Revision Test 4 (Vectors, Graphing, Diff & App, Maclaurin) 17 Feb 2014, 30 Min

1. A family of curves is defined by the equation sinye x xy c , where 0 x and c is an

arbitrary constant.

(i) Show that when c = 0, there is no point on the curve at which the tangent is horizontal. [5]

(ii) The tangent to the curve at the point , ln 2

2

intersects the x-axis at the point B. Find

the coordinates of B in exact form. [2]

2. It is given that ln cosy x , where π π

4 4x .

(i) Prove that

3 2

3 2

d d d2 0

d d d

y y y

x x x .

[2]

(ii) Find the Maclaurin’s series for ln cos x , up to and including the term in 4x . [3]

(iii) Deduce that the Maclaurin’s series for ln 1 cos2x up to and including the term in 4x

is 2 41ln 2 .

6x x

[2]

3. (a) The graph of f( )y x undergoes in succession, the following transformations:

(I) A translation of 1 unit in the positive x-direction.

(II) A scaling parallel to the x-axis with a factor of 2.

The equation of the resulting curve is 2 4-e .xy Obtain an expression for f(x). [3]

(b) The diagram below shows a sketch of the graph of f( ).y x The curve passes through the points

A(0,1), B(1,0) and C(5,0).

x

y

A

2 B C O

Sketch, on separate clearly labelled diagrams, the graphs of

(i) 2f 1 ,y x

[3]

(ii) f 1 ,y x [3]

showing clearly all axial intercepts and asymptotes.

4. (a) Refer to the origin O, the position vectors of points A and B are a and b respectively. Find an

expression for the position vector of point S such that S divides the line segment AB externally

in the ratio 3:2. The point C, with position vector c is such that O is the midpoint of CS. Show

that 2a + mb + nc = 0 where m and n are constants to be determined. [4]

(b)The line l has equation 1 2 , 5z x y , and the plane 1 has equation

r

1 2 1

0 1 0 , where ,

1 3 1

. Show that the position vector of B, the point of

intersection between l and 1 , is – 24i + 5j + 49k. [3]

The point A has coordinates ( 7, 10, 3) .

(i) Find the length of projection of AB onto 1 . [3]

(ii) Find an equation of the plane 2 that contains lines BA and 'BA , where 'A is the reflection

of A about 1 . State the angle between 1 and 2 . [3]

SAJC/MYE2010/Q 7, 8

1(i) Differentiating w.r.t. x,

d dsin cos 0

d d

y yy ye x e x x y

x x

d cos

d sin

y

y

y y e x

x e x x

For tangent to be horizontal,d

0d

y

x

cosyy e x

Substitute into equation of curve,

sin cos 0

sin cos 0

y ye x xe x

x x x

Can use GC e.g. Sketch sin cosy x x x for 0 x and see there is no zero.

No solution i.e. There is no point on the curve at which the tangent is horizontal.

(ii) Equation of tangent is

ln 2ln 2

22

2

y x

When y = 0,

ln 2ln 2

22

2

22 2

2

x

x

x

The coordinates of B are 2,0 .

2(i) ln cosy x

d sintan

d cos

y xx

x x

22

2

dsec

d

yx

x

3

3

d

d

y

x 2sec sec tanx x x

2

2

d d2 sec sec tan 2

d d

y yx x x

x x

(ii) 4

4

d

d

y

x

22 3

2 3

d d d2 2

d d d

y y y

x x x

At 0x , d

0d

yy

x ,

2

2

d1

d

y

x ,

3

3

d

d

y

x 0 and

4

4

d2

d

y

x

ln cos x = 2 3 41 20 0 0

2 4!x x x x

, i.e. 2 41 1

2 12x x

(iii) 2

2

2 4

2 4

ln(1 cos 2 ) ln(1 2cos 1)

ln 2cos

ln 2 2ln cos

1 1ln 2 2

2 12

1ln 2

6

x x

x

x

x x

x x

DHS/JC2MYE/2010/Q6

3(a) -2x+4ey

-4x+4ey

-4( +1)+4

-4

e

e

x

x

y

-4f ( ) e xx

ALTERNATIVE

f(x) f ( 1)2

x

Let P = 12

x x = 2 P +2

-2 +4e xy -2(2 2) 4 -4e eP Py

-4f e xx

Replace ‘x’ by ‘2x’ Replace ‘x’ by ‘x+1’

(b)(i)

(b)(ii)

3

3 O

B’ C’ x

y y = 3-

1 4

y = 2f(x+1)

x

O

B’ C’

A’

y

1 4

-3 -6

y= y=

-2

2011Promo/HCI/I/Q12

4(a)

By Ratio Theorem, 1 2

3 3OB OS OA

OS 3b – 2a

CO OS

– c = 3b – 2a

2a – 3b – c = 0

3, 1m n

(b) l: r

1/ 2 1

5 0

0 2

Equate l with vector equation of 1 :

11 2

2 …………(1)

5

1 3 2 ……………..(2)

Solving (1) and (2) gives 33 and 49

2

position vector of B is

24

5

49

(i)

2 1 1

1 0 5

3 1 1

Students need to be

reminded of the correct

way to write vectors

Students do not pay

attention to the word

'externally'. Many

students take it as internal

ratio.

A B S

1 2

24 7 17

5 10 5

49 3 46

AB

Length of projection of AB onto the plane 1

=

AB ×

–1

5

–1

27 =

–225

–63

–90

27 = 48.187

48.2 (correct to 3 significant figures).

(ii) r

7 17 1

10 5 5

3 46 1

Angle between 1 and 2 is 90

Length of projection of

AB onto the plane 1

was given as

AB

–1

5

–1

27 instead of

taking the cross product.

Read the question

carefully. If the question

does not specify the form

of equation of plane, they

may want to give the

answer with the least

working. There is no need

to always convert to

scalar product form.