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

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G.C.E. (A.L.) Examination August 2000 August 2000 Combined Mathematics I Combined Mathematics I (Q1) (Q1) Model Solutions Model 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.) Examination G.C.E. (A.L.) Examination

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Page 1: G.C.E. (A.L.) Examination

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

Page 2: G.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.

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Page 3: G.C.E. (A.L.) Examination

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)

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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)

Page 4: G.C.E. (A.L.) Examination

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) …

Page 5: G.C.E. (A.L.) Examination

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.

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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) …

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(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)

Page 7: G.C.E. (A.L.) Examination

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)

Page 8: G.C.E. (A.L.) Examination

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) …

Page 9: G.C.E. (A.L.) Examination

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) …

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(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)

Page 11: G.C.E. (A.L.) Examination

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)

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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) …

Page 13: G.C.E. (A.L.) Examination

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) …

Page 14: G.C.E. (A.L.) Examination

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