Edexcel AEA 2020 June — Question 5 22 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2020
SessionJune
Marks22
PaperDownload PDF ↗
TopicProjectiles
TypeProjectile passing through given point
DifficultyChallenging +1.8 This AEA question combines proof critique (identifying three specific errors in trigonometric manipulation) with a challenging projectile problem requiring finding two angles where trajectory passes through a point, then computing their difference using the arctan identity from part (a). The connection between parts and the need to apply the proven identity elevates this beyond standard mechanics questions, though the individual components are accessible with careful work.
Spec1.01a Proof: structure of mathematical proof and logical steps1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05l Double angle formulae: and compound angle formulae3.02i Projectile motion: constant acceleration model

5.(a)The box below shows a student's attempt to prove the following identity for \(a > b > 0\) $$\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }$$ Let \(x = \arctan a\) and \(y = \arctan b\) ,so that \(a = \tan x\) and \(b = \tan y\) $$\begin{aligned} \text { So } \tan ( \arctan a - \arctan b ) & \equiv \tan ( x - y ) \\ & \equiv \frac { \tan x - \tan y } { 1 - \tan ^ { 2 } ( x y ) } \\ & \equiv \frac { a - b } { 1 - ( a b ) ^ { 2 } } \\ & \equiv \frac { a - a b + a b - b } { ( 1 - a b ) ( 1 + a b ) } \\ & \equiv \frac { a ( 1 - a b ) - b ( 1 - a b ) } { ( 1 - a b ) ( 1 + a b ) } \\ & \equiv \frac { a - b } { 1 + a b } \end{aligned}$$ Taking arctan of both sides gives \(\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }\) as required. There are three errors in the proof where the working does not follow from the previous line.
  1. Describe these three errors.
  2. Write out a correct proof of the identity.
    (b)[In this question take \(g\) to be \(9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) ] \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{4d5b914c-28b2-4485-a42e-627c95fa16e2-22_244_1267_1870_504} \captionsetup{labelformat=empty} \caption{Figure 3}
    \end{figure} Balls are projected,one after another,from a point,\(A\) ,one metre above horizontal ground. Each ball travels in a vertical plane towards a 6 metre high vertical wall of negligible thickness,which is a horizontal distance of \(10 \sqrt { 2 }\) metres from \(A\) . The balls are modelled as particles and it is assumed that there is no air resistance.
    Each ball is projected with an initial speed of \(28 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and at a random angle \(\theta\) to the horizontal,where \(0 < \theta < 90 ^ { \circ }\) Given that a ball will pass over the wall precisely when \(\alpha \leqslant \theta \leqslant \beta\)
  3. find, in degrees, the angle \(\beta - \alpha\)
  4. Deduce that the probability that a particular ball will pass over the wall is \(\frac { 2 } { 3 }\)
  5. Hence find the probability that exactly 2 of the first 10 balls projected pass over the wall. You should give your answer in the form \(\frac { P } { Q ^ { k } }\) where \(P , Q\) and \(k\) are integers and \(P\) is not a multiple of \(Q\).
  6. Explain whether taking air resistance into account would increase or decrease the probability in (b)(iii).
  7. find, in degrees, the angle \(\beta - \alpha\)

5.(a)The box below shows a student's attempt to prove the following identity for $a > b > 0$

$$\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }$$

Let $x = \arctan a$ and $y = \arctan b$ ,so that $a = \tan x$ and $b = \tan y$

$$\begin{aligned}
\text { So } \tan ( \arctan a - \arctan b ) & \equiv \tan ( x - y ) \\
& \equiv \frac { \tan x - \tan y } { 1 - \tan ^ { 2 } ( x y ) } \\
& \equiv \frac { a - b } { 1 - ( a b ) ^ { 2 } } \\
& \equiv \frac { a - a b + a b - b } { ( 1 - a b ) ( 1 + a b ) } \\
& \equiv \frac { a ( 1 - a b ) - b ( 1 - a b ) } { ( 1 - a b ) ( 1 + a b ) } \\
& \equiv \frac { a - b } { 1 + a b }
\end{aligned}$$

Taking arctan of both sides gives $\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }$ as required.

There are three errors in the proof where the working does not follow from the previous line.
\begin{enumerate}[label=(\roman*)]
\item Describe these three errors.
\item Write out a correct proof of the identity.\\
(b)[In this question take $g$ to be $9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ ]

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{4d5b914c-28b2-4485-a42e-627c95fa16e2-22_244_1267_1870_504}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

Balls are projected,one after another,from a point,$A$ ,one metre above horizontal ground. Each ball travels in a vertical plane towards a 6 metre high vertical wall of negligible thickness,which is a horizontal distance of $10 \sqrt { 2 }$ metres from $A$ .

The balls are modelled as particles and it is assumed that there is no air resistance.\\
Each ball is projected with an initial speed of $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and at a random angle $\theta$ to the horizontal,where $0 < \theta < 90 ^ { \circ }$

Given that a ball will pass over the wall precisely when $\alpha \leqslant \theta \leqslant \beta$
\item find, in degrees, the angle $\beta - \alpha$
\item Deduce that the probability that a particular ball will pass over the wall is $\frac { 2 } { 3 }$
\item Hence find the probability that exactly 2 of the first 10 balls projected pass over the wall.

You should give your answer in the form $\frac { P } { Q ^ { k } }$ where $P , Q$ and $k$ are integers and $P$ is not a multiple of $Q$.
\item Explain whether taking air resistance into account would increase or decrease the probability in (b)(iii).
\item find, in degrees, the angle $\beta - \alpha$
\end{enumerate}

\hfill \mbox{\textit{Edexcel AEA 2020 Q5 [22]}}