HKALE Pure Maths 2001 Paper01

Hence px x x q x for some polynomial qx 2

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Unformatted text preview: 7 −6−8 3−3 2 r − 7 − 4 r − 8 − 5r − 3 −1 −3 2 5 −7 =0 0 7(2r − 7) − 5(4r + 8) − 11(5r + 3) = 0 r = −2 (ii) Let e be the position vector of E and BE : EA = λ1 : µ1 . Since E lies on CD , e is also a linear combination of c and d . Using (b) , λ1 : µ1 = 3 : 2 . e= λ1a + µ 1b 3 2 = (7i + 8 j + 3k ) + (2i − 7 j + 13k ) = 5i + 2 j + 7k 5 5 λ1 + µ1 2001 Paper 1 Section B 13. (a) (i) P(α ) = (α ) 4 + a(α ) 3 + b(α ) 2 + cα + d = α 4 + aα 3 + bα 2 + cα + d = P(α ) ∴ (ii) =0 if α is a complex root of P(x) = 0 , α is also a root of P(x) = 0 . ( x − α )( x − α ) = x 2 − (α + α ) x + αα = x 2 − 2 Re(α ) x + α 2 ∴ ( x − α )( x − α ) 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 − a1 )( x − a 2 )( x − a 3 )( x − a 4 ) for some a1 , a 2 , a 3 , a 4 ∈ R = [ x 2 − (a1 + a 2 ) x + a1 a 2 ][ x 2 − (a 3 + a 4 ) x + a 3 a 4 ] (II) If there is a complex root α , then α is also a root. Hence P(x) = ( x − α )( x − α ) Q( x) for some polynomial Q(x) 2 = [ x 2 − 2 Re(α ) x + α ] Q( x) Since P(x) is a polynomial with real coefficients of degree 4 2 and [ x 2 − 2 Re(α ) x + α ] is a quadratic polynomial with real coefficients, ∴ 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 = −2 Hence g(x) = ( x − 2) 4 + 8( x − 2) 3 + 23( x − 2) 2 + 26( x − 2) + 7 = ( x 4 − 8 x 3 + 24 x 2 − 32 x + 16) + 8( x 3 − 6 x 2 + 12 x − 8) + 23( x 2 − 4 x + 4) + 26( x − 2) + 7 = x 4 − x 2 − 2x −1 (ii) Suppose g(x) = ( x 2 + px + q)( x 2 + rx + s) . p + r = 0 q + s + pr = −1 By comparing coefficients, ps + qr = −2 qs = −1 Clearly p ≠ 0 , otherwise r = 0 and hence ps + qr = 0 ≠ −2 . 2 − 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.

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