C4-partials-binom - Binomial Expansion PDF

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Binomial Expansion...

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C4: QUESTIONS FROM PAST EXAM PAPERS – PARTIAL FRACTIONS AND BINOMIAL EXPANSION 2 x 2  5x  10 B C A   x  1 x  2 x 1 x  2

1. (a)

Find the values of the constants A, B and C. (4)

(b)

2 x2  5 x  10 in ascending powers of x, as far as the term in x  1x  2  x2. Give each coefficient as a simplified fraction.

Hence, or otherwise, expand

(7) (Total 11 marks)

f (x ) 

2.

27x 2  32x  16 2 ,x 2 3 (3x  2) (1  x)

Given that f(x) can be expressed in the form

f (x )  (a)

A B C ,   2 (3 x  2) (3x  2) (1  x)

find the values of B and C and show that A = 0. (4)

(b)

Hence, or otherwise, find the series expansion of f(x), in ascending powers of x, up to and including the term in x2. Simplify each term. (6)

(c)

Find the percentage error made in using the series expansion in part (b) to estimate the value of f (0.2). Give your answer to 2 significant figures. (4) (Total 14 marks)

f ( x) 

3.

Given that, for x  12 , (a)

3 x 1

x  21 .

(1  2 x) 2 3 x 1 (1  2 x)

2



A B  , where A and B are constants, (1 2x ) (1  2x )2

find the values of A and B. (3)

(b)

Hence, or otherwise, find the series expansion of f(x), in ascending powers of x, up to and including the term in x3, simplifying each term. (6) (Total 9 marks)

City of London Academy

1

4.

(a)

Find the binomial expansion of

(1  8 x) ,

1 x , 8

in ascending powers of x up to and including the term in x3, simplifying each term. (4)

(b)

Show that, when x =

1 , the exact value of √(1 – 8x) is 100

23 . 5 (2)

(c)

1 into the binomial expansion in part (a) and hence obtain an 100 approximation to √23. Give your answer to 5 decimal places.

Substitute x 

(3) (Total 9 marks)

1

f(x) =

5.

(4  x )

,

│x│ < 4

Find the binomial expansion of f (x) in ascending powers of x, up to and including the term in x3. Give each coefficient as a simplified fraction. (Total 6 marks)

6.

(a)

Expand

1 ( 4  3x )

, where x 

4 , in ascending powers of x up to and including the term 3

2

in x . Simplify each term. (5)

(b)

Hence, or otherwise, find the first 3 terms in the expansion of

x 8 (4  3x )

as a series in

ascending powers of x. (4) (Total 9 marks)

7.

(a)

Use the binomial theorem to expand 1

(8  3x ) 3 , x 

8 , 3

in ascending powers of x, up to and including the term in x3, giving each term as a simplified fraction. (5)

(b)

Use your expansion, with a suitable value of x, to obtain an approximation to your answers to 7 decimal places.

3

(7.7) Give (2) (Total 7 marks)

City of London Academy

2

3 x . 2

f ( x )  (3  2x )  3 ,

8.

Find the binomial expansion of f(x), in ascending powers of x, as far as the term in x3. Give each coefficient as a simplified fraction. (Total 5 marks)

f(x) = (2 – 5x)–2,

9.

|x | <

2 . 5

Find the binomial expansion of f(x), in ascending powers of x, as far as the term in x3, giving each coefficient as a simplified fraction. (Total 5 marks)

10.

(a)

Find the first four terms in the expansion, in ascending powers of x, of

1  2 x 

1 2

2 x 1,

giving each term in its simplest form. (4)

(b)

Hence write down the first four terms in the expansion, in ascending powers of x, of

100 200x 

1 2

, 2x  1. (2) (Total 6 marks)

11. f (x ) 

(a)

3x 2  16 (1 3x )(2  x)

2



A B C   , (1 3x ) (2  x) (2  x) 2

x  13

Find the values of A and C and show that B = 0. (4)

(b)

Hence, or otherwise, find the series expansion of f(x), in ascending powers of x, up to and including the term in x3. Simplify each term. (7) (Total 11 marks)

12.

Use the binomial theorem to expand (4 – 9x),

x <

4 , 9

in ascending powers of x, up to and including the term in x3, simplifying each term. (Total 5 marks)

City of London Academy

3

f(x) =

13. (a)

1 – (1 + x),  (1  x )

–1 < x < 1.

Find the series expansion of f(x), in ascending powers of x, up to and including the term in x 3. (6)

(b)

Hence, or otherwise, prove that the function f has a minimum at the origin. (4) (Total 10 marks)

14.

