Questions - SPS - A-Level Maths

SPS SPS SM Mechanics 2026 January Q4

4. A toy car is travelling in a straight horizontal line. One model of the motion for $0 \leqslant t \leqslant 8$, where $t$ is the time in seconds, is shown in the velocity-time graph Fig. 6. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{fb36606d-0ee3-4050-af31-1642e5f67a03-10_481_1226_374_424} \captionsetup{labelformat=empty} \caption{Fig. 6}

\end{figure}

Calculate the distance travelled by the car from $t = 0$ to $t = 8$.
How much less time would the car have taken to travel this distance if it had maintained its initial speed throughout?
What is the acceleration of the car when $t = 1$ ? From $t = 8$ to $t = 14$, the car travels 58.5 m with a new constant acceleration, $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
Find $a$. A second model for the velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of the toy car is $$v = 12 - 10 t + \frac { 9 } { 4 } t ^ { 2 } - \frac { 1 } { 8 } t ^ { 3 } , \text { for } 0 \leqslant t \leqslant 8$$ This model agrees with the values for $v$ given in Fig. 6 for $t = 0,2,4$ and 6. [Note that you are not required to verify this.] Use this second model to answer the following questions.
Calculate the acceleration of the car when $t = 1$.
Initially the car is at A. Find an expression in terms of $t$ for the displacement of the car from A after the first $t$ seconds of its motion. Hence find the displacement of the car from A when $t = 8$. \includegraphics[max width=\textwidth, alt={}, center]{fb36606d-0ee3-4050-af31-1642e5f67a03-13_2688_1886_118_118}

SPS SPS SM Mechanics 2026 January Q5

5. A toy sledge of mass 4 kg is being pulled in a straight line by a light string. The resistance to its motion is 6 N . \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{fb36606d-0ee3-4050-af31-1642e5f67a03-14_104_716_312_733} \captionsetup{labelformat=empty} \caption{Fig. 6.1}

\end{figure} At one time, the string is horizontal and the sledge is on horizontal ground, as shown in Fig. 6.1. The acceleration of the sledge is $3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ forwards.

Calculate the tension in the string. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fb36606d-0ee3-4050-af31-1642e5f67a03-14_190_718_813_733} \captionsetup{labelformat=empty} \caption{Fig. 6.2}
\end{figure} At another time, the sledge is again on horizontal ground but the string is now at $40 ^ { \circ }$ to the horizontal, as shown in Fig. 6.2. The tension in the string is 25 N .
Calculate the acceleration of the sledge. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fb36606d-0ee3-4050-af31-1642e5f67a03-16_364_465_283_479} \captionsetup{labelformat=empty} \caption{Fig. 6.3}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fb36606d-0ee3-4050-af31-1642e5f67a03-16_414_463_233_1226} \captionsetup{labelformat=empty} \caption{Fig. 6.4}
\end{figure} In another situation the sledge is on a slope inclined at $35 ^ { \circ }$ to the horizontal, as shown in Fig. 6.3. It is held in equilibrium by the light string parallel to the slope. The resistance to motion of 6 N acts up the slope.
Calculate the tension in the string. The sledge is now held in equilibrium with the light string inclined at $\theta ^ { \circ }$ to the slope, as shown in Fig. 6.4. The tension in the string is 25 N and the resistance to motion remains 6 N acting up the slope.
(A) Show all the forces acting on the sledge.
(B) Calculate the angle $\theta$.
(C) Calculate the normal reaction of the slope on the sledge. \includegraphics[max width=\textwidth, alt={}, center]{fb36606d-0ee3-4050-af31-1642e5f67a03-17_2688_1886_118_118}

SPS SPS SM Mechanics 2026 January Q6

6. A box of weight 147 N is held by light strings AB and BC . As shown in Fig. 7.1, AB is inclined at $\alpha$ to the horizontal and is fixed at $\mathrm { A } ; \mathrm { BC }$ is held at C . The box is in equilibrium with BC horizontal and $\alpha$ such that $\sin \alpha = 0.6$ and $\cos \alpha = 0.8$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{fb36606d-0ee3-4050-af31-1642e5f67a03-18_396_570_365_760} \captionsetup{labelformat=empty} \caption{Fig. 7.1}

\end{figure}

Calculate the tension in string AB .
Show that the tension in string BC is 196 N . As shown in Fig. 7.2, a box of weight 90 N is now attached at C and another light string CD is held at D so that the system is in equilibrium with BC still horizontal. CD is inclined at $\beta$ to the horizontal. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fb36606d-0ee3-4050-af31-1642e5f67a03-20_394_714_340_671} \captionsetup{labelformat=empty} \caption{Fig. 7.2}
\end{figure}
Explain why the tension in the string BC is still 196 N .
Draw a diagram showing the forces acting on the box at C . Find the angle $\beta$ and show that the tension in CD is 216 N , correct to three significant figures. The string section CD is now taken over a smooth pulley and attached to a block of mass $M \mathrm {~kg}$ on a rough slope inclined at $40 ^ { \circ }$ to the horizontal. As shown in Fig. 7.3, the part of the string attached to the box is still at $\beta$ to the horizontal and the part attached to the block is parallel to the slope. The system is in equilibrium with a frictional force of 20 N acting on the block up the slope. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fb36606d-0ee3-4050-af31-1642e5f67a03-22_451_1070_422_495} \captionsetup{labelformat=empty} \caption{Fig. 7.3}
\end{figure}
Calculate the value of $M$. \section*{End of Examination} [BLANK PAGE]
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SPS SPS FM Pure 2020 February Q1

1 In this question you must show detailed reasoning.

Determine the value of $$\int _ { 3 } ^ { 5 } \frac { 1 } { \sqrt { x ^ { 2 } - 9 } } d x$$ giving your answer as a single logarithm.
Determine the exact value of $$\int _ { 0 } ^ { \ln 3 } \sinh x \cosh x d x$$

SPS SPS FM Pure 2020 February Q2

2

Given that $$f ( x ) = \tan ^ { - 1 } ( x + 1 )$$ find $f ( 0 )$ and $f ^ { \prime } ( 0 )$, and show that $f ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 2 }$.
Hence find the first three terms in the Maclaurin series for $f ( x )$

SPS SPS FM Pure 2020 February Q3

3

Find the inverse of the matrix $\left( \begin{array} { l l l } 2 & 0 & 1
0 & 1 & a
1 & 3 & 0 \end{array} \right)$ in terms of $a$.
State the value of $a$ for which the matrix is singular.

SPS SPS FM Pure 2020 February Q4

4 The equation of a curve in polar coordinates is $r = \frac { 1 } { 2 } - \cos \theta$.

Sketch the polar graph of the curve.
Find the exact area enclosed by the curve between $\theta = \frac { 1 } { 3 } \pi$ and $\theta = \frac { 5 } { 3 } \pi$.

SPS SPS FM Pure 2020 February Q5

5 The roots of the equation $\quad 3 x ^ { 3 } + 2 x ^ { 2 } - 7 x - 6 = 0$ are $\alpha , \beta$ and $\gamma$.

Find the value of $\alpha \beta \gamma$.
Hence, by making use of a suitable substitution, or otherwise, find a cubic equation whose roots are $\alpha \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) , \beta \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) , \gamma \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right)$.

