G.C.E. (A.L.) Examination

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Scholastic home.video.tutor @gmail.com. G.C.E. (A.L.) Examination. G.C.E. (A.L.) Examination. August 2000 Combined Mathematics I ‌ (Q1) Model Solutions. We conduct individual classes upon request. Contact us at: home.video.tutor@gmail.com for more information. Scholastic - PowerPoint PPT Presentation

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  • G.C.E. (A.L.) Examination

    August 2000Combined Mathematics I(Q1)Model Solutions

    *We conduct individual classes upon request.Contact us at: home.video.tutor@gmail.com for more informationG.C.E. (A.L.) Examination

  • G.C.E. (A.L.) Examination Combined Mathematics I - August 2000

    Question No 1(a)(a) a and b are the roots of the equation x2 px + q = 0. Find the equation, whose roots are a(a + b) b(a + b). *

  • The equation, whose roots are a and b (x - a) (x - b) = 0 Expandingx2 xa xb + ab = 0 x2 (a + b)x + ab = 0 --- (1)x2 px + q = 0 --- (2)Comparing the coefficients of (1) and (2)p= (a + b) and q = ab --- (3)

    *G.C.E. (A.L.) Examination Combined Mathematics I - August 2000 Question No 1(a) (Model Solutions)

  • The equation, whose roots are a(a + b) and b(a + b) {x - a(a + b)}{(x -b(a + b) )}= 0 Expandingx2 xb(a + b) xa(a + b) + ab(a + b)2} = 0 x2 x{(a + b) (a + b)} + ab(a + b)2 = 0 -- (4)*G.C.E. (A.L.) Examination Combined Mathematics I - August 2000 Question No 1(a) (Model Solutions)

  • Substituting the value of p= (a + b) in the above equation (3) and the value of q = ab in the above equation (4) . We can obtain x2 p2x + qp2 = 0, whose roots are a(a + b) b(a + b).

    *G.C.E. (A.L.) Examination Combined Mathematics I - August 2000 Question No 1(a) (Model Solutions)

  • *(b) In order for the function f(x,y) = 2x2 + lxy + 3y2 - 5y - 2 to be written as a product of two linear factors, find the values of l. G.C.E. (A.L.) Examination Combined Mathematics I - August 2000 Question No 1(b)

  • L.S.= 2x2 + lxy + 3y2 - 5y - 2 R.S.= (ax + by + c)(lx + my + n)Substituting x = 0 in L.S. and R.S. 3y2 - 5y - 2 = (by + c)(my + n) (3y + 1)(y - 2) = (by + c)(my + n) Therefore b=3, c=1, m=1, n=-2Substituting y = 0 in L.S. and R.S.2x2 2 = (ax + c)(lx + n) 2x2 2 = alx2 +(an + cl)x + cn*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=cnSubstitute c=1, n = -2 in 0=an+cl 0=an+cl = a(-2)+1(l)=>l=2aSubstitute l=2a in 2=al2=a(2a) => and L.S. = 2x2 + lxy + 3y2 - 5y - 2 R.S. = (ax + by + c)(lx + my + n)*G.C.E. (A.L.) Examination Combined Mathematics I - August 2000 Question No 1(b) (Model Solutions)

  • Comparing the coefficients of L.S and R.S.l=am+blSubstituting m=1, and in above equationTherefore *G.C.E. (A.L.) Examination Combined Mathematics I - August 2000 Question No 1(b) (Model Solutions)

  • *(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. *G.C.E. (A.L.) Examination Combined Mathematics I - August 2000 Question No 1(c) (Model Solutions)

  • *G.C.E. (A.L.) Examination Combined Mathematics I - August 2000 Question No 1(c) (Model Solutions)

  • Comparing the coefficients of L.S and R.S. A+B=4, -2A-B+C=-3, A=3B=4-A=1, C=B+2A-3=1+6-3=4Therefore

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

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