(a)

Expand (1 + 3x)–2, x < simplifying each term.

1 3

, in ascending powers of x up to and including the term in x3, (4)

(b)

Hence, or otherwise, find the first three terms in the expansion of

x4 as a series in (1  3x ) 2

ascending powers of x. (4) (Total 8 marks) 3

15.

The binomial expansion of (1  12 x) 4 in ascending powers of x up to and including the term in x3 is 1 + 9x + px2 + qx3, 12x< 1. (a)

Find the value of p and the value of q. (4)

(b)

Use this expansion with your values of p and q together with an appropriate value of x 3

to obtain an estimate of (1.6) 4 . (2) 3

(c)

Obtain (1.6) 4 from your calculator and hence make a comment on the accuracy of the estimate you obtained in part (b). (2) (Total 8 marks)

16.

Given that

3 5x A B   , (1 3 x)(1 x) 1  3 x 1  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

3 5x . (1 3x )(1 x ) (5) City of London Academy

4

(c)

State, with a reason, whether your series expansion in part (b) is valid for x = 12 . (2) (Total 10 marks)

17.

The function f is given by f(x) = (a)

3(x  1) , x  , x  2, x  1. (x  2)(x 1)

Express f(x) in partial fractions. (3)

(b)

Hence, or otherwise, prove that f(x) < 0 for all values of x in the domain. (3) (Total 6 marks)

==================================================================================== MARK SCHEME

1.

A2

(a)

B1

2x 2  5x  10  A x 1 x  2  B x  2  C x 1

x 1 x  2

3  3B  B  1

M1 A1

12  3C  C  4

A14

2x 2  5x  10 x 1  2  1  x   2  1    x  1 x  2  2

(b)

1 x 

1

1

M1

 1 x  x 2  ...

B1

1

x x2 1 x      1    ... 2 4  2

B1

2 x2  5 x  10  1   2 1  2   1 1 x   1   x 2  ...  x 1 x  2   2

M1

 5  ...  ... 

3 2 x  ... 2

ft their A  B  12 C

A1 ft

0x stated or implied

A1 A17 [11]

2.

(a)

27x2 + 32x + 16 ≡ A(3x + 2)(1 – x) + B(1 – x) + C(3x + 2)2Forming this identity Substitutes either x  – 23 or x  – 23 ,12 – 64 16  53 B  20  35 B  B  4 3 3

City of London Academy

M1

x = 1 into 5

their identity or equates 3 terms or substitutes in values to write down three x = 1, 27 + 32 + 16 = 25C  75 = 25C  C =3 simultaneous equations. Both B = 4 and C = 3 (Note the A1 is dependent on both method marks in this part.)

M1

A1

27 = – 3A + 9C  27 = – 3A + 27  0 = –3A Equate x2:

 A=0 Compares coefficients or substitutes in a third x-value or uses simultaneous

x = 0, 16 = 2A + B + 4C

equations to show A = 0.

B14

 16 = 2A + 4 + 12  0 = 2A  A = 0

(b)

f(x) =

4 3  2 (1 – x) (3x  2)

= 4(3x + 2)–2 + 3(1 – x)–1 Moving powers to top on any one of the two expressions



M1



–2 –1 = 4 2(1  32 x) 3(1 – x)

–1 = 1  32 x   3(1 – x) –2

(–2)(–3) 3 x 2   ( 2 )  ... = 11 (–2)( 32x);  Either 1 (–2)(3x2 ); or 2!   1 ± (–1)(–x) from either first or second expansions respectively Ignoring 1 and 3, any one (–1)(–2)    3 1  (–1)(–x);  (– x) 2  ... correct {............} 2!   expansion. Both {............} correct.



 

A1 A1



2 2 = 1 – 3x  274 x  ...  3 1 x  x  ...

= 4  0 x ;  394 x 2 (c)

Actual = f (0.2) =

4 + (0x) ;

1.08  6.4  16 (6 .76)(0 .8)

39 4

x2

A1; A16

Attempt to find the actual value of f(0.2)

=

2935 23.48  4.341715976 ...  or seeing awrt 676 5 .408 4.3 and believing it is candidate‟s actual f(0.2).

M1

Or City of London Academy

6

Actual = f (0.2) =

4 3  (3( 0.2)  2) 2 (1 – 0. 2) Candidates can also attempt to find the actual value by using A B C   2 x ( 1 – (3  2) (3 x  2) x)

=

2935 4  3. 75  4.341715976 ... 676 6.76

with their A, B and C.

