Mathematics M (A)

Marking Scheme Trial Semester 1 STPM 2022

No. Marks

1a. ( ) = 4 + 3 3 + 2 2 + 6 + M1

1 M1
(2) = 0 M1
14 13 12 1 A1
(2) + 3 (2) + 2 (2) + 6 (2) + = 0 B1
−31 M1
16 + = 8 … … … . . ( ) M1
M1
(1) = 9 B1
A1
+ 3 + 2 + 6 + = 9
B1
+ = −2 … … … ( ) M1

(ii) – (i), 15 = 15 A1
16 8 M1
M1
= 2 A1
B1
= −2 − 2 = −4 M1
A1
b. ( ) = 2 4 + 3 3 + 2 2 + 6 − 4 B1
M1
2 4 + 3 3 + 2 2 + 6 − 4 < 0 M1
A1
(2 − 1)( 3 + 2 2 + 2 + 4) < 0
M1
(2 − 1)( + 2)( 2 + 2) < 0 A1

Since ( 2 + 2) >0, then (2 − 1)( + 2)< 0

∴ The solution set is { : −2 < < 1}

2

2a. = 1 − 3−2
−1 = 1 − 3−2( −1)
= − −1
= 1 − 3−2 − (1 − 3−2( −1))

= 3−2 +1 − 3−2

= 3−2 (32 − 1)

= 8(3−2 )

b. 1
+1 = 9
+1 1
= 9 ( a constant independent of n)

Thus, this is a geometric sequence.

c. 8
9
∞ = − 1 = 1
9
1

3a. 3 − 4 −3

| 1 4 − −2 | = 0

2 10 −5 − |21 |
(3 − ) |41−0 −2 −2 |21 4−
−5 − | − 4 −5 − | − 3 10 = 0

− 3 + 2 2 + − 2 = 0

b. 3 − 2 2 − + 2 = 0
( − 1)( 2 − − 2) = 0
( − 1)( − 2)( + 1) = 0

∴ = 1, = 2, = −1

4. 2 + 2 − 2 2 = 4

Differentiate w.r.t x,

(2 + 2 + 2 − 2 2 + (−4 ) = 0 M1
) M1

2 ( − ) + 2 + 2 − 4 = 0 M1
A1
When = 1, 2 − 2 − 3 = 0 A1

( + 1)( − 3) = 0 B1
M1
= −1 or = 3
M1
(1, −1) and (1,3) A1
M1
At (1, −1), 2(1) (−1 − 1) + (−1)2 + 2(1) − 4(1)(−1) = 0 M1

= 7 A1
4 M1
2(1) (3 − 1) + (3)2 + 2(1) − 4(1)(3) = 0
At (1,3), M1
A1
= 1
4

5. = ln

1
= ⇒ =
When x = 1, = 0

When = , = 1

∫ 1 = ∫ 11

1 (ln + 1)(ln + 4) 0 ( + 1)( + 4)

Let 1 ≡ A + B
( +1)( +4) +1 +4

1 ≡ A( + 4) + B( + 1)

When = −4, B = −1
3
1
When = −1, A = 3

1 11
∴ ( + 1)( + 4) ≡ 3( + 1) − 3( + 4)

11 11 1 1
∫ = ∫ ( + 1 − + 4)
( + 1)( + 4) 3
0 0
4)]01
= 1 [ln( + 1) − ln( +
3

= 1 [ln 2 − ln 5 − (ln 1 − ln 4)]
3
= 1 ln 2×4
35
= 1 ln 8
35

6. = + 2

= 1 + 2 …….(1)



+ 2 + 1
= 3 − 2 − 4

= +1 ……(2) ( +1 )

3−2 3−2

Sub. (2) into (1), = 1 + 2

= 3−2 + 2 +2
3−2 3−2
=5
3−2

∫(3 − 2 ) = ∫ 5 M1
M1
3 − 2 = 5 + M1A1
A1
(3 − ) = 5 +
B1
( + 2 )(3 − − 2 ) = 5 + M1
M1
When x = 2, y = 1, 4(−1) = 10 +
M1
= −14

( + 2 )(3 − − 2 ) = 5 − 14

∴ 2 + 2 + 4 2 + 4 − 6 + 14 = 0

7a. 2 − 1 2 2

Given A=(2 − 1 )

4 − 4 + 2 − 1

A is a symmetric matrix.

2 = 2 − 1….(i)

2 = 4 − 4 ….(ii)

= + ….(iii)

From (i), 2 − 2 + 1 = 0

( − 1)( − 1) = 0

= 1

2 = 4 − 4

2 − 4 + 4 = 0

( − 2)( − 2) = 0

= 2

+ =

When = 2, 2 + = 2 ⇒ = 2

b. 3 1 4

A=(1 1 4)

443
3 1 41 0 0 1 1 40 1 0
(1 1 4|0 1 0) → 2⟷ 1 (3 1 4|1 0 0)

4 4 30 0 1 4 4 30 0 1
3 1− 2⟶ 2
1 1 40 1 0
→4 1− 3⟶ 3 (0 2 8 |−1 3 0)

0 0 13 0 4 −1
1 1443|−2001 10
1 1
2→ 2→ 2 (0 1 3 0)
0 2
0 4 −1

1 −1 0
1 0 02 2
−→ 2+ 1→ 1 0 1 4 |−1
3 0
(0 0 13 2 2
0 4 −1)

1 1 0 1 −1 0
0 1 041||−221 2
→13 3→ 3 0 0 3 0
0 2 −1
( 4 13 )
13

1 0 1 − 1 0
0 1 001||−212 2
→−4 3+ 2→ 2 0 7 4
0 A1
0 26 13 A1

( 4 − 1 B1
13) M1M1
11 13
A1
2 −2 0
1 7 4 A1
−1 = −2 26 13 D1D1

41
(0 − 13)
3 14 13 3

(1 1 4) ( ) = (11)

4 4 3 8
11
−2 0
2 7
1 4 3
( ) = −2 26 (11)
4 13
1 8

( 0 13 − 13)
−4

51

( ) = 13

36

( 13 )
51 36

∴ = −4, = 13 , = 13

8a. x= -3 (3, ln 6)

ln3

-2 0.5

b. (3, ln 6)

c. (0, ln 3)

d. (3 ln 2 − 21)unit2

e. (8 7 − 9 ln 2) unit3
8