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G.C.E. (A.L.) Examination August 2000 Combined Mathematics I Combined Mathematics I (Q1) Model Solutions 1 We conduct individual classes upon request. Contact us at: for more information G.C.E. (A.L.) Examination

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Question No 1(a) G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 Question No 1(a) (a) and are the roots of the equation x 2 – px + q = 0. Find the equation, whose roots are. 2

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The equation, whose roots are and x x Expanding x 2 – x – x + x 2 – ( + )x + (1) x 2 – px + q = 0 (2) Comparing the coefficients of (1) and (2) p= ( + ) and q = (3) 3 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(a) (Model Solutions) Question No 1(a) (Model Solutions)

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The equation, whose roots are and x x Expanding x 2 – x – x + x 2 – x{ + (4) 4 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(a)(Model Solutions) … Question No 1(a) (Model Solutions) …

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Substituting the value of p= ( + ) in the above equation (3) and the value of q = in the above equation (4) We can obtain x 2 – p 2 x + qp 2 whose roots are. 5 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 Question No 1(a) (Model Solutions) …

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6 (b) In order for the function f(x,y) = 2x 2 + xy + 3y 2 - 5y - 2 to be written as a product of two linear factors, find the values of G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(b) Question No 1(b)

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L.S.= 2x 2 + xy + 3y 2 - 5y - 2 R.S.= (ax + by + c)(lx + my + n) Substituting x = 0 in L.S. and R.S. 3y 2 - 5y - 2 by + c)(my + n) 3y + 1)(y - 2) by + c)(my + n) Therefore b=3, c=1, m=1, n=-2 Substituting y = 0 in L.S. and R.S. 2x 2 – 2 = (ax + c)(lx + n) 2x 2 – 2 = alx 2 +(an + cl)x + cn 7 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 Question No 1(b) (Model Solutions)

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Comparing above coefficients of L.S and R.S. 2=al, 0=an+cl, and -2=cn Substitute c=1, n = -2 in 0=an+cl 0=an+cl = a(-2)+1(l)=>l=2a Substitute l=2a in 2=al 2=a(2a) => and L.S. = 2x 2 + xy + 3y 2 - 5y - 2 R.S. = (ax + by + c)(lx + my + n) 8 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(b) (Model Solutions) … Question No 1(b) (Model Solutions) …

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Comparing the coefficients of L.S and R.S. am+bl Substituting m=1, and in above equation Therefore 9 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 Question No 1(b) (Model Solutions) …

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10 (c) Express in partial fractions. G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 Question No 1(c)

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The fraction Since the denominator and the numerator powers of this fraction are the same we need to divide numerator by the denominator. 11 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(c) (Model Solutions) Question No 1(c) (Model Solutions)

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12 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(c) (Model Solutions) … Question No 1(c) (Model Solutions) …

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Comparing the coefficients of L.S and R.S. A+B=4, -2A-B+C=-3, A=3 B=4-A=1, C=B+2A-3=1+6-3=4 Therefore 13 G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(c) (Model Solutions) … Question No 1(c) (Model Solutions) …

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