Estimate = f(0.2) = 4  394 (0.2)2

Attempt to find an estimate for f(0.2) using their answer to (b)

M1ft

= 4 + 0.39 = 4.39 %age error =

4.39 – 4.341715976... 4.341715976...

100

their estimate- actual  100 actual = 1.112095408... = 1.1%(2sf)

1.1%

M1 A1 cao4 [14]

3.

(a)

3x  1  A (1  2x) + B Let x =

1 2

;

1= B

 B  12 Considers this identity and either substitutes x = 12 , equates coefficients or solves simultaneous equations 3 2

Equate x terms; 3 = 2A  A =  32 A= 

M1

A1;A1 3

3 ; B  12 2

(No working seen, but A and B correctly stated  award all three marks. If one of A or B correctly stated give two out of the three marks available for this part.) (b)

f ( x)   32 (1  2 x) 1  12 (1  2 x) 2

M1

Moving powers to top on any one of the two expressions

  ( 1)( 2)( 3) ( 1)( 2) ( 2 x) 2    32 1  ( 1) ( 2 x);  ( 2 x) 3  ... 3! 2!   Either 1 ± 2x or 1 ± 4x from either first or second expansions respectively

City of London Academy

dM1

7

  ( 2)( 3)( 4) (2)(3)  12 1  ( 2) ( 2 x);  ( 2 x) 3 ... ( 2x )2  3 ! 2 !   Ignoring  32 and 12 , any one correct {…………} expansion. Both {………..} correct.

=

A1 A1

3 1 {1 + 2x + 4x2 + 8x3 + ...} + {1 + 4x + 12x2 + 32x3 + ...} 2 2

= 1  x ; + 0x2 + 4x3

A1; A1 6

1 x; (0x ) + 4x 2

3

[9]

Aliter Way 2 (b)

f(x) = (3x  1)(1  2x)2 Moving power to top

M1

(2)(3)   (2 x)2  1 ( 2)(2 x);   2 !   (3x  1)    ( 2)( 3)( 4) 3 ( 2 x) ...  3!   1 ± 4x; Ignoring (3x  1), correct (…………..) expansion

dM1;

A1

= (3x  1)(1 + 4x + 12x2 + 32x8 + ...) 3x + 12x2 +36x3  1  4x  12x2  32x3+... = 1  x ; + 0x2 + 4x3

Correct expansion

A1

1 x; (0x2) + 4x3

A1; A1 6

Aliter Way 3 (b)

Maclaurin expansion f ( x)   32 (1  2 x) 1  12 (1 2 x) 2

M1

Bringing (b) both powers to top f(x) =  3(1  2x) 2 + 2(1  2x)3 Differentiates to give a(1  2x) 2 ± b(1  2x)3; 3(1  2x) 2 + 2(1  2x) 3

M1;

A1 oe

f(x) = –12(1  2x)3 + 12(1  2x)4 f(x) =  72(1  2x)4 + 96(1  2x)5 City of London Academy

A1 8

Correct f (x) and f(x) f(0) = 1, f(0) = 1, f(0) = 0 and f(0) = 24 gives f(x) = 1  x;+ 0x2 + 4x3 + ...

A1; A1 6

1 x; (0x ) + 4x 2

3

Aliter Way 4 (b)

f (x) = 3(2  4x)1 + 12 (12x)2 Moving powers to top on any one of the two expressions

M1

  1 ( 1)( 2) 2 ( 2) 3 ( 4x ) 2  ( 2)  ( 1)(2) ( 4x );   2!   3  ( 1 )( 2 )( 3 )      ( 2) 4 ( 4x ) 3  ....    3! Either 12  x or 1 ± 4x from either first or second expansions respectively

dM1;

  ( 2)( 3)( 4) ( 2)( 3) ( 2x ) 2   12 1  ( 2)( 2 x);  ( 2 x) 3  ... 3! 2!   Ignoring  3 and 12 , any one correct {…………} expansion. Both {………..} correct.  3



 



 x  2 x2  4 x ...  12 1  4 x 12 x2 32 x3 ....

1 2

= 1  x; +0x2 + 4x3

4.

(a)

1 – 8 x

1 2



(b)

1 x; (0x2) + 4x3

1  12 – 8 x 

 12 – 12– 32 

 – 8x3

1 – 8x  

 12 – 12  2

A1; A1 6

– 8 x2

 ....

M1 A1

=1–4x –8x2;–32x3– …

A1 A14

3!

8    1 – 100  =

(c)

A1 A1

92  100

M1

23  25

23 5

*

cso

A12

1–4x–8x2–32x3=1–4(0.01)–8(0.01)2 – 32(0.01)3 =1– 0.04– 0.0008–0.000 032 =

City of London Academy

9

0.959168

M1

23  5×0.959168

M1

= 4.795 84

cao

A13 [9]

5.

