Q. ab+bc+ca=0 find value 1/(b2-ac ) + 1/(c2-ab) + 1/(a2-bc)Solution: ab+bc+ca=0 find value 1/(a2-bc ) + 1/(c2-ab) + 1/(a2-bc) Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac you get : 1/(a2-bc (ab + bc + ca)/[(a+b+c)(abc)] put, ab+bc+ca=0 you get value = 0/[(a+b+c)(abc)] = 0 Q. A class of 20 boys and 15 girls is divided into n groups so that each group has x boys and y girls. Find x ,y and n.
Sol: n(x+y) = 20 + 15 = 35 Total boys = nx = 20 Þ nx = 2 x 10 = 4 x 5 = 10 x 2 Similarly, Total girls = ny = 15 Þ 3 x 5 The possible value of n may be n = 5 x = 4 and y = 5 OR, Hcf of 20 and 15 is 5 Þ No. of group will be n =5 then Total girls = ny = 15 Þ 3 x 5 The possible value of n may be n = 5 x = 4 and y = 5 Q. The pair of equations y=0 and y= -5 has 1.one solution 2.two solutions 3.infinitely many solutions 4.no solution Ans: o.x + y = 0 0.x + y = -5 therefore, a1/a2 = b1/b2 ≠ c1/c2 Þ The pair of equations y=0 and y= -5 has no solution Q. If secA + tanA = 1/x find the value of SecA and tanA Q. There are 20 cars and motorcycle in a parking lot. If there are 56 wheels together find the no of cars and motorcycles Ans: Le t the no. of car is x then no. of motorcycle will be (20-x) A/Q, 4x + 2(20-x) = 56 Þ x = 8 the no. of car is x = 8 then no. of motorcycle will be (20-x) = 20-8 = 12 Q. If constant term in Quardatic polynomial is zero ,then prove that one of its zero is zero Sol: let the quadratic polynomial be P(x) = a x2 + bx + c Þ P(x) = a x2 + bx [since the constant term is zero] Now P(0) = 0 + 0 = 0 Thus, 0 is one zero of p(x). Q. If p(x) = ax2 + bx +c and a + c = b, then one of the zeroes is (a) b/a (b) c/a (c) -c/a (d) -b/a Sol: c/a Q. if asin2q + bcos2 q = c , then sow that cot2q = (c-a)/(b-c) ab + ac ab + ac (ab + bc + ca)/[(a+b+c)(abc)] put, ab+bc+ca=0 you get value = 0/[(a+b+c)(abc)] = 0 Q. A class of 20 boys and 15 girls is divided into n groups so that each group has x boys and y girls. Find x ,y and n. Sol: n(x+y) = 20 + 15 = 35 Total boys = nx = 20 Þ nx = 2 x 10 = 4 x 5 = 10 x 2 Similarly, Total girls = ny = 15 Þ 3 x 5 The possible value of n may be n = 5 x = 4 and y = 5 OR, Hcf of 20 and 15 is 5 Þ No. of group will be n =5 then Total girls = ny = 15 Þ 3 x 5 The possible value of n may be n = 5 x = 4 and y = 5 Q. The pair of equations y=0 and y= -5 has 1.one solution 2.two solutions 3.infinitely many solutions 4.no solution Ans: o.x + y = 0 0.x + y = -5 therefore, a1/a2 = b1/b2 ≠ c1/c2 Þ The pair of equations y=0 and y= -5 has no solution Q. If secA + tanA = 1/x find the value of SecA and tanA Q. There are 20 cars and motorcycle in a parking lot. If there are 56 wheels together find the no of cars and motorcycles Ans: Le t the no. of car is x then no. of motorcycle will be (20-x) A/Q, 4x + 2(20-x) = 56 Þ x = 8 the no. of car is x = 8 then no. of motorcycle will be (20-x) = 20-8 = 12 Q. If constant term in Quardatic polynomial is zero ,then prove that one of its zero is zero Sol: let the quadratic polynomial be P(x) = a x2 + bx + c Þ P(x) = a x2 + bx [since the constant term is zero] Now P(0) = 0 + 0 = 0 Thus, 0 is one zero of p(x). Q. If p(x) = ax2 + bx +c and a + c = b, then one of the zeroes is (a) b/a (b) c/a (c) -c/a (d) -b/a Sol: c/a Q. if asin2q + bcos2 q = c , then sow that cot2q = (c-a)/(b-c)
2 Comments
admin
28/9/2013 02:59:00 pm
Anyone can post Question from SA-1 Board paper
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jsunil
2/10/2013 06:12:59 am
ab+bc+ca=0 find value 1/(a2-bc ) + 1/(c2-ab) + 1/(a2-bc)
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