MathematicsT STPM Baharu PENANG 2012

2012 TRIAL STPM BAHARU MATHEMATICS T SMJK JIT SIN ,PENANG

954/1*This question paper is CONFIDENTIALCONFIDENTIALCONFIDENTIALCONFIDENTIAL until the examination is over.

CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*2

SectionSectionSectionSection AAAA [45 marks]Answer allallallall questions in this section.

1.1.1.1. Find the set values of x for which 21 3 .xx

− > [5[5[5[5marksmarksmarksmarks]]]]

2.2.2.2. (a) Given that 2logy x= and 2 2log log 8 log 2 log 4 0kx xx k− + + = , show that

2 2 3 0.y ky k+ + − = [3[3[3[3marksmarksmarksmarks]]]](b) Solve the equation 2 12 3 2 1.x x+ = ⋅ − [4[4[4[4marksmarksmarksmarks]]]]

3.3.3.3. Given that M =2 0 10 2 11 1 2

−⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

and N =15 1 4

1 15 44 4 16

− −⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

. Find the matrix N – 6M and

show that M(N – 6M) = kI where k is an integer and I is a 3 × 3 matrix.State the value of k and hence find the inverse of matrix M. [7[7[7[7marksmarksmarksmarks]]]]

4.4.4.4. Solve the following system of linear equations using Gaussian elimination:x – 2y + z = 0

2x + y – 3z = 54x – 7y + z = –1 [8[8[8[8marksmarksmarksmarks]]]]

5.5.5.5. The functions f and g are defined as:: 0f x x x→ ≥: 3ln 0g x x x→ >

(a) Sketch the graph of f and state whether 1f − exist. Give a reason for your answer.(b) Find 1g − and state its domain.(c) Find the composite function 1fg − and state its range. [9[9[9[9marksmarksmarksmarks]]]]

6.6.6.6. (a) Express2

2

6 7 8( 2)(1 3 )x x

x x− +

+ −in partial fractions. [4[4[4[4marksmarksmarksmarks]]]]

(b) The remainder obtained when 3 23 4 2x mx x+ − − is divided by 1x + is twice theremainder obtained when the same expression is divided by 2x − . Find the valueof m. [5[5[5[5marksmarksmarksmarks]]]]




SectionSectionSectionSection BBBB [15 marks]Answer any oneoneoneone question in this section.

7.7.7.7. Relative to a fixed origin O, the points A, B and C have position vectors givenrespectively by aaaa = 2iiii + 3jjjj – kkkk, bbbb = 5iiii – 2jjjj +3kkkk, cccc = 4iiii +jjjj – 2kkkkFind (i) the length of AB, correct to 3 significant figures,(ii) angle BAC, correct to the nearest degree,(iii) the area of triangle ABC, correct to 3 significant figures.Show that, for all the real values of the parameter t, the point P with position vectorlies on the line through A and B.Find pppp such that OP is perpendicular to AB. [15[15[15[15marksmarksmarksmarks]]]]

8.8.8.8. The points A and B have position vectors 3iiii + 2jjjj + kkkk and iiii + 2jjjj + 3kkkk, respectively,relative to the origin O. The point C is on the line OA produced and is such that AC =2OA. The point D is on OB produced and is such that BD = OB. The point X is suchthat OCXD is a parallelogram. Show that the line AX is parallel to the vector iiii + jjjj + kkkk.Find(i) in the form rrrr = uuuu + tvvvv, the equations of the line AX and CD.(ii) the position vector of the point of intersection between the lines AX and CD.(iii) the angle BAX.(iv) the area of the parallelogram OCXD. [15[15[15[15marksmarksmarksmarks]]]]




ANSWER SCHEMESMJKSMJKSMJKSMJK JITJITJITJIT SINSINSINSIN −−−− STPMSTPMSTPMSTPM TrialTrialTrialTrial ExaminationExaminationExaminationExamination 2012201220122012MarkingMarkingMarkingMarking schemeschemeschemescheme forforforfor MathematicsMathematicsMathematicsMathematics TTTT PaperPaperPaperPaper 1111

SectionSectionSectionSection AAAA [45 marks]NoNoNoNo Working/AnswerWorking/AnswerWorking/AnswerWorking/Answer PartialPartialPartialPartial marksmarksmarksmarks TotalTotalTotalTotal

marksmarksmarksmarks1111 21 3x

x− > , x ≠ 0

The set of values of x is {x | x∈R, x < 0 or x > 1}.