SPS SPS FM Pure 2020 February Q6

6 Throughout this question, the complex number $z$ satisfies $\left| z - z _ { 0 } \right| \leq \sqrt { 2 }$, where $z _ { 0 } = 3 - \mathrm { i }$.

Draw an Argand diagram to illustrate the locus of $z$.
In this question you must show detailed reasoning. Show that the greatest possible argument of $z$ can be written as $\tan ^ { - 1 } \left( \frac { 1 } { n } \right)$, where $n$ is a positive integer to be determined and $\arg z \in ( - \pi , \pi ]$.

SPS SPS FM Pure 2020 February Q7

6 marks

7

Find the shortest distance from the point (9, 5, 2) to the plane $3 x + 2 y - 4 z = - 3$.
The vector equation of a second plane is given by $$\boldsymbol { r } = \lambda \left( \begin{array} { l } 3
0
1 \end{array} \right) + \mu \left( \begin{array} { c } 1
- 1
- 2 \end{array} \right)$$ By finding the Cartesian equation of the plane, calculate the obtuse angle between this plane and the plane $3 x + 2 y - 4 z = - 3$.
[0pt] [6]

SPS SPS FM Pure 2020 February Q8

8

Express $\frac { 8 x ^ { 2 } - x + 2 } { x \left( 4 x ^ { 2 } + 1 \right) }$ in partial fractions.
Hence find the general solution to the differential equation $$\frac { d y } { d x } + \frac { 2 y } { x } = \frac { 8 x ^ { 2 } - x + 2 } { x ^ { 3 } \left( 4 x ^ { 2 } + 1 \right) }$$

SPS SPS FM Pure 2020 February Q9

9

Show that $\cos 5 \theta = 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta$.
Use the result of part (a) to prove that $\cos \frac { \pi } { 10 } = \sqrt { \frac { 5 + \sqrt { 5 } } { 8 } }$.

SPS SPS FM Pure 2020 February Q10

77 marks

10 The matrix $\left( \begin{array} { l l } 1 & 1
1 & 0 \end{array} \right)$ is denoted by $\mathbf { M }$.

Evaluate $\mathbf { M } ^ { 3 }$. The Fibonacci series $F _ { 0 } , F _ { 1 } , F _ { 2 } , F _ { 3 } , \ldots$ is defined by
$F _ { n + 1 } = F _ { n } + F _ { n - 1 }$ for $n \geq 1 , F _ { 0 } = 0 , F _ { 1 } = 1$.
Prove by mathematical induction that $\boldsymbol { M } ^ { n } = \left( \begin{array} { c c } F _ { n + 1 } & F _ { n }
F _ { n } & F _ { n - 1 } \end{array} \right)$ for $n \geq 1$.

Use the result of part (b) to find an expression for $F _ { n + 1 } F _ { n - 1 } - F _ { n } { } ^ { 2 }$ in terms of $n$, for $n \geq 1$.
\section*{ADDITIONAL ANSWER SPACE} If additional space is required, you should use the following lined page(s). The question number(s) must be clearly shown.

1(a)

$\left[ \ln \left\{ x + \sqrt { x ^ { 2 } - 9 } \right\} \right] _ { 3 } ^ { 5 }$ $= \ln ( 5 + 4 ) - \ln ( 3 ) = \ln ( 9 / 3 )$

$= \ln 3$

B1

M1

A1

[3]

Correct indefinite integral

Substitute and simplify to single logarithm ln 3 only

1(b)

\(\begin{gathered} { \left[ \frac { 1 } { 2 } \sinh ^ { 2 } x \right] _ { 0 } ^ { \ln 3 } \text { or } \left[ \frac { 1 } { 2 } \cosh ^ { 2 } x \right] _ { 0 } ^ { \ln 3 } }

\text { or } \left[ \frac { 1 } { 8 } \left( e ^ { 2 x } + e ^ { - 2 x } \right) \right] _ { 0 } ^ { \ln 3 }

= 1 / 2 \left[ 1 / 2 \left( \mathrm { e } ^ { \ln 3 } - \mathrm { e } ^ { - \ln 3 } \right) \right] ^ { 2 } - 0 \text { or equivalent }

= \frac { 1 } { 2 } \left[ \frac { 1 } { 2 } \left( 3 - \frac { 1 } { 3 } \right) \right] ^ { 2 } = \frac { 8 } { 9 } \end{gathered}\)

M1

A1

M1

A1

[4]

Correct method (e.g. factors can be wrong)

Correct indefinite integral

E.g. 25/9 - 1

Final answer 8/9, completely correct

2(a)

\(\begin{gathered} f ( 0 ) = \pi / 4

f ^ { \prime } ( x ) = \frac { 1 } { 1 + ( x + 1 ) ^ { 2 } }

f ^ { \prime } ( 0 ) = \frac { 1 } { 2 }

f ^ { \prime \prime } ( x ) = - \frac { 2 ( x + 1 ) } { \left( 1 + ( x + 1 ) ^ { 2 } \right) ^ { 2 } }

f ^ { \prime \prime } ( 0 ) = - \frac { 2 } { 4 } = - \frac { 1 } { 2 } \end{gathered}\)

B1

M1

A1ft

M1

A1

[5]

Correct f(0)

First derivative

Substitute $0 , \mathrm { ft }$ on their $f ^ { \prime } ( x )$

Second derivative (need not be expanded)

Substitute 0

2(b)

\(\begin{gathered} \frac { \pi } { 4 } + \frac { 1 } { 2 } x + \frac { 1 } { 2 } \left( - \frac { 1 } { 2 } \right) x ^ { 2 }

\frac { \pi } { 4 } + \frac { 1 } { 2 } x - \frac { 1 } { 4 } x ^ { 2 } \end{gathered}\)

M1

A1

Substitute into Maclaurin formula

CAO

3(a)

\(\frac { 1 } { 6 a + 1 } \left( \begin{array} { c c c } 3 a

- 3

1

- a

1

2 a

1

6

- 2 \end{array} \right)\)

M1

A1

M1

A1

[5]

Obtain cofactors

All cofactors correct and in right place

Divide by determinant

Determinant correct ( $- 6 a - 1$ )

Completely correct (expect to see all signs reversed)

3(b)

$- \frac { 1 } { 6 }$

A1ft

ft on their determinant

4(a)

\includegraphics[max width=\textwidth, alt={}]{6280d53b-3c1c-4dc9-a96b-2c58f2a7bf51-22_199_391_282_420}

B2

[2]

Give B1 if: just one obvious flaw in sketch, or if part of curve for negative $r$ given, e.g.