1

f (x) =

(4  x )

= ( 4)

–

 ( 4  x)

–

1 2

M1

1 2 (1  ....) ...

1 1 (1  ....)... or 2 2 (1  ...)

1 3   x  (– )(– 2 )   ...1 (– 12 )  2   4 2    

B1

2  (– 1 )(– 32 )(– 52 )  x 3 x   ...    2  4 3! 4  

M1 A1ft

x  ft their   4  5 3 2 1 1  – x,  x – x 3  ... 2048 256 2 16

A1, A16

Alternative 1

f (x ) =

(4  x )

= 4

=

–

1 2

 ( 4  x)

 (– 12 ) 4

–

–

3 2x

1 2

M1

– 12– 32  1.2

–

5

4 2 x2 

 – 12 – 32 – 52  1. 2.3

–

7

4 2 x 3  ....

5 3 2 1 1 – x,  x – x 3  ... 2048 256 2 16

B1 M1 A1

A1, A16 [6]

6.

** represents a constant (which must be consistent for first accuracy mark) 1

(a)

( 4  3 x)



 ( 4  3x )

1 2

 ( 4)



1 2



 3x   1  4  

1 2



1 3x    1  4  2

1 2

  1   1  3  1     (** x); 2 2 (** x ) 2  ... 2!   2   with **  1 

1 2

City of London Academy

10



1   1  3 x   12  1        2! 2   2  4 

3 2



2  3 x     ...  4  

27 2 1 3  x  ... 1  x;   128 2 8  27 2  1 3  x  ... Ignore subsequent working   x;  256  2 16 





( 4)

1 2

or

1 outside brackets 2

Expands (1  * * x)



1 2

B1

to give a simplified or an un-simplified

 1 1    (** x ) ;  2

M1

A correct simplified or an un-simplified [...........] expansion with candidate‟s followed through (**x)

A1ft

 1  32  (** ) 2  1 x Award SC M1 if you see    (** x)  2 2!  2   3 1  8 x ;...   27 2   3 x  ... SC: K 1  x  8 128   1 2

1 2

(b)

A1 isw

27 2    .............;128 x   

A1 isw 5

27 2 1 3  x  ... (x + 8)   x  2 16 256   1 3 2  x  x  .... 2 16 3 27 2  4 x  x  ... 2 32 33 2  4  2x; x  ... 32 Writing (x + 8) multiplied by candidate‟s part (a) expansion.

M1

Multiply out brackets to find a constant term, two x terms and two x2 terms.

M1

Anything that cancels to 4 + 2x;

33 2 x 32

A1; A1 4 [9]

City of London Academy

11

7.

** represents a constant (which must be consistent for first accuracy mark) 1

(a)

1

1

1

3x  3  3x  3  (8  3 x) 3  (8) 3 1    2 1  8  8      1   2   1  2             1  3 3  3   3      2  2 1   (** x); (** x )  3! 2!  3  

5  3

   (** x) 3 ...  

with **  1 1  2   5   1  2              3  3  (** x) 2  3  3   3  (** x) 3 Award SC M1 if you see 3! 2!    1   2  5   1  2             2 3     1  3 x   3  3   3 x    3   3  3    3x      ...     21    ; 2! 3!  8  8    3  8    

1 2 5   1  2 1  x; x  x 3  ... 64 1536   8 1 5 3 1  2  x ; x 2  x  ... 32 768 4

1

Takes 8 outside the bracket to give any of (8) 3 or 2.

B1

1

Expands (1  * * x) 3 to give a simplified or an un-simplified

1 1    (** x ) ;  3 A correct simplified or an un-simplified {............} expansion with candidate‟s followed through (**x)

1  1  Either 2 1  x............... or anything that cancels to 2  x ; 4  8 

M1;

A1ft A1;

5

1

(b)

5 1 1 (0.1) 3  ... ( 0.1)2  ( 0.1)  768 32 4 = 2 – 0.025 – 0.0003125 – 0.0000065104166... = 1.97468099... (7.7) 3  2 

Attempt to substitute x = 0.1 into a candidate‟s binomial expansion.

M1

awrt 1.9746810

A1

You would award B1M1A0 for

City of London Academy

12

2

   1  2  5   1  2            2   x x   1  3  3 3  3   3  3  3  ( 3 x) 3 ...   2 !             3! 2!  8    3  8    because ** is not consistent.

If you see the constant term “2” in a candid...