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End of Unit - Algebraic Expression

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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‘Ž + 𝑏√5 where π‘Ž and 𝑏 are
integers.
..........................
(4 marks)
Question 3
The function 𝑓 is defined by
2
6
2
60
𝑓(π‘₯) = 2π‘₯+5 + 2π‘₯−5 + 4π‘₯ 2 −25
π‘₯>4
𝐴
Show that 𝑓(π‘₯) = 𝐡π‘₯+𝐢 where 𝐴, 𝐡 and 𝐢 are constants to be found.
..........................
(4 marks)
Question 4
The function 𝑓 is defined by
𝑓(π‘₯) =
6
2
60
+
+ 2
2π‘₯+5
2π‘₯−5
4π‘₯ −25
π‘₯ > 4It can be shown that 𝑓(π‘₯) =
8
2π‘₯−5
Find 𝑓 −1 (π‘₯).
𝑓 −1 (π‘₯) = ..........................
(3 marks)
Question 5
Given 𝑦 = 2π‘₯ , express the following in terms of 𝑦.
3
1
42π‘₯−3
Write your expression in its simplest form.
..........................
(2 marks)
Question 6
Solve the equation
22π‘₯+5 − 7(2π‘₯ ) = 0
giving your answer to 2 decimal places.
(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‘Ž√𝑏 where π‘Ž and 𝑏 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 𝐴𝐡𝐢𝐷𝐸𝐴 consists of a rectangular section 𝐴𝐡𝐷𝐸 joined to a
semicircular section 𝐡𝐢𝐷 as shown in Figure 4.
Given that 𝐴𝐸 = 2π‘₯ metres, 𝐸𝐷 = 𝑦 metres and the area of the pool is 250 m 2 , show that the
perimeter, 𝑃 metres, of the pool is given by
𝑃 = 2π‘₯ +
π‘Ž πœ‹π‘₯
+
π‘₯
2
where π‘Ž is a constant to be found.
..........................
(4 marks)
6
Answers
Question 1
1−π‘₯
(2π‘₯+1)(π‘₯+1)
Question 2
π‘Ž = −17, 𝑏 = 8
Question 3
𝐴 = 8, 𝐡 = 2, 𝐢 = −5
Question 4
𝑓 −1 (π‘₯) =
8+5π‘₯
2π‘₯
Question 5
64
𝑦4
7
Question 6
π‘₯ = −2.19
Question 7
π‘₯ = 5√2
Question 8
(9 + 4π‘₯ 2 )(3 − 2π‘₯)(3 + 2π‘₯)
Question 9
π‘Ž = 250
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