4(b)

$\frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 1 } { 2 } - \cos \theta \right) ^ { 2 } d \theta$ \(\begin{gathered} = \frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 1 } { 4 } - \cos \theta + \cos ^ { 2 } \theta \right) d \theta

= \frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 3 } { 4 } - \cos \theta + \frac { 1 } { 2 } \cos 2 \theta \right) d \theta

= \frac { 1 } { 2 } \left[ \frac { 3 } { 4 } \theta - \sin \theta + \frac { 1 } { 2 } \sin 2 \theta \right] _ { \pi / 3 } ^ { 5 \pi / 3 }

= \frac { 1 } { 2 } \pi + \frac { 3 } { 8 } \sqrt { 3 } \end{gathered}\)

M1

A1

[5]

Correct expression for area

Multiply out and use $\cos 2 \theta$

Correct integrand (ignore leading 1/2)

Correct indefinite integral (ignore leading 1/2)

Final answer, this or clear exact equivalent only

5(a)

2

A1

5(b)

Roots are $\alpha ( 1 + 1 / 2 ) , \beta ( 1 + 1 / 2 ) , \chi \left( 1 + \frac { 1 } { 2 } \right)$

Hence put $y = 3 x / 2$ $x = 2 y / 3$

$3 ( 2 y / 3 ) ^ { 3 } + 2 ( 2 y / 3 ) ^ { 2 } - 7 ( 2 y / 3 ) - 6 = 0$

$4 y ^ { 3 } + 4 y ^ { 2 } - 21 y - 27 = 0$

M1

A1ft

M1

A1

[5]

Use value of $\alpha \beta \gamma$ in $\alpha ( 1 + 1 / \alpha \beta \gamma )$ etc Correct new variable, ft on their $\alpha \beta \gamma$ Inverse function, ft Substitute inverse function

Complete equation including 0, any letter, any rational multiple

or

\(\begin{aligned}

\Sigma \alpha = - 2 / 3 , \Sigma \alpha \beta = - 7 / 3 , \alpha \beta \gamma = 2

\qquad \begin{array} { l } ( \alpha + \beta + \gamma ) \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ( = - 1 )

( \alpha \beta + \beta \gamma + \gamma \alpha ) \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ^ { 2 } \left( = - \frac { 21 } { 4 } \right) \end{array}

\qquad \alpha \beta \gamma \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ^ { 3 } \left( = \frac { 27 } { 4 } \right)

\text { Hence equation is } x ^ { 3 } + x ^ { 2 } - \frac { 21 } { 4 } x - \frac { 27 } { 4 } = 0 \end{aligned}\)

B1

M1

A1

All three correct

Correct formula for $\Sigma \alpha ^ { \prime }$ and substitute

Correct formula for $\Sigma \alpha ^ { \prime } \beta ^ { \prime }$ and substitute

Correct formula for $\alpha ^ { \prime } \beta ^ { \prime } \gamma ^ { \prime }$ and substitute

Correct final equation including 0, any letter, any rational multiple

\begin{table}[h]

\captionsetup{labelformat=empty} \caption{Further Mathematics Core (Pure) Mark Scheme (total 84)}

6(a)

Circle, centre $( 3 , - 1 )$

Radius $\sqrt { } 2$ indicated

Interior of circle indicated

M1

B1

A1

Allow +/- errors

Needs not to include origin

Any clear indication of interior

6(b)

Point of contact $( P )$ between circle and tangent through $O$ clearly identified

Form appropriate right-angled $\Delta$

$O C = \sqrt { 10 } , P C = \sqrt { 2 } \Rightarrow O P = 2 \sqrt { 2 }$ and $\angle P O C = \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$

$\angle x O C = \tan ^ { - 1 } \left( \frac { 1 } { 3 } \right)$

$\angle x O P = \tan ^ { - 1 } \left( \frac { \frac { 1 } { 2 } - \frac { 1 } { 3 } } { 1 + \frac { 1 } { 2 } \cdot \frac { 1 } { 3 } } \right) = \tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)$ or $n = 7$

B1

M1

A1

B1

M1

A1

[7]

Can be implied by correct working

Use exact method for a difference of angles Either $\tan ^ { - 1 } ( 1 / 7 )$ or $n = 7$.

If exactly 1/7 from calculator, give M0A0

or

Point of contact ( $P$ ) between circle and tangent through $O$ clearly identified

$y = m x$ meets $( x - 3 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 2$

$( x - 3 ) ^ { 2 } + ( m x + 1 ) ^ { 2 } = 2$ has double roots $\left( m ^ { 2 } + 1 \right) x ^ { 2 } + 2 ( m - 3 ) x + 8 = 0$ has double roots

$4 ( m - 3 ) ^ { 2 } = 4.8 \left( m ^ { 2 } + 1 \right)$

$7 m ^ { 2 } + 6 m - 1 = 0$, so $m = \frac { 1 } { 7 }$ or - 1

But $m$ is not - 1 as this gives minimum, not maximum, argument

B1

M1

A1

M1

A1

Can be implied by correct working, provided final answer does not include - 1

Line and circle equations used

Discriminant set to zero

Obtain $\tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)$ or $n = 7$

Reject - 1 , with reason. Not enough to say " $m = \frac { 1 } { 7 }$ or - 1 so $n = 7$ "

or

Point of contact $( P )$ between circle and tangent through $O$ clearly identified
Appropriate right-angled $\Delta$ with centre $C$
$O C = \sqrt { 10 } , P C = \sqrt { 2 } \Rightarrow O P = 2 \sqrt { 2 }$
Circles, centre $O , r = 2 \sqrt { 2 }$, \	$C , r = \sqrt { 2 }$
$x ^ { 2 } + ( 3 x - 8 ) ^ { 2 } = 8,5 x ^ { 2 } - 24 x + 28 = 0$
$\Rightarrow P = \left( 2 \frac { 4 } { 5 } , \frac { 2 } { 5 } \right)$
$\theta = \tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)$ or $n = 7$

B1

M1

B1

M1

A1

Can be implied by correct working

$x ^ { 2 } + y ^ { 2 } = 8$ and $( x - 3 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 2$

Eliminate $y$, solve quadratic in $x$

Correct $P$, both cords. allow decimals

Either $\tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)$ or $n = 7$

\end{table}

7(a)

Distance is $\frac { | \mathbf { b } \mathbf { . n } - \mathbf { p } | } { | \mathbf { n } | }$ $\mathbf { b } = ( 9,5,2 ) , \mathbf { n } = ( 3,2 , - 4 )$

Scalar product is 29

Distance is $32 / \sqrt { } 29$

M2

A1

[4]

Use correct formula with correct $\mathbf { b }$ and $\mathbf { n }$

Correct scalar product

Answer (accept 5.94 (3s.f.))

7(b)

\(\begin{aligned}

( 3 \mathbf { i } + \mathbf { k } ) \times ( \mathbf { i } - \mathbf { j } - 2 \mathbf { k } )

= \mathbf { i } + 7 \mathbf { j } - 3 \mathbf { k } \end{aligned}\) \(\begin{gathered} x + 7 y - 3 z = 0

\left( \begin{array} { c } 1

7

- 3 \end{array} \right) \cdot \left( \begin{array} { c } 3

2

- 4 \end{array} \right) \end{gathered}\) \(\begin{gathered} \cos \theta = \frac { 29 } { \sqrt { 59 } \sqrt { 29 } }

134.5 ^ { \circ } \text { or } 2.35 \text { radians } \end{gathered}\)

M1

A1

M1

A1ft

A1

[6]

Find perpendicular vector

Any multiple of this

Cartesian equation of second plane

Scalar product of normal vectors of 2 planes

Correct scalar product and moduli (ft their vectors)

Correct obtuse angle (awrt $135 ^ { \circ }$ or 2.35 rad )

8(a)

\(\begin{gathered} \frac { A } { x } + \frac { B x + C } { 4 x ^ { 2 } + 1 }

A \left( 4 x ^ { 2 } + 1 \right) + ( B x + C ) x \equiv 8 x ^ { 2 } - x + 2

A = 2 , B = 0 , C = - 1

\frac { 2 } { x } - \frac { 1 } { 4 x ^ { 2 } + 1 } \end{gathered}\)

