7/28/2019 C4 PF`s and Binomial Test A

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C4 Test: Partial Fractions and Binomial. Test A. Time allowed: 25 minutes

1. Given that

,131)1)(31(

53

 x

 B

 x

 A

 x x

 x

−+

+≡

−++

(a) find the values of the constants A and B.(3)

 

(b) Hence, or otherwise, find the series expansion in ascending powers of  x, up to and

including the term in x2, of 

)1)(31(

53

 x x

 x

−++

. (5)

 

(c) State, with a reason, whether your series expansion in part (b) is valid for  x = 21

.(2)

(Total 10 marks)

 

2.

f( x) =22

)1)(31(

31

 x x

 x

−+

−

,  x ≠ 1.

(a) Find the constants A, B, C and D such that

f( x) ≡ 22 )1(131 x

 D

 x

C 

 x

 B Ax

−

+

−

+

+

+

.

(5)

 

(b) Find a series expansion for f( x) in ascending powers of  x, up to and including the

term in x4.

(4)

(c) Find an equation of the tangent to the curve with equation y = f( x) at the point

where x = 0.(2)

(Total 11 marks)

Churston Ferrers Grammar School 1

7/28/2019 C4 PF`s and Binomial Test A

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Solutions: C4 Partial Fractions and Binomial Test A

1. (a) 3 + 5 x ≡  A(1 –  x) + B(1 + 3 x) Method for  A or  B M1

( x = 1) ⇒ 8 = 4 B   B = 2 A1

( x = –  31

) ⇒  34

= 34

 A   A = 1 A1 3

 

(b) 2(1 –   x) –1

= 2[1 + x + x2

+ … ] M1 [A1]

Use of binomial with n = –1 scores M1(×2)

(1 + 3 x) –1

= [1 – 3 x + !2

)2)(1( −−

(3 x)2

+ …] M1 [A1]

∴)31)(1(

53

 x x

 x

+−+

= 2 + 2 x + 2 x2

+ 1 – 3 x + 9 x2

= 3 –  x + 11 x 2

A1 5

 

(c) (1 + 3 x) –1

requires | x| < 31

, so expansion is not valid. M1, A1 2[10]

 

2.  N.B. Coeff n  x

2= 0 = –2 A + B + 3C + 3 D

(a) 1 – 3 x ≡ ( Ax + B)(1 –  x)2

+ C (1 + 3 x2)(1 –  x) + D(1 + 3 x

2) M1

extra (1 – x) is M0

 x = 1 ⇒ –2 ≡ 4 D ∴ D = – ½  B1

 x = 0 ⇒ 1 = B + C + D or  B + C = 2

3

1

coeff  x3: 0 = A – 3c or  A = 3c 2 M1

coeff  x: –3 = A – 2 B – C  3

condone missing term or sign error for M1

 sufficient suitable equations

extra (1 – x) can get this mark 

2 into 3 ⇒  2

3−

= C –  B, solving with 1 ⇒ C = 0, A = 0; B = 23

A1;A1 5

 

(b) f( x) ≡  2

3

(1 + 3 x2) –1

– 

++−−++ 432

......!2

)3)(2(21

2

1 x x x x

M1 A1

Use of B in (both [ ])

=   

   ++++− 

  

   +− ...

2

52

2

3

2

1...

2

27

2

9

2

3 43242 x x x x x x

M1

collecting terms up to x4

= 1 –  x – 6 x 2

– 2 x 3

+ 11 x 4… A1 cao 4

Churston Ferrers Grammar School 2  

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 Alternative (b)

Maclaurin: 1 –  x – 6 x2

M1, A1

 –2 x3

+ 11 x4

M1, A1

 

(c) f  ′( x) = –1 – 12 x…

∴f(0) = 1, f 

′(0) = 1 M1

 f(0), f ′    (0) (both)

∴equation of tangent is y = 1 –  x A1 ft 2

ft their expression only[11]

Churston Ferrers Grammar School 3