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G.C.E. (A.L.) Examination
August 2000August 2000Combined Mathematics ICombined Mathematics I
(Q1)(Q1)Model SolutionsModel Solutions
1
We conduct individual classes upon request.We conduct individual classes upon request.Contact us at: [email protected] for more information
G.C.E. (A.L.) ExaminationG.C.E. (A.L.) Examination
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
Question Question No 1(a)No 1(a)
(a) andare the roots of the equation x2 – px + q = 0. Find the equation, whose roots are හාහා.
2
The equation, whose roots are and xx Expanding x2 – x– x+ x2 – (+ )x+ (1) x2 – 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
QuestionQuestion No 1(a) No 1(a) (Model Solutions)(Model Solutions)
The equation, whose roots are and
xx Expanding x2 – x – x+
x2 – x{+ (4)
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G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
QuestionQuestion No 1(a)No 1(a) (Model Solutions) …(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 x2 – p2x+ qp2 whose roots are හාහා.
5
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
QuestionQuestion No 1(a) No 1(a) (Model Solutions) …(Model Solutions) …
6
(b) In order for the function f(x,y) = 2x2 + xy + 3y2 - 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
QuestionQuestion No 1(b)No 1(b)
L.S.= 2x2 + xy + 3y2 - 5y - 2 R.S.= (ax + by + c)(lx + my + n) Substituting x = 0 in L.S. and R.S. 3y2 - 5y - 2by + 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. 2x2 – 2 = (ax + c)(lx + n) 2x2 – 2 = alx2 +(an + cl)x + cn 7
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
Question Question No 1(b) No 1(b) (Model Solutions)(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. = 2x2 + xy + 3y2 - 5y - 2 R.S. = (ax + by + c)(lx + my + n)
1a 2l
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G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
QuestionQuestion No 1(b) No 1(b) (Model Solutions) …(Model Solutions) …
Comparing the coefficients of L.S and R.S. am+bl Substituting m=1, and in
above equation Therefore
2311 blam
7
2l1a
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G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
Question Question No 1(b) No 1(b) (Model Solutions) …(Model Solutions) …
10
(c) Express in partial fractions. 2
3
1
32
xx
xx
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
Question Question No 1(c)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
xxx
xx
xxx
xx
xx
xx
23
3
2
3
2
3
2
32
12
32
1
32
2322 323 xxxxx
334
2422
23
xx
xxx 2
2
2
3
1
3342
1
32
xx
xx
xx
xx
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
QuestionQuestion No 1(c) No 1(c) (Model Solutions)(Model Solutions)
12
22
2
111
334
x
C
x
B
x
A
xx
xx
CxxBxxAxx )1(1334 22
CxxxBxxAxx )(12334 222
ACBAxBAxxx )2(334 22
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
QuestionQuestion No 1(c) No 1(c) (Model Solutions) …(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
22
3
1
4
1
132
1
32
xxxxx
xx
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000
QuestionQuestion No 1(c) No 1(c) (Model Solutions) …(Model Solutions) …
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