M1

A1

[5]

Attempt at partial fractions

Set up identity

Compare coefficients or substitute x -values

All correct values

Final answer

8(b)

IF $\exp \left( \int 2 / x \mathrm {~d} x \right) = x ^ { 2 }$ \(\begin{gathered} y x ^ { 2 } = \int \frac { 2 } { x } - \frac { 1 } { 4 x ^ { 2 } + 1 } d x

y x ^ { 2 } = 2 \ln x - \frac { 1 } { 2 } \tan ^ { - 1 } ( 2 x ) + c

y = \frac { 4 \ln x - \tan ^ { - 1 } ( 2 x ) + c ^ { \prime } } { 2 x ^ { 2 } } \end{gathered}\)

M1

A1

M1

A1

[6]

Method for IF

$x ^ { 2 }$

Multiply through by IF and spot link to (a)

Use $\tan ^ { - 1 }$

$2 \ln x$ and $+ c$

Final answer, completely correct, any equivalent form, any correct way of writing constant (e.g. c or 2c)

\section*{Further Mathematics Core (Pure) Mark Scheme (total 84)}

9(a)

\(\begin{aligned}

( c + \mathrm { is } ) ^ { 5 } = c ^ { 5 } + 5 \mathrm { is } ^ { 4 } - 10 s ^ { 2 } c ^ { 2 } - 10 \mathrm { is } ^ { 3 } c ^ { 2 } + 5 s ^ { 4 } c + \mathrm { is } ^ { 5 }

\cos 5 \theta = \cos ^ { 5 } \theta - 10 \sin ^ { 2 } \theta \cos ^ { 3 } \theta + 5 \sin ^ { 4 } \theta \cos \theta

= c ^ { 5 } - 10 \left( 1 - c ^ { 2 } \right) c ^ { 3 } + 5 \left( 1 - c ^ { 2 } \right) ^ { 2 } c

= 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta \quad \text { AG } \end{aligned}\)

M1

A1

M1

A1

[4]

Use de Moivre (can ignore im parts if clear)

Correct equating of real parts

Use $\sin ^ { 2 } \theta = 1 - \cos ^ { 2 } \theta$

Correctly obtain given answer

9(b)

Using $\cos 5 \theta = 0$ gives $16 c ^ { 5 } - 20 c ^ { 4 } + 5 c = 0$ $c \neq 0 \text { so } 16 c ^ { 4 } - 20 c ^ { 2 } + 5 = 0$

$c ^ { 2 } = ( 20 \pm \sqrt { } 80 ) / 32$ $c = \pm \sqrt { \frac { 5 \pm \sqrt { 5 } } { 8 } }$

$\cos \pi / 10 > 0$ so first $\pm$ is +

$\cos \pi / 10 > \cos ( 3 \pi / 10 )$ so second $\pm$ is +

M1

B1

M1

A1

[6]

Use $\theta = \pi / 10$ and $\cos ( \pi / 2 ) = 0$

Justify cancellation of $c$

Solve quadratic in $c ^ { 2 }$ by exact method

Correct expression for $c$, any combination of $\pm$ and + signs

Justify outer +

Justify inner +

10(a)

\(\left( \begin{array} { l l } 3

2

1 \end{array} \right)\)

B2

[2]

If B0, give B1 if \(\mathbf { M } ^ { 2 } = \left( \begin{array} { l l } 2

1

1 \end{array} \right)\) seen

10(b)

Base case: \(n = 1 , \mathbf { M } = \left( \begin{array} { l l } 1	1
1	0 \end{array} \right) = \left( \begin{array} { l l } F _ { 2 }	F _ { 1 }
F _ { 1 }	F _ { 0 } \end{array} \right)\)
Inductive step: assume \(\mathbf { M } ^ { k } = \left( \begin{array} { c c } F _ { k + 1 }	F _ { k }
F _ { k }	F _ { k - 1 } \end{array} \right)\)
\(\Rightarrow \mathbf { M } ^ { k + 1 } = \left( \begin{array} { c c } F _ { k + 1 }	F _ { k }
F _ { k }	F _ { k - 1 } \end{array} \right) \left( \begin{array} { l l } 1	1
1	0 \end{array} \right)\) \(\begin{aligned}	= \left( \begin{array} { c c } F _ { k + 1 } + F _ { k }	F _ { k } + F _ { k - 1 }
F _ { k } + F _ { k - 1 }	F _ { k } \end{array} \right)
	\left( \begin{array} { c c } F _ { k + 2 }	F _ { k + 1 }
F _ { k + 1 }	F _ { k } \end{array} \right) \text { as required } \end{aligned}\)
Base case and inductive step both true, so statement true for all $n \in \mathbb { N }$ by mathematical induction

B1

M1

A1

[5]

Fully correct

Correct "assume" statement, allow $n$

Consider $\mathbf { M } \times$ hypothesised $\mathbf { M } ^ { k }$

Correctly show this result

Award only if all 4 other marks gained

10(c)

Det $\left( \mathbf { M } ^ { n } \right) = F _ { n + 1 } F _ { n - 1 } - F _ { n } { } ^ { 2 }$

$\operatorname { Det } \left( \mathbf { M } ^ { n } \right) = ( \operatorname { Det } \mathbf { M } ) ^ { n }$ $= ( - 1 ) ^ { n }$

M1

A1

[3]

Use determinant of $\mathbf { M } ^ { n }$

Relate $\operatorname { det } \mathbf { M } ^ { \boldsymbol { n } }$ to $\operatorname { det } \mathbf { M }$

Correct answer

SPS SPS SM Mechanics 2021 January Q1

1. Which of the options below best describes the correlation shown in the diagram below?
\includegraphics[max width=\textwidth, alt={}, center]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-05_719_1198_408_340} Tick ( $\checkmark$ ) one box.
moderate positive □
strong positive
\includegraphics[max width=\textwidth, alt={}, center]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-05_103_109_1407_881}
moderate negative
\includegraphics[max width=\textwidth, alt={}, center]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-05_103_109_1535_881}
strong negative □

SPS SPS SM Mechanics 2021 January Q2

1 marks

2. Lenny is one of a team of people interviewing shoppers in a town centre.
He is asked to survey 50 women between the ages of 18 and 29
Identify the name of this type of sampling.
Circle your answer.
[0pt] [1 mark]
simple random
stratified
quota
systematic

SPS SPS SM Mechanics 2021 January Q3

3. The Venn diagram shows the probabilities associated with four events, $A , B , C$ and $D$
\includegraphics[max width=\textwidth, alt={}, center]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-06_524_897_351_625}

Write down any pair of mutually exclusive events from $A , B , C$ and $D$ Given that $\mathrm { P } ( B ) = 0.4$
find the value of $p$ Given also that $A$ and $B$ are independent
find the value of $q$ Given further that $\mathrm { P } \left( B ^ { \prime } \mid C \right) = 0.64$
find
1. the value of $r$
2. the value of $s$

SPS SPS SM Mechanics 2021 January Q4

4. Each member of a group of 27 people was timed when completing a puzzle.
The time taken, $x$ minutes, for each member of the group was recorded.
These times are summarised in the following box and whisker plot.
\includegraphics[max width=\textwidth, alt={}, center]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-08_357_1454_523_335}

Find the range of the times.
Find the interquartile range of the times. For these 27 people $\sum x = 607.5$ and $\sum x ^ { 2 } = 17623.25$
calculate the mean time taken to complete the puzzle,
calculate the standard deviation of the times taken to complete the puzzle. Taruni defines an outlier as a value more than 3 standard deviations above the mean.
State how many outliers Taruni would say there are in these data, giving a reason for your answer. Adam and Beth also completed the puzzle in $a$ minutes and $b$ minutes respectively, where $a > b$.
When their times are included with the data of the other 27 people
- the median time increases
- the mean time does not change
- Suggest a possible value for $a$ and a possible value for $b$, explaining how your values satisfy the above conditions.
- Without carrying out any further calculations, explain why the standard deviation of all 29 times will be lower than your answer to part (d).

