \]
\[ |Z| > 1 \]

\( P(|Z| > 1) = 2 \cdot P(Z > 1) = 2 \cdot (1 – \Phi(1)) \approx 2 \cdot (0.1587 )= 0.3174 \).

(h) Probability that \( X_1 > 0.5 \) and \( X_2 > 0.5 \).

Roots: \( X_1 = -Z – \sqrt{Z^2 – 1} \), \( X_2 = -Z + \sqrt{Z^2 – 1} \). Given two roots, \( Z^2 > 1 \). \( X_2 > X_1 \), so if \( X_1 > 0.5 \), check \( X_2 \):

\[ -Z – \sqrt{Z^2 – 1} > 0.5 \]
\[ -\sqrt{Z^2 – 1} > 0.5 + Z \]

For \( Z < -1 \), both roots positive when \( Z < -1.5 \). Compute \( P(Z < -1.5 | |Z| > 1) = \frac{\Phi(-1.5)}{2 \cdot (1 – \Phi(1))} \approx 0.4338 \).

————Markscheme—————–

Solution: –

(a) \(6^{3}\) OR \(6\times6\times6\)

= 216

(b) EITHER

attempts to find \(\Delta\)

\[\Delta \left [ =\left ( 4^{2}-4(1)(4) \right ) \right ]=0\]

OR

attempts to solve \(x^{2} + 4x + 4 = 0\)

\((x+2)^{2} = 0 \Rightarrow x = -2\)

OR

attempts to express \(x^{2} + 4x + 4 = 0\) as a perfect square

\((x+2)^{2} = 0\) is a perfect square

OR

a graph of \(y = x^{2} + 4x + 4\) with the vertex touching the x-axis at \(x = -2\)

THEN

graph of \(f\) has only one x-intercept

(c) recognizes that $b^{2}-4ac=0$ (or equivalent)

EITHER

attempts to use $\frac{b^{2}}{ac}=4 \left(\frac{b^{2}}{4}=ac\right)$

determines one value of $b$ from $b=2,4$ or 6 only OR one value of $ac$ from $ac=1,4$ or 9 only

OR

attempts to find a possible value of $b$

determines one value of $b$ from $b=2,4$ or 6 only

OR

recognizes that $b^{2}$ must be a multiple of 4 OR $b$ must be a multiple of 2

determines one value of $b$ from $b=2,4$ or 6 only

OR

attempts to find a possible value of $ac$

determines one value of $ac$ from $ac=1,4$ or 9 only

THEN

b=2 and ac=1:

(a,b,c)=(1,2,1) OR 1 possible way  OR $\frac{1}{216}$

b=4 and ac=4:

(a,b,c)=(1,4,4),(4,4,1),(2,4,2) OR 3 possible ways OR $\frac{3}{216}$

b=4 and ac=9:

(a,b,c)=(3,6,3) OR 1 possible way OR $\frac{1}{216}$

Therefore, the required probability  is $\frac{1}{216}+\frac{3}{216}+\frac{1}{216}=\frac{5}{216}$

(d) recognizes that $b^{2}-4ac>0$ (or equivalent  eg. $\frac{b^{2}}{4}>ac$)

maximum value of $b^{2}$ is 36 OR maximum value of $ac$ is 8

Note: The above (A1) is independent of the (M1).

$ac=1,2,3,4,5,6,8$

(e) (i) $ac=1(b^{2}>4)$

$b=3,4,5,6$ OR $1\times4$ (quadratics) OR 6-2 (quadratics)
there are four quadratic functions

(ii) $ac=2(b^{2}>8)$

$b=3,4,5,6$

Note: Award (A1) for referencing their result shown in part (e) (i).

EITHER

$(a,b,c)=(1,3,2),(1,4,2),(1,5,2),(1,6,2),(2,3,1),(2,4,1),(2,5,1),(2,6,1)$

OR

2 \times 4 (quadratics)