OR

x2x31 >− , x ≠ 0

⇔x2x31 >− orororor

x2x31 −<−

0x2x31 >−−

0x

2x3x 2

>−−

×(−1), 0x

2xx3 2

<+−

Graph: V shape &

( 1 ,03

): D1D1D1D1

Reciprocal :D1D1D1D1

Point A : B1B1B1B1

Ans : M1M1M1M1 A1A1A1A1

M1

5

y

x

1

0

2yx

=

13

3 1y x= −1 3y x= −

1

• A(1, 2)




3x2 − x + 2 = 3(x2 −3x ) + 2

= 2)61(3)

61x(3 22 +−−−

=12111)

61x(3 2 +− > 0

Since 3x2 − x + 2 > 0⇒ x < 0

x2x31 −<− , x ≠ 0

0x2x31 <+−

0x

2x3x 2

<+−

×(−1), 0x

2xx3 2

>−−

0x

)1x)(2x3(>

−+

Let 3x + 2 > 0, x − 1 > 0, x > 0

x >32

− , x > 1, x > 0

use number line,…

32

− < x < 0 or x > 1

∴ the set of values of x is {x | x∈R, x < 0 or x > 1}

M1 (either)

A1

A1

A1

2(a)2(a)2(a)2(a) Given that 2logy x=

2 2log log 8 log 2 log 4 0kx xx k− + + =

2 22 2

2 2

log 8 log 4log log 2 0log log

x k kx x

⎛ ⎞− + + =⎜ ⎟

⎝ ⎠3 2 0y k ky y

⎛ ⎞− + + =⎜ ⎟

⎝ ⎠,

32log8log 322 == , 22log4log 2

22 ==2 3 2 0y ky k− + + =2 2 3 0y ky k+ + − =

M1M1M1M1 (changing base)

M1M1M1M1(subst. correct y,2 2log 8 3, log 4 2= = )

A1A1A1A1

3




2(b)2(b)2(b)2(b) 2 12 3 2 1.x x+ = ⋅ −22(2 ) 3 2 1 0x x− ⋅ + =

(2 2 1)(2 1) 0x x⋅ − − =Then 2 2 1 0x⋅ − = and 2 1 0x − =

122

x = 2 1x =

12 2x −= 02 2x =1x = − 0x =

∴ x = −1, 0

M1M1M1M1(quadratic form)

M1M1M1M1(factorize)A1A1A1A1(both)

A1A1A1A1(both)

4

3333

Given that M =2 0 10 2 11 1 2

−⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

and N =15 1 4

1 15 44 4 16

− −⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

.

N – 6M15 1 4

1 15 44 4 16

− −⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟−⎝ ⎠

– 62 0 10 2 11 1 2

−⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

3 1 21 3 2

2 2 4

−⎛ ⎞⎜ ⎟= − −⎜ ⎟⎜ ⎟−⎝ ⎠

M(N – 6M)2 0 10 2 11 1 2

−⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠

3 1 21 3 2

2 2 4

−⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟−⎝ ⎠

4 0 00 4 00 0 4

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

100010001

4

∴ M(N – 6M) = 4I shown

∴ k = 4

M(N – 6M) = 4I

I)4

M6N(M =−

M1M1M1M1

A1

M1M1M1M1

A1A1A1A1A1A1A1A1

M1M1M1M1

7




∴ 1 1M (N 6M)4

− = −

3 1 21 1 3 24

2 2 4

−⎛ ⎞⎜ ⎟= − −⎜ ⎟⎜ ⎟−⎝ ⎠3 1 14 4 21 3 14 4 2

1 1 12 2

⎛ ⎞−⎜ ⎟⎜ ⎟⎜ ⎟= − −⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

A1A1A1A1

4444 Given that x – 2y + z = 02x + y – 3z = 54x – 7y + z = –1

1 2 1 02 1 3 54 7 1 1

⎛ − ⎞⎜ ⎟−⎜ ⎟⎜ ⎟− −⎝ ⎠

2 1 2( 2 )R R R+ − →

3 1 3( 4 )R R R+ − →

2 3R R↔

3 2 3( 5 )R R R+ − →[ echelon form ]

Thus, 10z = 10 ……..(1)y – 3z = –1 ……..(2)

x – 2y + z = 0 ………(3)

1 2 1 00 5 5 50 1 3 1

⎛ − ⎞⎜ ⎟−⎜ ⎟⎜ ⎟− −⎝ ⎠

1 2 1 00 1 3 10 5 5 5

⎛ − ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟−⎝ ⎠

1 2 1 00 1 3 10 0 10 10

⎛ − ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎝ ⎠

B1B1B1B1

M1M1M1M1 (one(one(one(one operation)operation)operation)operation)M1M1M1M1 (one(one(one(one operation)operation)operation)operation)

M1A1M1A1M1A1M1A1 (one(one(one(oneoperation)operation)operation)operation)

A1A1A1A1

8




from (1), z = 1

subst. z = 1 into (2), y – 3(1) = –1y = 2

subst. z = 1 and y = 2 into (3), x – 2(2) + (1) = 0x = 3

Therefore, x = 3, y = 2 and z = 1

M1M1M1M1

A1A1A1A15(a)5(a)5(a)5(a) Given that : 0f x x x→ ≥

Since any horizontal line y = k for k ≥ 0 cuts the graphy = f(x) at only one point, therefore y = f(x) is one toone function as such f −1 exists.