SPS SPS SM Mechanics 2021 January Q5

5. Patrick is practising his skateboarding skills. On each day, he has 30 attempts at performing a difficult trick. Every time he attempts the trick, there is a probability of 0.2 that he will fall off his skateboard. Assume that the number of times he falls off on any given day may be modelled by a binomial distribution.

1. Find the mean number of times he falls off in a day.
(ii) Find the variance of the number of times he falls off in a day.
1. Find the probability that, on a particular day, he falls off exactly 10 times.
(ii) Find the probability that, on a particular day, he falls off 5 or more times.
Patrick has 30 attempts to perform the trick on each of 5 consecutive days.
1. Calculate the probability that he will fall off his skateboard at least 5 times on each of the 5 days.
(ii) Explain why it may be unrealistic to use the same value of 0.2 for the probability of falling off for all 5 days.

SPS SPS SM Mechanics 2021 January Q6

6. The discrete random variable $D$ has the following probability distribution

$d$	10	20	30	40	50
$\mathrm { P } ( D = d )$	$\frac { k } { 10 }$	$\frac { k } { 20 }$	$\frac { k } { 30 }$	$\frac { k } { 40 }$	$\frac { k } { 50 }$

where $k$ is a constant.

Show that the value of $k$ is $\frac { 600 } { 137 }$ The random variables $D _ { 1 }$ and $D _ { 2 }$ are independent and each have the same distribution as $D$.
Find P $\left( D _ { 1 } + D _ { 2 } = 80 \right)$ Give your answer to 3 significant figures. A single observation of $D$ is made.
The value obtained, $d$, is the common difference of an arithmetic sequence.
The first 4 terms of this arithmetic sequence are the angles, measured in degrees, of quadrilateral $Q$
Find the exact probability that the smallest angle of $Q$ is more than $50 ^ { \circ }$

SPS SPS SM Mechanics 2021 January Q7

1 marks

7. A health centre claims that the time a doctor spends with a patient can be modelled by a normal distribution with a mean of 10 minutes and a standard deviation of 4 minutes.

Using this model, find the probability that the time spent with a randomly selected patient is more than 15 minutes. Some patients complain that the mean time the doctor spends with a patient is more than 10 minutes. The receptionist takes a random sample of 20 patients and finds that the mean time the doctor spends with a patient is 11.5 minutes.
Stating your hypotheses clearly and using a $5 \%$ significance level, test whether or not there is evidence to support the patients' complaint. The health centre also claims that the time a dentist spends with a patient during a routine appointment, $T$ minutes, can be modelled by the normal distribution where $T \sim \mathrm {~N} \left( 5,3.5 ^ { 2 } \right)$
Using this model,
1. find the probability that a routine appointment with the dentist takes less than 2 minutes
2. find $\mathrm { P } ( T < 2 \mid T > 0 )$
3. hence explain why this normal distribution may not be a good model for $T$. The dentist believes that she cannot complete a routine appointment in less than 2 minutes.
  She suggests that the health centre should use a refined model only including values of $T > 2$

Find the median time for a routine appointment using this new model, giving your answer correct to one decimal place. Name: □ \section*{U8th A LEVEL Single Mathematics Assessment Mechanics $7 ^ { \text {th } }$ January 2021 } Instructions

Answer all the questions
Write your answer to each question in the space provided under each question. The question number(s) must be clearly shown.
Use black or blue ink. Pencil may be used for graphs and diagrams only.
You should clearly write your name at the top of this page and on any additional sheets that you use.
You are permitted to use a scientific or graphical calculator in this paper.
Final answers should be given to a degree of accuracy appropriate to the context.

Information

The total mark for this paper is 55 marks.
The marks for each question are shown in brackets ( ).
You are reminded of the need for clear presentation in your answers.
You should allow approximately 60 minutes for this section of the test

\section*{Formulae} \section*{A Level Mathematics A (H240)} \section*{Arithmetic series} $S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}$ \section*{Geometric series} $S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$
$S _ { \infty } = \frac { a } { 1 - r }$ for $| r | < 1$ \section*{Binomial series} $( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )$, where ${ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }$
$( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )$ \section*{Differentiation}

$\mathrm { f } ( x )$	$\mathrm { f } ^ { \prime } ( x )$
$\tan k x$	$k \sec ^ { 2 } k x$
$\sec x$	$\sec x \tan x$
$\cot x$	$- \operatorname { cosec } ^ { 2 } x$
$\operatorname { cosec } x$	$- \operatorname { cosec } x \cot x$

