Unformatted text preview: ˆ’ 7 âˆ’6âˆ’8 3âˆ’3 2 r âˆ’ 7 âˆ’ 4 r âˆ’ 8 âˆ’ 5r âˆ’ 3
âˆ’1 âˆ’3 2 5 âˆ’7 =0 0 7(2r âˆ’ 7) âˆ’ 5(4r + 8) âˆ’ 11(5r + 3) = 0
r = âˆ’2
(ii) Let e be the position vector of E and BE : EA = Î»1 : Âµ1 .
Since E lies on CD , e is also a linear combination of c and d .
Using (b) , Î»1 : Âµ1 = 3 : 2 .
e= Î»1a + Âµ 1b 3
2
= (7i + 8 j + 3k ) + (2i âˆ’ 7 j + 13k ) = 5i + 2 j + 7k
5
5
Î»1 + Âµ1 2001 Paper 1
Section B 13. (a) (i) P(Î± ) = (Î± ) 4 + a(Î± ) 3 + b(Î± ) 2 + cÎ± + d
= Î± 4 + aÎ± 3 + bÎ± 2 + cÎ± + d
= P(Î± )
âˆ´ (ii) =0
if Î± is a complex root of P(x) = 0 , Î± is also a root of P(x) = 0 . ( x âˆ’ Î± )( x âˆ’ Î± ) = x 2 âˆ’ (Î± + Î± ) x + Î±Î±
= x 2 âˆ’ 2 Re(Î± ) x + Î± 2 âˆ´ ( x âˆ’ Î± )( x âˆ’ Î± ) is a quadratic polynomial in x with real coefficients.
Since P(x) is a polynomial of degree 4, therefore the equation P(x) = 0
has 4 roots.
(I) If all the 4 roots are real, then P(x) can be factorized into 4 linear
factors with real coefficients. Hence
P(x) = ( x âˆ’ a1 )( x âˆ’ a 2 )( x âˆ’ a 3 )( x âˆ’ a 4 ) for some a1 , a 2 , a 3 , a 4 âˆˆ R
= [ x 2 âˆ’ (a1 + a 2 ) x + a1 a 2 ][ x 2 âˆ’ (a 3 + a 4 ) x + a 3 a 4 ] (II) If there is a complex root Î± , then Î± is also a root. Hence
P(x) = ( x âˆ’ Î± )( x âˆ’ Î± ) Q( x)
for some polynomial Q(x)
2 = [ x 2 âˆ’ 2 Re(Î± ) x + Î± ] Q( x)
Since P(x) is a polynomial with real coefficients of degree 4
2 and [ x 2 âˆ’ 2 Re(Î± ) x + Î± ] is a quadratic polynomial with
real coefficients,
âˆ´ Q(x) is also a quadratic polynomial with real coefficients.
Hence P(x) is a product of two quadratic polynomials
with real coefficients. 2001 Paper 1
Section B (b) (i) g(x) = ( x + k ) 4 + 8( x + k ) 3 + 23( x + k ) 2 + 26( x + k ) + 7
Coefficient of x 3 = 4k + 8 = 0
k = âˆ’2
Hence g(x) = ( x âˆ’ 2) 4 + 8( x âˆ’ 2) 3 + 23( x âˆ’ 2) 2 + 26( x âˆ’ 2) + 7
= ( x 4 âˆ’ 8 x 3 + 24 x 2 âˆ’ 32 x + 16) + 8( x 3 âˆ’ 6 x 2 + 12 x âˆ’ 8) + 23( x 2 âˆ’ 4 x + 4) + 26( x âˆ’ 2) + 7
= x 4 âˆ’ x 2 âˆ’ 2x âˆ’1
(ii) Suppose g(x) = ( x 2 + px + q)( x 2 + rx + s) .
p + r = 0
q + s + pr = âˆ’1 By comparing coefficients, ps + qr = âˆ’2
qs = âˆ’1 Clearly p â‰ 0 , otherwise r = 0 and hence ps + qr = 0 â‰ âˆ’2 .
2
âˆ’ p + q...
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This note was uploaded on 06/07/2012 for the course MATHS 2002 taught by Professor Lo during the Fall '02 term at Wisc Oshkosh.
 Fall '02
 lo
 Math

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