Maths Sa.msa Revision Package Solutions

St. Andrew’s Junior College H2 Mathematics/JC1/2009 MSA Revision Package Solutions Partial Fractions 1.

3x + 1 3x + 1 = 2 2 x − x − 1 (2 x + 1)( x − 1) A B + = 2x +1 x −1 1 3(− ) + 1 3(1) + 1 4 1 2 = = , B= Using Cover-Up Rule, A = 1 2(1) + 1 3 3 − −1 2 . 3x + 1 1 1 4  ∴ 2 =  +  2 x − x − 1 3  2 x + 1 x −1 

2.

3.

2 + 5 x + 15 x 2 A Bx + C = + 2 (2 − x )(1 + 2 x ) 2 − x 1 + 2 x2 2 + 5(2) + 15(2) 2 Using Cover-Up Rule, A = = 8. 1 + 2(2) 2 2 + 5 x + 15 x 2 = A(1 + 2 x2 ) + ( Bx + C )(2 − x ) x = 0 : 2 = A + 2C ⇒ C = −3 x : 5 = 2B − C ⇒ B = 1 2 + 5 x + 15 x 2 8 x−3 ∴ = + 2 (2 − x)(1 + 2 x ) 2 − x 1 + 2 x2 2 x2 − x + 1 A B C = + 2+ 2 x (1 − x) x x 1− x 2(1) 2 − 1 + 1 =2 12 2 x 2 − x + 1 = Ax + B (1 − x) + Cx2

Using Cover-Up Rule, C = x 0 :1 = B x : −1 = A − B ⇒ A = 0 2 x2 − x + 1 1 2 ∴ 2 = 2+ x (1 − x) x 1− x 4.

3x 2 + 23x + 45 14 x + 45 = 3+ x( x + 3) x ( x + 3) 14 x + 45 A B = + x( x + 3) x x + 3 45 14(−3) + 45 = 15, B = = −1 3 −3 3 x 2 + 23 x + 45 15 1 ∴ = 3+ − . x ( x + 3) x x+3 Using Cover-Up Rule, A =

5. 2 x 3 − x − 16 x = 2− 3 2 x −8 ( x − 2)( x + 2 x + 4) x A Bx + C = + 2 2 ( x − 2)( x + 2 x + 4) x − 2 x + 2 x + 4 2 1 A= 2 = 2 + 2(2) + 4 6 x = A( x 2 + 2 x + 4) + ( Bx + C )( x − 2) 1 x = 0 : 0 = 4 A − 2C ⇒ C = 2 A = Using Cover-Up Rule, 3 1 x2 : 0 = A + B ⇒ B = − A = − 6 3 2 x − x − 16 1 1 2− x  ∴ = 2−  + 2  3 x −8 6  x − 2 x + 2x + 4  6. 3x3 + 1 Ax + B Cx + D = 2 + 2 2 ( x + 1) x + 1 ( x2 + 1)2 3 x 3 + 1 = ( Ax + B)( x2 + 1) + Cx + D x3 : 3 = A x2 : 0 = B x : 0 = A + C ⇒ C = −3 x 0 :1 = B + D ⇒ D = 1 ∴

3x3 + 1 3x 1 − 3x = 2 + 2 2 2 ( x + 1) x + 1 ( x + 1)2

7. x3 + 3x + 1 15 x + 17 = x−4+ 2 ( x + 2) ( x + 2)2 15 x + 17 A B = + 2 ( x + 2) x + 2 ( x + 2)2 15 x + 17 = A( x + 2) + B x = −2 :15(−2) + 17 = B ⇒ B = −13 x = 0 :17 = 2 A + B ⇒ A = 15 ∴

x 3 + 3x + 1 15 13 = x−4+ − 2 ( x + 2) x + 2 ( x + 2)2

2 x 3 + 5 x 2( x3 + 3 x + 1) − x − 2 = Since ( x + 2) 2 ( x + 2)2  x3 + 3x + 1  x+2 2 − 2  =  ( x + 2)  ( x + 2)2  15 13  1 − − = 2 x − 4 + 2  x + 2 ( x + 2)  x + 2  29 26 − = 2 ( x − 4) + x + 2 ( x + 2) 2

.

.

Trigonometry θ is obtuse, so cos θ and tan θ are both negative. cos 2 θ + sin 2 θ = 1 cos 2 θ + x 2 = 1 cos 2 θ = 1 − x 2 cos θ = − 1 − x 2

tan θ =

.

sin 2θ = 2 sin θ cos θ

.

7 sin( A − B ) = 5 sin( A + B )

= −2 x 1 − x 2

7 sin A cos B −7 sin B cos A

cos 2θ = cos θ − sin θ 2

.

2

=1 − x − x 2

=5 sin A cos B +5 sin B cos A

2

=1 − 2x 2

2 sin A cos B =12 sin B cos A tan A = 6 tan B

RHS =

. =

1 + cos x csc x + cot x + tan x

= sin x cos x = LHS

RHS =

=

RHS =

cos x + cos 2 x + sin 2 x

cos x + 1

=

h=6

( sin x cos x )(1 + cos x )

( sin x cos x )(1 + cos x ) =

.