Quotient rule $y = \frac { u } { v } , \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { v \frac { \mathrm {~d} u } { \mathrm {~d} x } - u \frac { \mathrm {~d} v } { \mathrm {~d} x } } { v ^ { 2 } }$ \section*{Differentiation from first principles} $\mathrm { f } ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }$ \section*{Integration} $\int \frac { \mathrm { f } ^ { \prime } ( x ) } { \mathrm { f } ( x ) } \mathrm { d } x = \ln | \mathrm { f } ( x ) | + c$
$\int \mathrm { f } ^ { \prime } ( x ) ( \mathrm { f } ( x ) ) ^ { n } \mathrm {~d} x = \frac { 1 } { n + 1 } ( \mathrm { f } ( x ) ) ^ { n + 1 } + c$
Integration by parts $\int u \frac { \mathrm {~d} v } { \mathrm {~d} x } \mathrm {~d} x = u v - \int v \frac { \mathrm {~d} u } { \mathrm {~d} x } \mathrm {~d} x$ Small angle approximations
$\sin \theta \approx \theta , \cos \theta \approx 1 - \frac { 1 } { 2 } \theta ^ { 2 } , \tan \theta \approx \theta$ where $\theta$ is measured in radians \section*{Trigonometric identities} $\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B$
$\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B$
$\tan ( A \pm B ) = \frac { \tan A \pm \tan B } { 1 \mp \tan A \tan B } \quad \left( A \pm B \neq \left( k + \frac { 1 } { 2 } \right) \pi \right)$ \section*{Numerical methods} Trapezium rule: $\int _ { a } ^ { b } y \mathrm {~d} x \approx \frac { 1 } { 2 } h \left\{ \left( y _ { 0 } + y _ { n } \right) + 2 \left( y _ { 1 } + y _ { 2 } + \ldots + y _ { n - 1 } \right) \right\}$, where $h = \frac { b - a } { n }$
The Newton-Raphson iteration for solving $\mathrm { f } ( x ) = 0 : x _ { n + 1 } = x _ { n } - \frac { \mathrm { f } \left( x _ { n } \right) } { \mathrm { f } ^ { \prime } \left( x _ { n } \right) }$ \section*{Probability} $\mathrm { P } ( A \cup B ) = \mathrm { P } ( A ) + \mathrm { P } ( B ) - \mathrm { P } ( A \cap B )$
$\mathrm { P } ( A \cap B ) = \mathrm { P } ( A ) \mathrm { P } ( B \mid A ) = \mathrm { P } ( B ) \mathrm { P } ( A \mid B )$ or $\mathrm { P } ( A \mid B ) = \frac { \mathrm { P } ( A \cap B ) } { \mathrm { P } ( B ) }$ \section*{Standard deviation} $\sqrt { \frac { \sum ( x - \bar { x } ) ^ { 2 } } { n } } = \sqrt { \frac { \sum x ^ { 2 } } { n } - \bar { x } ^ { 2 } }$ or $\sqrt { \frac { \sum f ( x - \bar { x } ) ^ { 2 } } { \sum f } } = \sqrt { \frac { \sum f x ^ { 2 } } { \sum f } - \bar { x } ^ { 2 } }$ \section*{The binomial distribution} If $X \sim \mathbf { B } ( n , p )$ then $\mathbf { P } ( X = x ) = \binom { n } { x } p ^ { x } ( 1 - p ) ^ { n - x }$, Mean of $X$ is $n p$, Variance of $X$ is $n p ( 1 - p )$ \section*{Hypothesis test for the mean of a normal distribution} If $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$ then $\bar { X } \sim \mathrm {~N} \left( \mu , \frac { \sigma ^ { 2 } } { n } \right)$ and $\frac { \bar { X } - \mu } { \sigma / \sqrt { n } } \sim \mathrm {~N} ( 0,1 )$ \section*{Percentage points of the normal distribution} If $Z$ has a normal distribution with mean 0 and variance 1 then, for each value of $p$, the table gives the value of $z$ such that $\mathrm { P } ( Z \leqslant z ) = p$.

$p$	0.75	0.90	0.95	0.975	0.99	0.995	0.9975	.0 .999	0.9995
$z$	0.674	1.282	1.645	1.960	2.326	2.576	2.807	3.090	3.291

\section*{Kinematics} Motion in a straight line
$v = u + a t$
$s = u t + \frac { 1 } { 2 } a t ^ { 2 }$
$s = \frac { 1 } { 2 } ( u + v ) t$
$v ^ { 2 } = u ^ { 2 } + 2 a s$
$s = v t - \frac { 1 } { 2 } a t ^ { 2 }$ Motion in two dimensions
$\mathbf { v } = \mathbf { u } + \mathbf { a } t$
$\mathbf { s } = \mathbf { u } t + \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }$
$\mathbf { s } = \frac { 1 } { 2 } ( \mathbf { u } + \mathbf { v } ) \boldsymbol { t }$
$\mathbf { s } = \mathbf { v } t - \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }$
[0pt] [BLANK PAGE]
1. A vehicle is driven at a constant speed of $12 \mathrm {~ms} ^ { - 1 }$ along a straight horizontal road. Only one of the statements below is correct. Identify the correct statement.
Tick ( $\checkmark$ ) one box. The vehicle is accelerating □ The vehicle's driving force exceeds the total force resisting its motion □ The resultant force acting on the vehicle is zero □ The resultant force acting on the vehicle is dependent on its mass □
2. A number of forces act on a particle such that the resultant force is $\binom { 6 } { - 3 } \mathrm {~N}$
One of the forces acting on the particle is $\binom { 8 } { - 5 } \mathrm {~N}$
Calculate the total of the other forces acting on the particle.
Circle your answer.
[0pt] [1 mark] $$\binom { 2 } { - 2 } \mathrm {~N} \quad \binom { 14 } { - 8 } \mathrm {~N} \quad \binom { - 2 } { 2 } \mathrm {~N} \quad \binom { - 14 } { 8 } \mathrm {~N}$$ 3. A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$
A brick $P$ of mass $m$ is placed on the plane.
The coefficient of friction between $P$ and the plane is $\mu$
Brick $P$ is in equilibrium and on the point of sliding down the plane.
Brick $P$ is modelled as a particle.
Using the model,

find, in terms of $m$ and $g$, the magnitude of the normal reaction of the plane on brick $P$
show that $\mu = \frac { 3 } { 4 }$ For parts (c) and (d), you are not required to do any further calculations.
Brick $P$ is now removed from the plane and a much heavier brick $Q$ is placed on the plane. The coefficient of friction between $Q$ and the plane is also $\frac { 3 } { 4 }$
Explain briefly why brick $Q$ will remain at rest on the plane. Brick $Q$ is now projected with speed $0.5 \mathrm {~ms} ^ { - 1 }$ down a line of greatest slope of the plane.
Brick $Q$ is modelled as a particle.
Using the model,
describe the motion of brick $Q$, giving a reason for your answer.
4. A particle $P$ moves with acceleration $( 4 \mathbf { i } - 5 \mathbf { j } ) \mathrm { ms } ^ { - 2 }$
At time $t = 0 , P$ is moving with velocity $( - 2 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$
Find the velocity of $P$ at time $t = 2$ seconds. At time $t = 0 , P$ passes through the origin $O$.
At time $t = T$ seconds, where $T > 0$, the particle $P$ passes through the point $A$.
The position vector of $A$ is $( \lambda \mathbf { i } - 4.5 \mathbf { j } ) \mathrm { m }$ relative to $O$, where $\lambda$ is a constant.
Find the value of $T$.
Hence find the value of $\lambda$
5.
1. At time $t$ seconds, where $t \geqslant 0$, a particle $P$ moves so that its acceleration $\mathbf { a } \mathrm { ms } ^ { - 2 }$ is given by $$\mathbf { a } = ( 1 - 4 t ) \mathbf { i } + \left( 3 - t ^ { 2 } \right) \mathbf { j }$$ At the instant when $t = 0$, the velocity of $P$ is $36 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$
Find the velocity of $P$ when $t = 4$
Find the value of $t$ at the instant when $P$ is moving in a direction perpendicular to $\mathbf { i }$
(ii) At time $t$ seconds, where $t \geqslant 0$, a particle $Q$ moves so that its position vector $\mathbf { r }$ metres, relative to a fixed origin $O$, is given by $$\mathbf { r } = \left( t ^ { 2 } - t \right) \mathbf { i } + 3 t \mathbf { j }$$ Find the value of $t$ at the instant when the speed of $Q$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-28_529_993_374_529} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A small ball is projected with speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ at the top of a vertical cliff.
The point $O$ is 25 m vertically above the point $N$ which is on horizontal ground.
The ball is projected at an angle of $45 ^ { \circ }$ above the horizontal.
The ball hits the ground at a point $A$, where $A N = 100 \mathrm {~m}$, as shown in Figure 2 .
The motion of the ball is modelled as that of a particle moving freely under gravity.
Using this initial model,
show that $U = 28$
find the greatest height of the ball above the horizontal ground $N A$. In a refinement to the model of the motion of the ball from $O$ to $A$, the effect of air resistance is included. This refined model is used to find a new value of $U$.
How would this new value of $U$ compare with 28 , the value given in part (a)?
State one further refinement to the model that would make the model more realistic.
7. Block $A$, of mass 0.2 kg , lies at rest on a rough plane.
The plane is inclined at an angle $\theta$ to the horizontal, such that $\tan \theta = \frac { 7 } { 24 }$
A light inextensible string is attached to $A$ and runs parallel to the line of greatest slope until it passes over a smooth fixed pulley at the top of the slope. The other end of this string is attached to particle $B$, of mass 2 kg , which is held at rest so that the string is taut, as shown in the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-32_424_1070_815_486}
$\quad B$ is released from rest so that it begins to move vertically downwards with an acceleration of $\frac { 543 } { 625 } \mathrm {~g} \mathrm {~m} \mathrm {~s} ^ { - 2 }$ Show that the coefficient of friction between $A$ and the surface of the inclined plane is 0.17
In this question use $g = 9.81 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ When $A$ reaches a speed of $0.5 \mathrm {~ms} ^ { - 1 }$ the string breaks.
1. Find the distance travelled by $A$ after the string breaks until first coming to rest.
(ii) State an assumption that could affect the validity of your answer to part (b)(i).

