# G.C.E. (A.L.) Examination August 2000 Combined Mathematics I Combined Mathematics I (Q1) Model Solutions 1 We conduct individual classes upon request.

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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: home.video.tutor@gmail.com for more information G.C.E. (A.L.) Examination

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

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)

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) …

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) …

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)

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)

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) …

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) …

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

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)

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

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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