Mathematics T Penggal 1 STPM 2014 ANSWERS.pdf

CONFIDENTIAL*

PEPERIKSAAN PERCUBAAN PEPERIKSAAN PERCUBAAN PEPERIKSAAN PERCUBAAN


















Skema Pemarkahan Mathematics T Paper 1 (954/1)

PEPERIKSAAN PERCUBAAN PENGGAL 1

SIJIL TINGGI PERSEKOLAHAN MALAYSIA 2014

954/1 STPM 2014

One and a half hours

MATHEMATICS T

PAPER 1

CONFIDENTIAL*

2

1. The functions f and g are defined by :

f : x x + 4, x bxa , ,

g : x x2 – 4x, x , 4x

(a) By finding the values of )1(g and )3(g , explain why the inverse function g-1

(x) is not

defined [2 marks]

(b) Given that the function )(xfg is defined, find the minimum value of a and the

maximum value of b. [3 marks]

1.(a) 3)1( g , 3)3( g

3)3()1( gg

g is not one-to-one

function )(1 xg not defined

M1

A1

Get )1(g and )3(g and

conclude

1.(b) For function )(xfg defined

gf DR

gDxf )( fDx

]4,4[)( xf

4)(4 xf

444 x

08 x

But bxa

minimum a = –8 , maximum b = 0

M1

A1

A1

2.(a) Evaluate

100

1

133r

rr , give your answer in the standard form nA 10 . [3 marks]

(b) Express xx 31)2(

4

in the form of increasing power of x including the term of x

3.

Determine the range of values of x so that this expansion is valid. [6 marks]

2(a)

M1

A1

A1

(b)

2

1

312

1231)2(

41

x

x

xx

=

....)

2(

!3

)3)(2)(1()

2(

!2

)2)(1()

2)(1(12 32 xxx

...)3(

!3

))()(()3(

!2

))(()3)(

2

1(1 32

523

21

223

21

xxx

M1

M1

See

2

1

1

2)31()1(

xk x

Either one expanded

CONFIDENTIAL*

3

=

...

8

1

4

1

2

112 32 xxx

...

16

135

8

27

2

31 32 xxx

= ...)78

231(2 32 xxx

= ...144

2322 32 xxx

A1

A1

correctly

Either one series

correct

The expansion is valid when

12

x and 13 x

2x and 3

1x

3

1x or

3

1

3

1 x

M1

A1

3. Matrix P is invertible if 0P where P is the determinant of P.

The matrix

452

301

143

A has an inverse 1A because 10

452

301

143

.

(a) Find 1A by using the method of elementary row operations. [3 marks]

(b) Solve the following system of linear equations by method of matrices that involves A

and 1A .

23

3452

143

zx

zyx

zyx

[4 marks]

(c) State the value of

(i)

454

302

146

(ii)

301

452

143

[2 marks]

3(a)

IA =

100

010

001

452

301

143

=

100

001

010

452

143

301

21 RR

=

120

031

010

1050

1040

301,

M1

M1

A1

Idea from IA to

get 1AI

See two ERO carried out

correctly 2213 RRR

3312 RRR

CONFIDENTIAL*

4

= 1AI

A1

3.(b)

B1

M1

A1

A1

3.(c) (i) –20

(ii) 10

B1

B1

4.(a) Given p(1 + 5i) – 2q = 3 + 7i, find the values of p and q if p and q are both real numbers.

[3 marks]

4.(b) Express the complex number i3 in the form )sin(cos ir , where r is the modulus and

is the argument of the complex number.

Hence, simplify 5( 3 ) .i [5 marks]

4.(a) p – 2q = 3 and 5p = 7 M1

7

5p A1

4

5q A1

4.(b)

, M1

Either one correct

CONFIDENTIAL*

5

r = 2 ,

6

M1

Both correct

3 2 cos sin

6 6i i

A1

5 532 cos sin

6 6i

M1

See )sin(cos6

56

5 ik

i16316 A1

5. Show that the two curves 4x2 + 9y

2 = 36 and 4x

2 – y

2 = 4 have the same foci.

For the hyperbola, state the equations of the asymptotes. [6 marks]

Sketch the curves 4x2 + 9y

2 = 36 and 4x

2 – y

2 = 4 on the same axes, showing clearly the

asymptotes of the hyperbola. [4 marks]

5 4x

2 + 9y

2 = 36 or 4x

2 – y

2 = 4

2 2

369 4

x y or

2 2

11 4

x y

M1

c2 = 9 - 4 c

2 = 1 + 4 M1 either one correct

= 5 = 5 A1 both correct

Centre is (0, 0) Foci is ( 5 , 0) and ( 5 , 0)

Centre is (0, 0) Foci is ( 5 , 0) and ( 5 , 0) A1

The asymptotes are y = 2x and y = - 2x B1

B1

D1

D1

D1

D1

Shape of ellipse

Vertices and foci

shown

Shape of hyperbola

Vertices and

asymptotes shown

6. The points A, B and C have position vectors kjia 2 , kjib 423 and

kjic 44 respectively.