SPS SPS SM 2021 January Q1

4 marks

1. The graph below shows the velocity of an object moving in a straight line over a 20 second journey.
\includegraphics[max width=\textwidth, alt={}, center]{82e828ab-0812-4de1-b066-720f3632046f-04_581_1678_395_191}

Find the maximum magnitude of the acceleration of the object.
The object is at its starting position at times $0 , t _ { 1 }$ and $t _ { 2 }$ seconds. Find $t _ { 1 }$ and $t _ { 2 }$
[0pt] [4 marks]

SPS SPS SM 2021 January Q2

8 marks

2. In this question use $g = 9.8 \mathrm {~ms} ^ { - 2 }$
A boy attempts to move a wooden crate of mass 20 kg along horizontal ground. The coefficient of friction between the crate and the ground is 0.85

The boy applies a horizontal force of 150 N . Show that the crate remains stationary.
[0pt] [3 marks]
Instead, the boy uses a handle to pull the crate forward. He exerts a force of 150 N , at an angle of $15 ^ { \circ }$ above the horizontal, as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{82e828ab-0812-4de1-b066-720f3632046f-04_245_915_2161_575} Determine whether the crate remains stationary.
Fully justify your answer.
[0pt] [5 marks]

SPS SPS SM 2021 January Q3

4 marks

3. A driver is road-testing two minibuses, A and B , for a taxi company.
The performance of each minibus along a straight track is compared.
A flag is dropped to indicate the start of the test.
Each minibus starts from rest.
The acceleration in $\mathrm { ms } ^ { - 2 }$ of each minibus is modelled as a function of time, $t$ seconds, after the flag is dropped: $$\begin{aligned} & \text { The acceleration of } \mathrm { A } = 0.138 t ^ { 2 }
& \text { The acceleration of } \mathrm { B } = 0.024 t ^ { 3 } \end{aligned}$$

Find the time taken for A to travel 100 metres. Give your answer to four significant figures.
The company decides to buy the minibus which travels 100 metres in the shortest time. Determine which minibus should be bought.
[0pt] [4 marks]

SPS SPS SM 2021 January Q4

4. In this question use $g = 9.81 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
A particle is projected with an initial speed $u$, at an angle of $35 ^ { \circ }$ above the horizontal.
It lands at a point 10 metres vertically below its starting position.
The particle takes 1.5 seconds to reach the highest point of its trajectory.
Find $u$.

SPS SPS SM 2021 January Q5

8 marks

5. A buggy is pulling a roller-skater, in a straight line along a horizontal road, by means of a connecting rope as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{82e828ab-0812-4de1-b066-720f3632046f-06_243_1004_390_497} The combined mass of the buggy and driver is 410 kg
A driving force of 300 N and a total resistance force of 140 N act on the buggy.
The mass of the roller-skater is 72 kg
A total resistance force of $R$ newtons acts on the roller-skater.
The buggy and the roller-skater have an acceleration of $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$

Find $R$.

Find the tension in the rope. Name: □ \section*{U8th A LEVEL Single Mathematics Assessment
Statistics
18 ${ } ^ { \text {th } }$ January 2021 } \section*{Instructions}

Answer all the questions
Write your answer to each question in the space provided under each question. The question number(s) must be clearly shown.
Use black or blue ink. Pencil may be used for graphs and diagrams only.
You should clearly write your name at the top of this page and on any additional sheets that you use.
You are permitted to use a scientific or graphical calculator in this paper.
Final answers should be given to a degree of accuracy appropriate to the context.

Information

The total mark for this paper is $\mathbf { 3 0 }$ marks.
The marks for each question are shown in brackets.
You are reminded of the need for clear presentation in your answers.
You should allow approximately 30 minutes for this section of the test

\section*{Formulae} \section*{A Level Mathematics A (H240)} \section*{Arithmetic series} $S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}$ \section*{Geometric series} $S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$
$S _ { \infty } = \frac { a } { 1 - r }$ for $| r | < 1$ \section*{Binomial series} $( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )$, where ${ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }$
$( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )$ \section*{Differentiation}

$\mathrm { f } ( x )$	$\mathrm { f } ^ { \prime } ( x )$
$\tan k x$	$k \sec ^ { 2 } k x$
$\sec x$	$\sec x \tan x$
$\cot x$	$- \operatorname { cosec } ^ { 2 } x$
$\operatorname { cosec } x$	$- \operatorname { cosec } x \cot x$

