2019 TJC Promotional Examination H2 Mathematics [Solution]

1 [Solution](i)

(ii)

21

x xx

Solve 21

x xx

2 2 2 0x x

2 4 8 1 2 12

x

1 2 1x (Larger value of x)

From graph, 1 2 1x , 1x

(or equivalent form 1x or 1 2 11 x )

y

2y x

1 1xy

x x

x

y

O

2

2

1x

2

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2 [Solution] Vector equation r = a + m:

The equation gives the set of position vectors of the points on the line which passes through the point with position vector a and is parallel to m. Equation of line AB: r = a + (b a) = (1 )a + b Since C lies on line AB, c = (1 )a + b for some OR using ratio theorem, c = (1 )a + b for some Area of triangle OBC

121 [(1 ) ]21 (1 )21 (1 )21 (1 ) 1 4 sin 302(1 )

c b

a b b

a b b b

a b

(1 ) 6

1 = 6 or 1 = 6 = 5 or 7

6 5OC a b6OC 6OC or 7 6OC b a7OC 7 Alternatively

Area of triangle OAB 1 1 sin302 2

a b a b =1

Since A, B, C are collinear and Area triangle OBC = 6, Area of 6Area of 11

6 621 1 12

OBCOAB

h BC BC

ABh AB

a of O

BC 6BC 61

61AB

since b b = 0

3

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By ratio Theorem, Case 1:

6 7 67

OA OCOB OC b a6 7OA OCOB OC6OA OC 7OCOA OC

Case 2:

5 6 56

OB OCOA OC a b5 6 5OB OCOA OC5OB OC 6OC 5OB OC

O

A B

C

B1

6

O

A

B A

BB1

5

C

4

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3 [Solution] (a)(i)

(ii) (iii)

3f ( )y x a

y = f2x

1f ( )

yx

(b)

y = 2 + e x

y = 2 + e x 5 = 3 + e x

y = ( 3 + e x) = 3 e x

y = 3 e x + 5 y = 8 e x

x O

( 2a, 0) B2(2a, 0)

y

x

y

O x

B1(2a, 0) C1(–2a, 0)

A1(0, 9a)

y

x O OA3(–a, )

x = –3a x = a

y = 0

5

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4 [Solution] (i)

(ii) 4 4

0 1

1 12 f ( ) d 2 6 (1) d2

x x xx x

Let d 1 1 d 22uu xx ux

(giving dx = 2u du)

When x = 1, u = 1

When x = 4, u = 2

4 221 1

2

1

21

12 1 d 12 2 d

124 d1

24 ln( 1)

324ln2

x u uu ux x

uuu

4

0

3 f ( ) d 4 24ln2

x x

Alternative method

y = 2 + e x

y = (2 + e x)

y = (2 + e x) +10 y = 8 e x

O

y

x 1 4 5 6 2

2

6

6

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5 [Solution] (a) sin cos3 d cos3 sin d

1 sin 4 sin 2 d2

1 1cos 4 cos 28 4

x x x x x x

x x x

x x c

(b)

2 2

12 2

12

2

1 1 2(1 ) d d21 2 1 2

1 212

1 2

x xx xx x x x

x xc

x x c

2

2

2

1For to be defined, 1 2

1 2 0 ( 1) 2 0

1 2 1 2 0

1 2 1 2

xx x

x xx

x x

x

Greatest integer value of b is 2

(c) cos d sin sin d

sin cos

x x x x x x x

x x x c

32 2

2

32

32 2

32 2

2

0 0

0

2

2

cos d cos d cos d cos d

sin cos sin cos

sin cos (using above result)

3 31 12 2 2 2

4

x x x x x x x x x x x x

x x x x x x

x x x

7

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6 [Solution] (a) 8 7 20T a d --- (1)

27 15 24T T

( 26 ) ( 14 ) 2412 24

2

a d a dd

d

Subst into (1): 20 7(2) 6a

40

7

8 40 40 7

40 12 39(2) 18002

7 12 6(2) 842

Sum of to 1800 84 1716

S

S

T T S S

2

1218 (

0

12 ( 1)(2) t2

more han 1716

5 293

1218)

4

n n

n n

Method 1 n 2 5 2934n n 51 78 < 0 52 30 > 0

Using GC, the smallest value of is 52n Method 2

2 5 2934 0n n

56.7 or 51.7n n smallest value of is 52n

(b) Let the first term be a and the common ratio be r.

Given S Sn = 2Tn

11

21 1

nn

a ra arr r

121

nnr ar

ra 2

1r

r23

r

3213

aS a

Hence, sum to infinity is 3 times of the first term.

51.7 56.7 n

8

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7 [Solution] (i)(a)

Radius of cylinder 2 2a h

Internal cylindrical area 2 2 2 22 2 4a h h h a h 2 2

2 2

4 4

4 (shown)

S ah h a h

h a a h

(b) 2 2

2 2

2 2 2

2 2

2 2

2 2

d 1 24 4d 2

4

24

S ha a h hh a h

a h haa h

a haa h

2 2

2 2

22 2 2 2 2

4 2 2 4 4 2 2

4 2 2

2 2 2

dLet 0d

2

2

4 44 3 0

(4 3 ) 0

Sh

h a aa h

h a a a h

h h a a a a hh h a

h h a

22 3Since 0,

43 ( 0)

2

ah h

h a h( 0)((

Max value of S 2 23 342 4

a a a a

2

12 32

3 3

a a a

a

y

x O a h h a

a

9

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(ii) Equation of circle is 2 2 2 2 2 2 x y a y a x

22 2 2 2

0

32 2 2

32 2 3

3

0

2 d (2 )

