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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 home.video.tutor @gmail.com G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000Question No 1(a) (a) a andb are the roots of the equation x2 – px + q = 0. Find the equation, whose roots are a(a + b) හා b(a + b).
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(a) (Model Solutions) Scholastic home.video.tutor @gmail.com • The equation, whose roots area andb • (x - a) (x - b) = 0 • Expanding • x2 – 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)andq = ab --- (3)
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(a)(Model Solutions) … Scholastic home.video.tutor @gmail.com • The equation, whose roots area(a + b) andb(a + b) • {x - a(a + b)}{(x -b(a + b))}= 0 • Expanding • x2 – 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 QuestionNo 1(a) (Model Solutions) … Scholastic home.video.tutor @gmail.com • 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) හා b(a + b).
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(b) Scholastic home.video.tutor @gmail.com (b) In order for the function f(x,y) = 2x2 + lxy + 3y2 - 5y - 2to 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) (Model Solutions) Scholastic home.video.tutor @gmail.com • 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=-2 • Substituting 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 QuestionNo 1(b) (Model Solutions) … Scholastic home.video.tutor @gmail.com • 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 in2=al • 2=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) … Scholastic home.video.tutor @gmail.com • Comparing the coefficients of L.S and R.S. • l=am+bl • Substituting m=1,and in above equation • Therefore
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 Question No 1(c) Scholastic home.video.tutor @gmail.com (c) Express in partial fractions.
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(c) (Model Solutions) Scholastic home.video.tutor @gmail.com • 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 QuestionNo 1(c) (Model Solutions) … Scholastic home.video.tutor @gmail.com •
G.C.E. (A.L.) Examination – Combined Mathematics I - August 2000 QuestionNo 1(c) (Model Solutions) … Scholastic home.video.tutor @gmail.com • 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
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