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Title: Cambridge International AS and A level Mathematics Pure mathematics 1 Sophie Goldie Series Editor: Roger Porkess Ex 3C shahbaz ahmed May 2023
Description: Cambridge International AS and A level Mathematics Pure mathematics 1 Sophie Goldie Series Editor: Roger Porkess Ex 3C shahbaz ahmed May 2023
Description: Cambridge International AS and A level Mathematics Pure mathematics 1 Sophie Goldie Series Editor: Roger Porkess Ex 3C shahbaz ahmed May 2023
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Cambridge International AS and A level
Mathematics
Pure mathematics 1
Sophie Goldie
Series Editor: Roger Porkess
Ex 3C
shahbaz ahmed
May 2023
Q12
The first three terms in the expansion of (2 + ax)n , in ascending powers of x,
are 32 − 40x + bx2
...
The first three terms in the expansion of the binomial theorem are
(A + B)n =
n
0
An +
n
1
n
2
An−1 B +
An−2 B 2
...
eq(2)
Putting A = 2, B = ax in equation (1)
(2 + ax)n =
n
0
2n +
n
1
2n−1 (ax) +
n
2
2n−2 (ax)2
...
(ii) Find the value of k for which there is no term in x2 in the expansion of
(1 + kx)(2 − x)6
...
eq(1) Solution PutA=2,B=-
26 +
6
1
26−1 (−x) +
6
2
26−2 (−x)2
...
= 6 Hint(1! = 1) by definition
...
Putting the above values in eq(2)
(2 − x)6 = 26 + 6(25 )(−x) + 15(2)4 (−x)2
Or (2 − x)6 = 64 − 192x + 240x2
(ii) Find the value of k for which there is no term in x2 in the expansion of
(1 + kx)(2 − x)6
...
From equation eq(2)
3
(2 − x)6 = 64 − 192x + 240x2
Hence
...
Simplyfing and neglecting the term involving x3
64 + 64kx − 192x − 192kx2 + 240x2
64 + (64k − 192)x − (192k − 240)x2
Putting the coefficient of x2 = 0
192k − 240 = 0
192k = 240
k=
240
192
=
5
4
Q14
(i) Find the first three terms in the expansion of (1 + ax)5 in ascending powers
of x
...
(iii) For this value of a, find the coefficient of x2 in the expansion of (1 −
2x)(1 + ax)5
...
Eq(1) By definition
⇒
4
By definition
5
0
=
5
0
=1
5
1
=
5
1
=5
5
2
=
5!
0!(5−0)!
5!
1!(5−1)!
5!
2!(5−2)!
=
5!
1×5!
=
5!
5!
=
5!
5×4!
=
5×4!
4!
=
5!
2× 3!
=
=1⇒
=5⇒
5×4×3!
2×3!
= 10 ⇒
Putting in eq(1)
(1 + ax)5 =
5
0
+
5
1
(ax)1 +
5
2
(ax)2
(1 + ax)5 = 1 + 5ax + 10a2 x2
(ii)First three terms in the expansion of (1 + ax)5 are:
(1 + ax)5 = 1 + 5ax + 10a2 x2
Putting in
(1 − 2x)(1 + ax)5 = (1 − 2x)(1 + 5ax + 10a2 x2 )
On simplifaction and neglecting the term involving x2 and higher powers of x
1 − 2x + 5ax
Or 1 − (2 − 5a)x
Putting coefficient of x = 0
⇒ 2 − 5a = 0
Or
-a=
2
5
(iii)Neglecting the term x3 and higher powers of x in the expression(1 −
2x)(1 + ax)5 :
(1 − 2x)(1 + ax)5 = (1 − 2x)(1 + 5ax + 10a2 x2 )
5
The term involving x2 in the above expression
Title: Cambridge International AS and A level Mathematics Pure mathematics 1 Sophie Goldie Series Editor: Roger Porkess Ex 3C shahbaz ahmed May 2023
Description: Cambridge International AS and A level Mathematics Pure mathematics 1 Sophie Goldie Series Editor: Roger Porkess Ex 3C shahbaz ahmed May 2023
Description: Cambridge International AS and A level Mathematics Pure mathematics 1 Sophie Goldie Series Editor: Roger Porkess Ex 3C shahbaz ahmed May 2023