2 2 ( )3

2 2 23

4 which is independent of 3

h

h

V a x x a h h

xa x h a h

ha h ha h

h a

8 [Solution] (i) 1f ( ) 7

1y x x

x

Rf = , 10 6,

y

x O

y = x 7

(0, 6)

( 2, 10)

x = 1

10

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(ii) 1 1g( ) ax abx ax b x b

Rg = \ a\ a

Df = \ 1\ 1 For fg to exist, Rg Df i.e \ a\ a \ 1\ 1 Hence, a = 1

Rf = , 10 6,

Dg = \ b\ b For gf to exist, Rf Dg i.e , 10 6, \ b\ b

Hence 10 < b < 6

6 < b < 10

(iii) Let 1axy

x b

1

1xy by ax

byxy a

1 1 g ( ) bxxx a

1Given g ( ) g( ) for all real values of ,

1 1

x x xbx ax

x a x b

By comparison, b = a

y

x y = a

x = b

11

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9 [Solution]

(i)0 20 00 , 0 , 1020 0 0

OA OB OCOA0 20 0

10O OC0 000 100 000 100 00000

200 ,20

01020

AB OB OA

AC OC OA

AB OB OA20OB OA 00

AC OC OA0

1OC OA 11

200 1400 200 2200 1

AB ACAB AC2040AC 404040

1 20 12 0 2 201 0 1

r 202 0 22 0 20 22 0 20 20 20 20 20 2

Equation of the face ABC is x + 2y + z = 20

Shortest distance from O to ABC

=2 2 2

20 2061 2 1

(ii) Method 1 (Using Unit vector)

Using (i),1 10

20 1 12 2036 6 1 10

ON 20ON

Coordinates of N is 10 20 10, ,3 3 3

Alternative Solution

Line 1

: 2 ,1

ON r

1 1 12 20 2 2 201 1 1

106 103

ONON1

2020 2 22 220 2 22 22 20 2 220 2 220 2 22 22 2222

Coordinates of N is 10 20 10, ,3 3 3

3

n

: ,,,,,,,,,,,,N ,11111111111111111112222222 ,222222222222222222222222222221111111111111111111111111111111

0

12

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B

Oi j

k

A

C

Floor

Wall Wall

20

i20

10

B

Oi j

k

A

C

Floor

Wall Wall

20

i20

10

(iii) Let be the required acute angle.1 02 01 1 1cos

6 1 665.9 (1 d.p.)

0001 2

1 2

cosn nn n

(iv)

Method 1(Use concept of point R is a point on plane OBR)

Using ratio theorem, 0

(1 ) 1020(1 )

OR OC OA)))1 )

Normal of plane OBR =1 1 00 2 10 1 2

Equation of plane OBR is 01 0

2r

1

R

ppoin

heoeoeooeooooooeooreeeeeeeeeeeeeeeeem,mm,m,m,m,m,mm,m,mmm,m,m,m,mm OOOOOOOOOOOOOOOOOOOORRRRRRRRRRRRRRRRRRRRRRRRR (OOCC (OOCC (1((1OOOOOOOOOOOOORRRRRRRRRRRRRRRRRR OOOOCCCC ((1OOCC (1OOCC (1

l

13

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Since point R is a point on plane OBR,0 0

10 1 020(1 ) 2

----- (*)

410 40 40 05

45

45

0 010 8

20(1 ) 4OROR 1111

Therefore, coordinates of R is (0, 8, 4)

Method 2 (Use concept of point N is a point on line l)

Using ratio theorem, 0

(1 ) 1020(1 )

OR OC OA(1 )OR OC (1 )OCOC (1 )OC (1 )

2010

20(1 )BR OR OBBR OR OBOR OB

Equation of line l:20 20 ,0 2(1 )

r

Given N lies on line l,103 20 220 03

0 2(1 )103

10 2520 23 320 43 5

10 2 (1 )3

45

45

0 010 8

20(1 ) 4OROR 1111

Therefore, coordinates of R is (0, 8, 4)

225533

14

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10 [Solution](i) d2 sin 2 2 2cos 2

ddcos 2 2sin 2d

xx t t ttyy t tt

2

d 2sin 2 sin 2d 2 2cos 2 1 cos 2

2sin cos1 (2cos 1)sincostan (shown)

y t tx t t

t tt

ttt

As2

t , d tan is undefinedd 2yx

The tangent to the curve is parallel to the y-axis.

(ii)When y = 0, cos 2 0t

For 0 t , 32 or 2 2

t

3 or 4 4

t

1 23 1 or 1

2 2x x

(iii)0, 0 & 1t x y

, 2 & 1t x y

, & 12

t x y

(iv) Area = 2

1

dx

xy x

cos 2 (2 2cos 2 ) dt t t3π4

π434

4

22cos 2 + 2cos 2 dt t t34

42cos 2 + 1 cos 4 dt t t

y

O

(2 , 1)(0, 1)

( , 1)

x

22222 dddddddddddddxxxxx2222 yyyyyyyyyyyyy xxxxxxxxxxxxxx ddddddddddddddddd

((22 2((22

OO

15

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343

4

44

sin 4sin 24

3 3 sin 3 sinsin sin2 2 4 4 4 4

22

22

tt t

(v)From (ii), 1

d1, , tan 12 4 d 4

yx tx

Equation of tangent: 12

y x

When the tangent intersects C again,

cos 2 2 sin 2 12

t t t

i.e. cos 2 sin 2 2 12

t t t

Using GC, t = 1.99 (3 s.f) , t = 0.785 (i.e. /4)

Thus the tangent to C intersects C again at t = 1.99 (3 s.f).

End of Paper