1 exist because for the given domain ( )f f x− is one toone, and defined for all values of x.

Graph : D1D1D1D1

B1B1B1B1

2

5(b)5(b)5(b)5(b) g(x) = 3 ln x, Dg = (0, ∞), Rg = (−∞, ∞)Let 1( )y g x−=∴ x = g( y )

= 3 lny

ln3xy =

∴ 1 3( )x

g x e− =Domain of 1( )g x− = Rg = {x | x ∈ℜ}

M1M1M1M1

A1A1A1A1

A1A1A1A1

3

5(c)5(c)5(c)5(c)1 3( ) ( )

x

fg x f e− = M1M1M1M14

y

x0

( )f x x=




3x

e=

6x

e= , }Rx|x{D 1fg∈=−

The range of 1( )fg x− is {y : y > 0 }

A1A1A1A1A1A1A1A1

A1A1A1A1

6(a)6(a)6(a)6(a)Let

2

2 2

6 7 8( 2)(1 3 ) 1 3 2x x A Bx c

x x x x− + +

≡ ++ − − +

2 26 7 8 ( 2) (1 3 )( )x x A x x Bx C− + ≡ + + − +

Subst. 13

x = ,2 21 1 16 7 8 2

3 3 3A⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + = +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

2 7 18 23 3 9

A⎛ ⎞− + = +⎜ ⎟⎝ ⎠

19 193 9

A=

3A =

Comparing coefficients of 2x , 6 3A B= −3 3B− =

1B = −

Comparing the constant term : 8 = 2A + CC = 2

∴2

2 2

6 7 8 3 ( 2)( 2)(1 3 ) 1 3 2x x x

x x x x− + − +

≡ ++ − − +

M1M1M1M1

A1A1A1A1

M1M1M1M1

A1A1A1A1

4

6(b)6(b)6(b)6(b) 3 23 4 2x mx x+ − −

When x = –1,f(−1) = 3 2 3 23 4 2 3( 1) ( 1) 4( 1) 2x mx x m+ − − = − + − − − −

5




3 4 2m= − + + −1 m= − +

When x = 2,f(2) = 3 2 3 23 4 2 3(2) (2) 4(2) 2x mx x m+ − − = + − −

24 4 8 2m= + − −4 14m= +

f(–1) = 2f(2)1 2(4 14)m m− + = +1 8 28m m− + = +

29 7m− =297

m = −

B1B1B1B1

B1B1B1B1M1M1M1M1M1M1M1M1

A1A1A1A1

SectionSectionSectionSection BBBB [15 marks]NoNoNoNo Working/AnswerWorking/AnswerWorking/AnswerWorking/Answer PartialPartialPartialPartial marksmarksmarksmarks TotalTotalTotalTotal

marksmarksmarksmarks7 (i) AB b a= −

��

= (5iiii – 2jjjj +3kkkk) – (2iiii + 3jjjj – kkkk)= 3iiii – 5jjjj + 4kkkk

Length of 222 4)5(3AB +−+=

50== 7.07 units (3 sig. fig.)

(ii) AC c a= −��

= (4iiii + jjjj – 2kkkk) – (2iiii + 3jjjj – kkkk)= 2iiii – 2jjjj – kkkk

Length of 2 2 22 2 1AC = + +��

= 3

cos AB ACBACAB AC

⋅∠ =

��

(3i 5j 4k) (2i 2 j k)( 50)(3)

− + ⋅ − −=

6 10 415 2+ −

=

M1M1M1M1

M1M1M1M1

A1A1A1A1

B1B1B1B1

M1M1M1M1

M1M1M1M1

15




4 2 255 2

= =

56BAC∴∠ = � (nearest degree)

(iii) Area of 1 sin2

ABC AB AC BAC∆ = ∠��

21 2 2( 50)(3) 12 5

⎛ ⎞= − ⎜ ⎟⎜ ⎟

⎝ ⎠

1 175 2(3)2 5

=

3 342

=

= 8.75 (3 sig. fig.)

OROROROR

(iii) 3 5 42 2 1

AB AC× = −− −

i j ki j ki j ki j k��

=(5+8)iiii – (–3 – 8)jjjj + (–6 + 10)kkkk= 13iiii + 11jjjj + 4kkkk

Area of 12

ABC AB AC∆ = ×��

2 2 21 13 11 42

= + +

3062

=

= 8.75 (3 sig. fig.)

