# End of Unit - Algebraic Expression

```End Of Unit-Algebraic Expression
Total Marks: 30
Question 1
Express
3
2
−
2π₯ + 1 π₯ + 1
as a single algebraic fraction in its simplest form.
..........................
(2 marks)
Question 2
Solve the equation (2 + √5)π₯ = 6 − √5, giving π₯ in the form π + π√5 where π and π are
integers.
..........................
(4 marks)
Question 3
The function π is defined by
2
6
2
60
π(π₯) = 2π₯+5 + 2π₯−5 + 4π₯ 2 −25
π₯&gt;4
π΄
Show that π(π₯) = π΅π₯+πΆ where π΄, π΅ and πΆ are constants to be found.
..........................
(4 marks)
Question 4
The function π is defined by
π(π₯) =
6
2
60
+
+ 2
2π₯+5
2π₯−5
4π₯ −25
π₯ &gt; 4It can be shown that π(π₯) =
8
2π₯−5
Find π −1 (π₯).
π −1 (π₯) = ..........................
(3 marks)
Question 5
Given π¦ = 2π₯ , express the following in terms of π¦.
3
1
42π₯−3
Write your expression in its simplest form.
..........................
(2 marks)
Question 6
Solve the equation
22π₯+5 − 7(2π₯ ) = 0
(Solutions based entirely on graphical or numerical methods are not acceptable.)
..........................
(4 marks)
Question 7
Solve the equation
4
10 + π₯√8 =
6π₯
√2
Give your answer in the form π√π where π and π are integers.
π₯ = ..........................
(4 marks)
Question 8
Factorise fully
81 − 16π₯ 4
..........................
(3 marks)
5
Question 9
Figure 4 shows the plan view of the design for a swimming pool.
The shape of this pool π΄π΅πΆπ·πΈπ΄ consists of a rectangular section π΄π΅π·πΈ joined to a
semicircular section π΅πΆπ· as shown in Figure 4.
Given that π΄πΈ = 2π₯ metres, πΈπ· = π¦ metres and the area of the pool is 250 m 2 , show that the
perimeter, π metres, of the pool is given by
π = 2π₯ +
π ππ₯
+
π₯
2
where π is a constant to be found.
..........................
(4 marks)
6
Question 1
1−π₯
(2π₯+1)(π₯+1)
Question 2
π = −17, π = 8
Question 3
π΄ = 8, π΅ = 2, πΆ = −5
Question 4
π −1 (π₯) =
8+5π₯
2π₯
Question 5
64
π¦4
7
Question 6
π₯ = −2.19
Question 7
π₯ = 5√2
Question 8
(9 + 4π₯ 2 )(3 − 2π₯)(3 + 2π₯)
Question 9
π = 250
```