STPM Maths T Sem 1 Trial King George Answer

1314-1

SMK TINGGI PORT DICKSON



= 2   



x



A1

M1





M1 A1



B1 B1 M1 A1

M1



M1 A1

B1 B1

MARKING SCHEME

M1

2  5 B1

A1 A1

D1 (correct vertex (-1,0) and y-intercept (0, 1)) D1(V shape and above x-axis)

1. a) f -1(x) = x – 1 g has no inverse because it is not a one-to-one function. gof b) g o f = │x + 1 │







c) │x + 1 │= │x│, solution set is { x: x ≥ 0}

2. a)



= 10 – 







= 2  = constant → Un is the nth term of a GP. b) d = T2 – T1 







 

 

r = 2  

!  ∑  

3. a) (i) Show L.H.S. = R.H.S. = 26 (ii) z1z2 = –13 + 13 i Arg (z1z2) = tan-1(13/-13) 

"

= π

2ab = –30

b) √16  30( = a + bi a2 – b2 = 16

(a2 +9)(a2 – 25 ) = 0 M1 a2 = -9 and a = ±5, b=)3 A1 (reject) A1 (both answers given) √16  30( = 5 – 3i , –5 +3i

1 2 3 1 2 3 4. a) *2 6 11,. 0 →→do ERO →*0 2 5, .  2- 0 M1 A1 0 0 0 / 1 2.  51 2 7 /  for equation to have solutions r + 2q – 5p = 0. A1

5





B1(for both)

, t) where t 2 3 . A1

and x =1 – 5t + 3t= 1 – 2t.



5

M1

, √1  7  ) B1 B1



M1

Equations in systems are independent and there are infinitely many solutions. B1B1 b) x + 2y – 3z = 1 2y – 5z = 0

Let z = t where t 2 3,  y =



i.e. solutions of system is (1 – 2t,



and Y = √1  7 





6

5. a) For graph , D1 (correct vertices) ,D1(correct shape)

6

b) Midpoint of PN is ( Let X =



x = 2X-2 then Y = 81  27  2 



4(X – 1)2 + 4Y2 = 1 i.e. equation of locus of midpoint of PN when P moves on the ellipse is 4(x – 1)2 + 4y2 = 1 A1

radius ½.

M1 A1A1

≤

 6 >

=

=

9: :;:6 <

B1

M1

: :

 6?

=

A  

+ … B1

A1

=

:: :=@:

]x3 +…

(x – 1)2 + y2 = ()2 i.e. locus is equation of a circle with centre (1,0) and

6. (1+x2)p= 1 + px2+ …

=

= 1 + qx + [q(q+1)+p]x2 + [

(1 – x )-q = 1 + qx + ½ (q2+q)x2 +

M1

Substitute above into

: :

b) Coefficient of x3 is 0≤

Largest possible coefficient of x3 is 162. A1

Section B 7. 3 – 2p +7 = 0 M1 p=5 A1

(-1)3 + p(-1)2 + 7(-1)+ q = - 16 M1 q = –13 A1

a) f(1) = 0 → (x – 1 ) is a factor of f(x) B1 f(x) = (x – 1 )(x2+6x+13) =0 M1 A1 For x2+6x+13 =0, b2 - 4ac = –16 < 0 → it has no real roots. M1  f(x) = 0 has only one real rood, i.e., x = 1. A1 When f(x)>0, (x – 1 )(x2+6x+13) >0, M1 (get answer from sketching graph or number line) set of x is {x :x>1, x23} A1

6B

C6





6 



M1 A1

B1

6 =6 

6

A1

M1 b) x + 9 = A(x2 + 6x+13) + (Bx+C)(x – 1 ) Solve…. A= ½ , B= ½ and C = – 5/2 M1 A1 

M1

M1

√

DE

8. a) a + b+ c = 3i + 6k

A1

8"

 * 0.* 0 = √√" B

unit vector is

b) cos θ = = θ = 39.2o

c) a x b = (3 -2)i – (3 – 1)j + (2 – 1)k M1 = i – 2j + k A1 d) BD = BA + BC M1 OD = OA +OC – OB = i – 4j A1 e ) BA = -j – 2k and BC = -5j – k Area of parallelogram ABCD = │BA X BC│ = │- 9i │ = 9 unit2