Quotient rule $y = \frac { u } { v } , \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { v \frac { \mathrm {~d} u } { \mathrm {~d} x } - u \frac { \mathrm {~d} v } { \mathrm {~d} x } } { v ^ { 2 } }$ \section*{Differentiation from first principles} $\mathrm { f } ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }$ \section*{Integration} $\int \frac { \mathrm { f } ^ { \prime } ( x ) } { \mathrm { f } ( x ) } \mathrm { d } x = \ln | \mathrm { f } ( x ) | + c$
$\int \mathrm { f } ^ { \prime } ( x ) ( \mathrm { f } ( x ) ) ^ { n } \mathrm {~d} x = \frac { 1 } { n + 1 } ( \mathrm { f } ( x ) ) ^ { n + 1 } + c$
Integration by parts $\int u \frac { \mathrm {~d} v } { \mathrm {~d} x } \mathrm {~d} x = u v - \int v \frac { \mathrm {~d} u } { \mathrm {~d} x } \mathrm {~d} x$ Small angle approximations
$\sin \theta \approx \theta , \cos \theta \approx 1 - \frac { 1 } { 2 } \theta ^ { 2 } , \tan \theta \approx \theta$ where $\theta$ is measured in radians \section*{Trigonometric identities} $\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B$
$\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B$
$\tan ( A \pm B ) = \frac { \tan A \pm \tan B } { 1 \mp \tan A \tan B } \quad \left( A \pm B \neq \left( k + \frac { 1 } { 2 } \right) \pi \right)$ \section*{Numerical methods} Trapezium rule: $\int _ { a } ^ { b } y \mathrm {~d} x \approx \frac { 1 } { 2 } h \left\{ \left( y _ { 0 } + y _ { n } \right) + 2 \left( y _ { 1 } + y _ { 2 } + \ldots + y _ { n - 1 } \right) \right\}$, where $h = \frac { b - a } { n }$
The Newton-Raphson iteration for solving $\mathrm { f } ( x ) = 0 : x _ { n + 1 } = x _ { n } - \frac { \mathrm { f } \left( x _ { n } \right) } { \mathrm { f } ^ { \prime } \left( x _ { n } \right) }$ \section*{Probability} $\mathrm { P } ( A \cup B ) = \mathrm { P } ( A ) + \mathrm { P } ( B ) - \mathrm { P } ( A \cap B )$
$\mathrm { P } ( A \cap B ) = \mathrm { P } ( A ) \mathrm { P } ( B \mid A ) = \mathrm { P } ( B ) \mathrm { P } ( A \mid B )$ or $\mathrm { P } ( A \mid B ) = \frac { \mathrm { P } ( A \cap B ) } { \mathrm { P } ( B ) }$ \section*{Standard deviation} $\sqrt { \frac { \sum ( x - \bar { x } ) ^ { 2 } } { n } } = \sqrt { \frac { \sum x ^ { 2 } } { n } - \bar { x } ^ { 2 } }$ or $\sqrt { \frac { \sum f ( x - \bar { x } ) ^ { 2 } } { \sum f } } = \sqrt { \frac { \sum f x ^ { 2 } } { \sum f } - \bar { x } ^ { 2 } }$ \section*{The binomial distribution} If $X \sim \mathbf { B } ( n , p )$ then $\mathbf { P } ( X = x ) = \binom { n } { x } p ^ { x } ( 1 - p ) ^ { n - x }$, Mean of $X$ is $n p$, Variance of $X$ is $n p ( 1 - p )$ \section*{Hypothesis test for the mean of a normal distribution} If $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$ then $\bar { X } \sim \mathrm {~N} \left( \mu , \frac { \sigma ^ { 2 } } { n } \right)$ and $\frac { \bar { X } - \mu } { \sigma / \sqrt { n } } \sim \mathrm {~N} ( 0,1 )$ \section*{Percentage points of the normal distribution} If $Z$ has a normal distribution with mean 0 and variance 1 then, for each value of $p$, the table gives the value of $z$ such that $\mathrm { P } ( Z \leqslant z ) = p$.

$p$	0.75	0.90	0.95	0.975	0.99	0.995	0.9975	.0 .999	0.9995
$z$	0.674	1.282	1.645	1.960	2.326	2.576	2.807	3.090	3.291

\section*{Kinematics} Motion in a straight line
$v = u + a t$
$s = u t + \frac { 1 } { 2 } a t ^ { 2 }$
$s = \frac { 1 } { 2 } ( u + v ) t$
$v ^ { 2 } = u ^ { 2 } + 2 a s$
$s = v t - \frac { 1 } { 2 } a t ^ { 2 }$ Motion in two dimensions
$\mathbf { v } = \mathbf { u } + \mathbf { a } t$
$\mathbf { s } = \mathbf { u } t + \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }$
$\mathbf { s } = \frac { 1 } { 2 } ( \mathbf { u } + \mathbf { v } ) \boldsymbol { t }$
$\mathbf { s } = \mathbf { v } t - \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }$
1. The histogram below shows the heights, in cm, of male A-level students at a particular school.
\includegraphics[max width=\textwidth, alt={}, center]{82e828ab-0812-4de1-b066-720f3632046f-10_853_1095_402_429} Which class interval contains the median height?
2. A teacher in a college asks her mathematics students what other subjects they are studying. She finds that, of her 24 students: 12 study physics
8 study geography
4 study geography and physics
A student is chosen at random from the class. Determine whether the event 'the student studies physics' and the event 'the student studies geography' are independent.
[0pt] [2 marks]
3. Abu visits his local hardware store to buy six light bulbs.
He knows that $15 \%$ of all bulbs at this store are faulty.
(a) State a distribution which can be used to model the number of faulty bulbs he buys.
(b) Find the probability that all of the bulbs he buys are faulty.
(c) Find the probability that at least two of the bulbs he buys are faulty.
(d) Find the mean of the distribution stated in part (a).
(e) State two necessary assumptions in context so that the distribution stated in part (a) is valid.
4. A survey of 120 adults found that the volume, $X$ litres per person, of carbonated drinks they consumed in a week had the following results: $$\sum x = 165.6 \quad \sum x ^ { 2 } = 261.8$$ (a) (i) Calculate the mean of $X$.
(a) (ii) Calculate the standard deviation of $X$.
(b) Assuming that $X$ can be modelled by a normal distribution find
(b) (i) $\mathrm { P } ( 0.5 < X < 1.5 )$
(b) (ii) $\mathrm { P } ( X = 1 )$
(c) Determine with a reason, whether a normal distribution is suitable to model this data.
[0pt] [2 marks]
(d) It is known that the volume, $Y$ litres per person, of energy drinks consumed in a week may be modelled by a normal distribution with standard deviation 0.21 Given that $\mathrm { P } ( Y > 0.75 ) = 0.10$, find the value of $\mu$, correct to three significant figures.
[0pt] [4 marks]
5. In a region of England, the government decides to use an advertising campaign to encourage people to eat more healthily. Before the campaign, the mean consumption of chocolate per person per week was known to be 66.5 g , with a standard deviation of 21.2 g
(a) After the campaign, the first 750 available people from this region were surveyed to find out their average consumption of chocolate.
(a) (i) State the sampling method used to collect the survey.
(a) (ii) Explain why this sample should not be used to conduct a hypothesis test.
(b) A second sample of 750 people revealed that the mean consumption of chocolate per person per week was 65.4 g Investigate, at the $10 \%$ level of significance, whether the advertising campaign has decreased the mean consumption of chocolate per person per week. Assume that an appropriate sampling method was used and that the consumption of chocolate is normally distributed with an unchanged standard deviation.

\(d\)	10	20	30	40	50
\(\mathrm { P } ( D = d )\)	\(\frac { k } { 10 }\)	\(\frac { k } { 20 }\)	\(\frac { k } { 30 }\)	\(\frac { k } { 40 }\)	\(\frac { k } { 50 }\)

\(\mathrm { f } ( x )\)	\(\mathrm { f } ^ { \prime } ( x )\)
\(\tan k x\)	\(k \sec ^ { 2 } k x\)
\(\sec x\)	\(\sec x \tan x\)
\(\cot x\)	\(- \operatorname { cosec } ^ { 2 } x\)
\(\operatorname { cosec } x\)	\(- \operatorname { cosec } x \cot x\)

\(\mathrm { f } ( x )\)	\(\mathrm { f } ^ { \prime } ( x )\)
\(\tan k x\)	\(k \sec ^ { 2 } k x\)
\(\sec x\)	\(\sec x \tan x\)
\(\cot x\)	\(- \operatorname { cosec } ^ { 2 } x\)
\(\operatorname { cosec } x\)	\(- \operatorname { cosec } x \cot x\)

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\(p\)	0.75	0.90	0.95	0.975	0.99	0.995	0.9975	.0 .999	0.9995
\(z\)	0.674	1.282	1.645	1.960	2.326	2.576	2.807	3.090	3.291