A vector equation of the line passing through A andB is given by

rrrr = (2iiii + 3jjjj – kkkk) + λ( AB��

)= (2iiii + 3jjjj – kkkk) + λ(3iiii – 5jjjj + 4kkkk)= (2 + 3λ)iiii + (3 – 5λ)jjjj + (–1 +4λ)kkkk

pppp = (2 + 3t)iiii + (3 – 5t)jjjj + (–1 + 4t)kkkkThis has the form given for the position vector of P.Therefore, for all values of t, P lies on the line

A1A1A1A1

M1M1M1M1

A1A1A1A1

M1M1M1M1

A1A1A1A1

M1M1M1M1

A1A1A1A1




through A and B.

ORpppp = (2 + 3t)iiii + (3 – 5t)jjjj + (–1 + 4t)kkkkpppp = 2iiii + 3jjjj – kkkk ++++ t(3iiii – 5jjjj + 4kkkk) ... (1)since pppp = aaaa + t AB orsince OA = 2iiii + 3jjjj – kkkk and AB = (3iiii – 5jjjj + 4kkkk),pppp satisfies the vector equation of the line passes throughA and B for all values of t.Therefore for all values of t, P lies on the line throughA and B.

For OP to be perpendicular to AB,0OP AB⋅ =

��

[(2 + 3t)iiii + (3 – 5t)jjjj + (–1 +4t)k]k]k]k] ⋅ [[[[3iiii – 5jjjj + 4k]k]k]k] = 06 + 9t – 15 + 25t – 4 + 16t = 0

–13 + 50t = 0t = 0.26

∴ pppp = (2 + 0.78)iiii + (3 – 1.3)jjjj + (–1 + 1.04)kkkk==== 2.78iiii + 1.7jjjj + 0.04kkkk

M1M1M1M1

A1A1A1A1M1M1M1M1A1A1A1A1

8.

Given3 12 , 21 3

OA OB⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

��

2 2AX AC CX OA OB= + = +��

)OBOA(2 += M1M1M1M1

A1A1A1A1

15

O

1D X

• •

••

•

• C

1

1 A 2

B




]321

123

[2⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

3 12 2 2

1 3

+⎛ ⎞⎜ ⎟= +⎜ ⎟⎜ ⎟+⎝ ⎠

42 4

4

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

18 1

1

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

1 is parallel to 1 .

1AX

⎛ ⎞⎜ ⎟⇒ ⎜ ⎟⎜ ⎟⎝ ⎠

��(Shown)

(i) Equation of AX :3 12 11 1

λ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

rrrr

2 3CD OD OC OB OA= − = −��

2 9 74 6 26 3 3

−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∴ Equation of CD :2 74 26 3

µ−⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟= + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

rrrr

(ii) At point of intersection,3 2 72 4 21 6 3

λ µλ µλ µ

+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

3 2 7λ µ+ = − ⇒ 7 1λ µ+ = − ……(1)2 4 2λ µ+ = − ⇒ 2 2λ µ+ = ……(2)1 6 3λ µ+ = + ⇒ 3 5λ µ− = ……(3)

(1) – (2) : 5 3µ = − ⇒35

µ = −

From (1), 3 161 75 5

λ ⎛ ⎞= − − − =⎜ ⎟⎝ ⎠

Check (3), LHS = 16 33 3 55 5

λ µ ⎛ ⎞− = − − = =⎜ ⎟⎝ ⎠

RHS

A1A1A1A1

B1B1B1B1

M1M1M1M1

A1A1A1A1

M1M1M1M1

M1M1M1M1




∴position vector of point of intersection is

315265215

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

(iii)1 3 2 12 2 0 2 03 1 2 1

AB OB OA− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = − = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

��

1 10 11 1

cos1 1 1 1 1

AB AXBAXAB AX

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⋅⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⋅ ⎝ ⎠ ⎝ ⎠∠ = =+ + +

��

1 1 02 3− +

= =

90BAX∴∠ = �

(iv) 9 6 32 4 6

OC OD× =i j ki j ki j ki j k��

6 3 9 3 9 64 6 2 6 2 4

= − +i j ki j ki j ki j k

= (36 – 12)iiii –––– (54 – 6)jjjj + (36 – 12)kkkk= 24iiii –––– 48jjjj + 24kkkk

area of the parallelogram OCXD= OC OD×��

2 2 224 ( 48) 24= + − +

3456== 57.79

A1A1A1A1

M1M1M1M1

M1M1M1M1

A1A1A1A1

M1M1M1M1

M1M1M1M1

A1A1A1A1

MathematicsT STPM Baharu PENANG 2012

Documents

Transcript of MathematicsT STPM Baharu PENANG 2012