这个作业是完成一系列数学练习题
IFYME002 Mathematics Engineering

问题A1

解方程7?+2?=2

5?-6?=7

[四]

问题A2

包里有红色的珠子和蓝色的珠子。两颗珠子是从

包,一个接一个,没有替换。

根据?和?写下两个珠子都是红色的概率。(你知道

不用简化你的回答)。

[三]

问题A3

(a) 写?

2-10?+5的形式(?+?)

2+?,其中?和?是整数。[二]

(b) 用你对(a)部分的回答来解方程?

2-10?+5=0,给你

以surd形式回答。

[二]

问题A4

用二项式展开来表示

(4-)

1个

2个

?)

4个

=256-128?+24?

2-2?

3个+

1个

16个

?

4[3]

问题A5

如果log2

(8?)

2个

)=?,根据?写出log2?。[三]

问题A6

求解方程cos 2?=-0.53(0<?<180°)

问题A7

评价

∫(? − ?)

2??

?

0个

其中?是一个常数,用最简单的形式用?给出你的答案。

[四]

问题A8

证明

csc 2?-cot 2??tan?

[三]

问题A9

曲线有方程?=sin 3?cos 2?。

(a) 找到

??

.

(b) 求出??的值

??

当?=

?

6个

. 回答四个重要数字。

在这个问题中,正确使用有效数字得1分。

[二]

[三]

问题A10

函数?(?)定义为?(?)=?

2-6?(?>2)

函数?(?)定义为?(?)=2?−7(?>2)

假设?[?(?)]=27,求出?的值。

(记住展示你工作的每个阶段)

[三]

问题A11

使用代换式?=?

找到2-2?+5

∫(?-1)(?

2-2?+5)

4??。

[四]

IFYME002数学工程

1920 V1©2020英国北方财团有限公司第4页,共9页

B部分

只回答三个问题。这个部分有60分。

问题B1

(a) a点位于(-6,-2),B点位于(9,7)。AB垂直于

符合方程式10?+??−7=0。

求出?的值。[三]

(b) i.求解2?-10≥3。[一]

二。解?

2≤64。[二]

iii.说明满足2?-10≥3和?的值范围

2≤64。[一]

四、你对第三部分的回答是开放区间还是封闭区间?给一个

你回答的理由。[一]

(c) 当?

3-2?

2+??+2?除以(?-2),余数与

当?

3+2?

2-2??+?除以(?+3)。

利用余数定理求出?的值。[四]

(d) 3号

几何级数的rd项是1152和6

第三学期是3888。

找出公倍数和第一项。[五]

(e) 一个算术级数有公共差7和前20项之和

是1590。

显示6

第四学期是48。[三]

Section B
Answer THREE questions ONLY. This section carries 60 marks.
Question B1
(a) Point A lies at (−6, −2) and point B lies at (9, 7). AB is perpendicular to the
line with equation 10? + ?? − 7 = 0.
Find the value of ?. [ 3 ]
(b) i. Solve 2? − 10 ≥ 3. [ 1 ]
ii. Solve ?
2 ≤ 64. [ 2 ]
iii. State the range of values which satisfy both 2? − 10 ≥ 3 and ?
2 ≤ 64. [ 1 ]
iv. Is your answer to part iii an open interval or a closed interval? Give a
reason for your answer. [ 1 ]
(c) When ?
3 − 2?
2 + ?? + 2? is divided by (? − 2), the remainder is the same as
when ?
3 + 2?
2 − 2?? + ? is divided by (? + 3).
Use the Remainder Theorem to find the value of ?. [ 4 ]
(d) The 3
rd term of a geometric series is 1152 and the 6
th term is 3888.
Find the common ratio and the first term. [ 5 ]
(e) An arithmetic series has common difference 7 and the sum of the first 20 terms
is 1590.
Show that the 6
th term is 48. [ 3 ]
IFYME002 Maths Engineering
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Question B2
(a)
i.
The variables ? and ? are connected by the formula
? = ?
2? + ?
? − ?.
Find the exact value of ? when ? = ln 3, giving your answer in its
simplest form.
[ 2 ]
ii.
The graph of ? = ?
2? + ?
? − ? has one stationary value.
Find the exact value of the ? − coordinate where this stationary value
occurs. [ 4 ]
(b) Given that
log??
3 + log?3 + log?
(? − 6) = 3logy? (? > 0)
where ? is a positive constant, find the value of ?.
[ 4 ]
(c) Find the value of ? if

