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

date post

04-Feb-2016Category

## Documents

view

21download

0

Embed Size (px)

description

### Transcript of G.C.E. (A.L.) Examination

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)

Home Based Mathematics Tuition using the Internet We conduct individual classes upon request for International Baccalaureate (IB)GCE(O/L), GCE(A/L) , AQA, EDEXCELwithPast Exam Paper Discussions. University Foundation Year Mathematics

Flexible contact hours. Save time of travelling.A unique way of tutoring.

For more information please contact:home.video.tutor@gmail.com http://scholastictutors.webs.com/

*View more*