THEN

there are eight quadratic functions

(f) METHOD 1

varies $ac (ac \neq 1,2)$ and determines possible values of $b$ such that $\Delta > 0$

correctly determines one of the following five cases 

correctly determines a further two of the following five cases

correctly  determines a further two cases 

case 1: $ac = 3 (b^{2} > 12 \Rightarrow b = 4,5,6)$

$(a,b,c) = (1,4,3), (1,5,3), (1,6,3), (3,4,1), (3,5,1), (3,6,1)$ OR
6 possible ways OR
$\frac{6}{216}$

case 2: $ac = 4 (b^{2} > 16 \Rightarrow b = 5,6)$

$(a,b,c) = (1,5,4), (1,6,4), (2,5,2), (2,6,2), (4,5,1), (4,6,1)$ OR
6 possible ways OR
$\frac{6}{216}$

case 3: $ac = 5 (b^{2} > 20 \Rightarrow b = 5,6) $

$(a,b,c) = (1,5,5), (1,6,5), (5,5,1), (5,6,1)$ OR 4 possible ways OR
$\frac{4}{216}$

case 4: $ac = 6 (b^{2} > 24 \Rightarrow b = 5,6)$

$(a,b,c) = (1,5,6), (2,5,3), (3,5,2), (6,5,1), (1,6,6), (2,6,3), (3,6,2), (6,6,1)$ OR
8 possible ways OR $\frac{8}{216}$

case 5: $ac = 8 (b^{2} > 32 \Rightarrow b = 6)$

$(a,b,c) = (2,6,4), (4,6,2)$ OR 2 possible ways OR
$\frac{2}{216}$

adds their probabilities

$p = \frac{4}{216} + \frac{8}{216} + \frac{6}{216} + \frac{6}{216} + \frac{4}{216} + \frac{8}{216} + \frac{2}{216} $

$= (0.0185… + 0.0370… + 0.0277… + 0.0277… + 0.0185… + 0.0370… + 0.0092…)$

$= \frac{38}{216} = \frac{19}{108} = 0.176 $

METHOD 2

varies $b^{2}(+1,4)$ OR $b(\pm1,2)$ and determines possible values of $ac$ such that $\Delta>0$

correctly determines one of the following four cases

correctly determines another case from the following four cases

correctly determines the remaining two cases

case 1: $b^{2}=9(b=3)(ac=1,2)$

$(a,b,c)=(1,3,1),(1,3,2),(2,3,1)$ OR 3 possible ways OR $\frac{3}{216}$

case 2: $b^{2}=16(b=4)(ac=1,2,3)$

$(a,b,c)=(1,4,1),(1,4,2),(2,4,1),(1,4,3),(3,4,1)$ OR 5 possible ways OR $\frac{5}{216}$

case 3: $b^{2}=25(b=5)(ac=1,2,3,4,5,6)$

$(a,b,c)=(1,5,1),(1,5,2),(2,5,1),(1,5,3),(3,5,1),(1,5,4),(2,5,2),$
$(4,5,1),(1,5,5),(5,5,1),(1,5,6),(2,5,3),(3,5,2),(6,5,1)$

OR 14 possible ways OR $\frac{14}{216}$

case 4: $b^{2} = 36 (b = 6) (ac = 1,2,3,4,5,6,8)$

$(a,b,c) = (1,6,1), (1,6,2), (2,6,1), (1,6,3), (3,6,1), (1,6,4), (2,6,2), (4,6,1)$
$(1,6,5), (5,6,1), (1,6,6), (2,6,3), (3,6,2), (6,6,1), (2,6,4), (4,6,2)$

OR 16 possible ways OR $\frac{16}{216}$

adds their probabilities

$p = \frac{3}{216} + \frac{5}{216} + \frac{14}{216} + \frac{16}{216}$

$= (0.013889… + 0.023148… + 0.064815… + 0.074074…)$

$= \frac{38}{216} = \frac{19}{108} = 0.176$

(g) recognizes that $4Z^2-4>0 (Z^2>1)$

probability of two x-intercepts is

EITHER

$P(|Z|>1)$

OR

$P(Z<-1)$ or $P(Z>1)$ (can be shown on a labelled diagram)
$= 0.158655… + 0.158655…$

OR

$1-P(-1 \le Z \le 1)$ (can be shown on a labelled diagram)
$= 1 – 0.682689…$

THEN

$= 0.317310…$

$= 0.317$

(h) attempts to solve $X_{1} > 0.5$ for $Z$

$-1.25 < Z \leq -1$

attempts to calculate their $P(X_{1}, X_{2} \text{ both } > 0.5)$

$P(-1.25 < Z \leq -1) = 0.053005…$

attempts to calculate their $P(X_{1}, X_{2} \text{ both } > 0.5 \text{ x-intercepts}) $

$\frac{0.053005…}{0.317310…}$

$= 0.167$