# Stpm 2012 Kedah Maths T Trial

**Topics:**Random variable, Probability density function, Cumulative distribution function

**Pages:**66 (3913 words)

**Published:**October 16, 2012

STPM KEDAH 2012-MT PAPER 1

1.

ln 3

Given that log2 P = x, log3 P = y and x + y = 1, show that x = ln 6 .

n

2.

CHU/SMKK

Prove that

7.

The complex numbers z and w are given by z = 3 + 2i and w = –5 + 4i. (a)

[4 marks]

Find |w| in surd form and arg w in radians correct to three significant figures. [3 marks]

(b)

z

Express w in the form a + ib, where a and b are exact fractions.

(c)

n +1

∑ 4 r 2 − 1 = − 2n + 1 .

r =0

1

In an Argand diagram, the points Z and W represent the complex numbers z

[4 marks]

[2 marks]

and w respectively, whereas the point Z* represents z*, the conjugate of z. The point P is 3.

Show, by means of the substitution x = tan θ, that

1

∫ (1 + x )

0

such that ZWZ*P (in that order) is a parallelogram. Find the complex number p represented by point P.

π

1

22

dx =

∫

4

[3 marks]

2

cos θ dθ .

0

1

Hence, find the exact value of

∫ (1 + x ) dx .

1

[6 marks]

4.

22

8.

If y = 52 x , by taking logarithm to the base e for both sides of the equation, show

that

0

dy

= 52 x ln 25 .

dx

[2 marks]

1

In the same diagram, sketch the graph y = x − 2 , x ∈ ℜ and the graph

Hence, determine the integers m and n for which

y = x, x ≥ 0 .

∫ x5

2x

(ln 25)2 dx = m ln 5 − n

.

0

[7 marks]

Hence or otherwise solve the inequality x − 2 < x .

[6 marks]

Function f and g are defined by

9.

5.

Given that y =

cos (ax)

x3

x2

, where a is a constant and x ≠ 0 , show that

f : x → a( x + 3) − b for x ≤ −3 , where a and b are constants. 2

d2y

dx 2

+ 6x

dy

dx

+ (6 + a 2 x 2 )y = 0 .

[6 marks]

g:x→

x+5

2

(a)

6.

State the domain and range of g .

[2 marks]

(b)

Find the values of a and b if g o f ( x) = −( x + 3) , x ≤ −3 .

[3 marks]

A straight line l1 with gradient m, passes through the point (4, 0) is parallel to the line

l 2 , which passes through the point (-4, 0). l1 and l2 meet the line 4 x + 5 y = 25 at point F

(c) Sketch the graph of f and explain why the inverse function of f exists, hence

and at point G respectively.

find f

(a)

Find the coordinates of point F and point G , in terms of m.

(b)

If the distance of FG is 8 units, find the possible values of m .

−1

( x) .

[5 marks]

[5 marks]

[3 marks]

http://edu.joshuatly.com/

3

10.

1

Matrix P is given by P = 1

2

4

If P 2 + aP + bI = − 1

− 2

− 1

1 .

3

0

2

2

2

− 2 , where I is the 3 × 3 identity matrix, find

2

−2

5

−2

the values of a and b .

[4 marks]

Find P( P 2 + aP + bI ) .

[1 mark]

Hence, solve the simultaneous equations

x−z =3

x + 2y + z = 2

2 x + 2 y + 3z = 5

11.

[5 marks]

Find the value of a and of b if f ( x ) = x 4 + ax 3 + 13 x 2 − 12 x + b is exactly divisible

by g ( x) = x 2 − 3x + 2 .

[2 marks]

Hence,

(a)

find the solution of f ( x ) = 0 .

(b)

find the set of values of

(c)

using the substitution y =

4

3

which satisfy f ( x ) ≤ −3g ( x ) .

[5 marks]

1

, solve the equation

x

2

4 y − 12 y + 13 y − 6 y + 1 = 0

12.

[3 marks]

[3 marks]

5

Given that f(x) = 2x2 + ln(4x + 5), has domain {x : x ∈ R, – 4 < x ≤ 1}. (a)

State the asymptote of f.

[1 mark]

(b)

Find all (local) maximum and minimum points of f.

[7 marks]

(c)

Find the coordinate of the point of inflexion.

[3 marks]

(d)

Sketch the graph of f.

[4 marks]

(e)

State the maximum value of f for the given domain.

[1 mark]

http://edu.joshuatly.com/

2

STPM KEDAH 2012-MT PAPER 1-MARKING SCHEME

2x = 3 y

2 x = 31 – x

1.

4.

B1

D1: y = x

x ln 2 = (1 – x)ln 3

M1A1

D1 : shape of y = x − 2

x(ln 2 + ln 3) = ln 3

ln 3

x = ln 6 .

D1 : points (2, 0) , (0, 2)

A1

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