40?
7 ÷ 5?
3?
4 × 9?
5 =
1
16

[ 3 ]
(d)
i.
Figure 1 shows the acute-angled triangle PQR with PQ = ? metres,
PR = 12 metres and angle QPR = 60°.
The area of triangle PQR is 33√3 cm2
.
Find the value of ?.
[ 2 ]
ii. Find the length of QR, giving your answer in surd form. [ 3 ]
iii. Find angle PQR. [ 2 ]
P
60° NOT TO SCALE
? m
12 m
Q
R
Figure 1
IFYME002 Maths Engineering
1920 V1 © 2020 Northern Consortium UK Ltd Page 6 of 9
Question B3
(a)
In this question, you are given the volume of a hemisphere (half of
a sphere) with radius ? is 2
3
??
3
and its surface area is 2??
2
.
Figure 2 shows a solid piece of metal which is made up of a cylinder of
radius r cm and height ℎ cm with a hemisphere attached to its top.
The total surface area of the solid is 180? cm2
.
i. Find ℎ in terms of ?. [ 2 ]
ii. Show that the volume of the solid, ?, is given by
? = 90?? −
5
6
??
3
[ 3 ]
iii. Use calculus to find the value of ? which gives the maximum volume. [ 4 ]
iv. Confirm that your value of ? give a maximum.
Part (b) is on the next page.
[ 3 ]
ℎ cm
? cm
Figure 2
IFYME002 Maths Engineering
1920 V1 © 2020 Northern Consortium UK Ltd Page 7 of 9
(b)
Question B3 – (continued)
Figure 3 shows the curve ? =
1
2
?
2 − 4? + 11 and line ? which is a tangent
to the curve at (6, 5).
i. Find the equation of line ?, giving your answer in the form ? = ?? + ?. [ 3 ]
ii. Find the area, which is shaded on the diagram, that is bounded by the
curve ? =
1
2
?
2 − 4? + 11, line ? and both axes.
[ 5 ]
? NOT TO SCALE
? =
1
2
?
2 − 4? + 11
?
(6, 5)

O ?
Figure 3
IFYME002 Maths Engineering
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Question B4
(a)
i.
Line ?1 has equation ? = (3? − ? + 4?) + ?(2? + 4? − ?)
Line ?2 has equation ? = (−3? + 7? + 17?) + ?(3? + ? − 4?)
where ? and ? are scalars.
Show that lines ?1 and ?2 intersect and find the coordinates of point ?,
which is their point of intersection. [ 6 ]
ii. Find the acute angle between lines ?1 and ?2 . [ 3 ]
iii.
Point ? lies on ?2 and has coordinates (?, ?, −19).
Find the values of ? and ?. [ 1 ]
iv. Find the shortest distance from point ? to line ?1 . [ 3 ]
(b) i. Solve 3 cos 2? − cos ? + 1 = 0 (0° ≤ ? ≤ 360°) [ 4 ]
ii. If tan ? =
1
?
and tan ? =
1
?
, show that tan(? + ?) =
? + ?
?? − 1
.
[ 3 ]
IFYME002 Maths Engineering
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Question B5
(a) Use the Quotient Rule to differentiate ? =
?
1 − ?
2
.
Simplify your answer. [ 2 ]
(b) Solve the differential equation ?
2? (
??
?? ) = ? + 3
subject to ? = 2 when ? = 1.
Give your answer in the form ? = ?(?).
[ 4 ]
(c) i. Show that ∫ 6?
2
ln ? ?? = 2?
3
ln ? −
2
3
?
3 + ?
where ? is a constant
[ 3 ]
ii. Use the result in part i. to explain why
∫ 6?
2
ln (
1
?
) ?? = −2?
3
ln? +
2
3
?
3 − ?
[ 2 ]
(d) i. Show that 1
(?+1)(?+?)
can be expressed in partial fractions in the form
1
?
(
1
?+1

1
?+?
) where ? and ? are constants. Find ? in terms of ?. [ 4 ]
ii. Hence find
(? − 1) ∫
1
(? + 1)(? + ?)
1
0
??.
Give your answer as a single logarithm [ 5 ]
– This is the end of the Time-Controlled Assessment. –


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