.

sin θ x =− cos θ 1− x2

tan x − cot x sec x − csc x

sin 2 x − cos 2 x sin x − cos x

( sin x − cos x )( sin x + cos x ) sin x − cos x

x x  x  cos −  2 cos 2 − 1 − 1 2 2  2  = x x x   1 − 1 − 2 sin 2  − 2 sin cos 2 2 2  2 sin

x x cos − 2 cos 2 2 2 = x 2 x 2 sin − 2 sin cos 2 2 2 sin

x 2 x 2

x x x  sin − cos  2 2 2 = x x x 2 sin  sin − cos  2 2 2 2 cos

= sin x +cos x

3 solutions

sin x − cos x −1 1 − cos x − sin x

x 2 = x sin 2 cos

=

2 cos 2 2 sin

= =

x 2

x x cos 2 2 x −1 +1 2 sin x

2 cos 2

2 sin A cos A cos 2 A + sin 2 A

=

sin 2 A 1

= sin 2 A = LHS

cos x +1 = LHS sin x

2 tan A 1 + tan 2 A

.

RHS =

.

LHS = cos 4 A = 1 − 2 sin

.

=

= 1 − 2( sin 2 A)

2

2A

2

= 1 − 2( 2 sin A cos A)

2

= 1 − 8 sin 2 A cos 2 A = 1 − 8 sin 2 A(1 − sin 2 A) = 8 sin 4 A − 8 sin 2 A +1 = RHS

LHS = sin 3 x + sin 6 x − sin 9 x = sin 3x + 2 sin 3 x cos 3 x − sin 3 x cos 6 x − sin 6 x cos 3 x

(

)

= sin 3 x + 2 sin 3 x cos 3 x − sin 3 x 2 cos 2 3 x −1 − 2 sin 3 x cos 2 3 x

(

= sin 3 x 1 + 2 cos 3 x − 2 cos 2 3 x +1 − 2 cos 2 3 x

(

= sin 3 x 2 + 2 cos 3 x − 4 cos 2 3 x

(

= 2 sin 3 x 1 + cos 3 x − 2 cos 2 3 x

)

)

= 2 sin 3 x(1 − cos 3 x )(1 + 2 cos 3 x ) = RHS

sin 3 x + sin 6 x = sin 9 x sin 3 x + sin 6 x − sin 9 x = 0 2 sin 3 x(1 − cos 3 x )(1 + 2 cos 3 x ) = 0

)

0.

sin 3 x = 0  π 2π  x = 0, , ,π   3 3 

1 − cos 3 x = 0  2π  x = 0,   3 

or

or

1 + 2 cos 3 x = 0  2π 4π 8π  x= , ,  9 9 9 

 2π π 4π 2π 8π  , , , , ,π   9 3 9 3 9 

So the solution set is x = 0,

a. PS = 2 sin θ + cos θ b. 2 sin θ + cos θ = R cos (θ − α ) where R is positive and α is acute 2 sin θ + cos θ = R cos θ cos α + R sin θ sin α

Comparing the two terms, we see that R cos α = 1 and

R sin α = 2

So, R 2 cos 2 α + R 2 sin 2 α = 12 + 2 2 R 2 ( cos 2 α + sin 2 α ) = 5

R2 = 5 R= 5

c. The maximum value of cosine is 1, hence, the maximum value for This maximum value is at θ − 63 .4° = 0 , so θ = 63 .4° d.

2.15 = 5 cos (θ −63 .4°)

θ = cos −1

2.15 + 63 .4° ≈ 79 .4° 5

5 cos (θ − 63 .4°) = 5

Number Systems, Surds & Indices Solutions 1)

  

(

=

 2 + 2 − 2   2 + 2− 2

(

= 2− 2− 2

)(

 2 − 2− 2   2 − 2− 2

)

)

=42 2

(

2

) ( ) = 10 + 51 − 2 ( 10 + 51 ) ( 10 − = 20 − 2 ( 10 + 51 ) ( 10 − 51 )  10 + 51 − 

10 − 51  

= 20 − 2 (100 − 51) 3

4

=6 1 20 1 = 5 5 +5 7 −2 7 + 2 5 1 = 5 5 +3 7 + 5 10 51 = 5 +3 7 10 2 5+ 2 2 5+ 2 5+ 2 = g 5− 2 5− 2 5+ 2 125 + 175 − 28 +

10 + 10 + 2 10 + 2 5−2 12 + 3 10 = 3 = 4 + 10 ∴ a = 4, b = 10 2 2 log x +4)4) log(2(2xx++1)1) 44== 22 log x+ −−log x +(2 1) (2 (2 x(2+1) =

5(i)

log (2 x +1)

(2 x + 4) 2 =2 4

(2 x + 4) 2 = (2 x + 1) 2 4 2 4 x + 16 x + 16 = 4 x2 + 4x + 1 4 3x 2 − 3 = 0 3( x + 1)( x − 1) = 0 x = 1 or -1

)

51 + 10 − 51

Check: When x = -1 , the base of the logarithmic becomes negative, therefore we reject this solution. ∴ Solution set = { x ∈ ¡ : x = 1} 5(ii)

5(8e 2 x − 3)3 = 625 (8e 2 x − 3)3 = 125 8e 2 x − 3 = 5 e2 x = 1 2 x = ln1 x=0 ∴ Solution set = { x ∈ ¡ : x = 0}