Find : (a) ba [1 mark]

(b) )()( acab [3 marks]

6.(a) ba = 13 B1

6.(b) kjiab 22 ,

kjiac 632

B1

B1

● (-1, 0) x

y

(1, 0) ● ● ● ●

●

●

(3, 0) 5,0

(-3, 0)

(0, 2)

(0, - 2)

● 5,0

CONFIDENTIAL*

6

)()( acab =

632

212

kji

= kji 8812

M1

A1

Determinant shown or

any two components in

answer correct

7.(a) The polynomial Q(x) is defined by Q(x) = x³ + mx² + 5x – n .

If Q(x) has a quadratic factor (x² + 5) and a zero of –2 , find the values of m and n.

[3 marks]

7.(b) Express xx sin2cos in the form )sin( xR where R is positive and 2

0

.

Hence, state the maximum value of xx sin2cos as well as the corresponding value of x in

the range 2

0

x correct to two decimal places. [5 marks]

7.(c) Prove that 2cot2tancot . [3 marks]

Hence or otherwise, solve the equation 12tancot for 3600 . [4 marks]

7.(a) x³ + mx² + 5x – n = (x² + 5)(x + 2)

= x³ + 2x² + 5x + 10

By comparison,

m = 2 , n = –10

M1

M1

A1

7.(b) Let R sin(x + α) cos x + 2 sin x

R sin x cos α + R cos x sin α cos x + 2 sin x

R sin α = 1 -------- (1)

R cos α = 2 -------- (2)

(1)2 + (2)

2, R

2 = 1

2 + 2

2

R = 5 , R > 0

M1

Find R or α

,

)2(

)1( tan α =

21

α = 0.4636 0 < α < 2

A1

R or α correct

cos x + 2 sin x 5 sin(x + 0.4636) A1 CAO

Since the maximum value of sin(x + 0.4636) = 1

The maximum value of cos x + 2 sin x = 5

and it occurs when x + 0.4636 = 2

x = 1.1072

1.11

B1

B1

7.(c) cot θ – tan θ =

cos

sin

sin

cos

=

cossin

sincos 22

=

2sin

2cos

21

= 2 cot 2θ

M1

M1

A1

Using cot θ and tan θ

Using sin2θ

CONFIDENTIAL*

7

cot θ – tan θ = 12

2 cot 2θ = 12

2 cot 2θ = 32

cot 2θ = 3

tan 2θ = 3

1

M1

A1

See equation in one

variable term

For 3600 ,

72020

2θ = ,30 ,210 390 , 570

θ = 15 , 105 , 195 , 285

A1

A1

8. The line l has equation r =

0

1

2

5

0

5

, .

(a) Show that l lies in the plane whose equation is r . 5

0

2

1

. [3 marks]

(b) Find the position vector of A, the foot of the perpendicular from the origin O

to l. [4 marks]

(c) Find an equation of the plane containing O and l. [4 marks]

(d) Find the position vector of the point P where l meets the plane whose equation

is r . .11

2

2

1

[4 marks]

8(a) r = + λ and r • = -5

For any point R on l with position vector r

r =

• = -5 – 2λ + 2λ

= -5

r satisfies vector equation of plane,

M1

A1

Get r and try checking if

r satisfies equation of

plane

R is a point on the plane

∴ line l lies on the plane. A1

(b) = + λ1 , λ1 ∈ ℝ

• = 0

M1

Use idea of perpendicular

CONFIDENTIAL*

8

• = 0

M1

Scalar product

10 + 4 λ1 + λ1 = 0

λ1 = -2

A1

= - 2

=

A1

(c) n = ×

=

M1

Find normal vector

=

A1

∴ equation of plane is r • = •

r • = 0

or – 5x + 10y + 5z = 0

M1

A1

Accept r • = 0

(d) Since P is on line l,

= + λ2 , λ2 ∈ ℝ

Given that r . .11

2

2

1

. .11

2

2

1

M1

Get position vector of

point and substitute to

equation of plane

5 + 2 λ2 + 2 λ2 + 10 = 11 M1 Scalar product to get λ2

λ2 = - 1 A1

= - 1

=

A1

Mathematics T Penggal 1 STPM 2014 ANSWERS.pdf

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