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)
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1. What is the largest number that divides 245 and 1029 , leaving remainder 5 in each case? (a) 16 (b) 16 (c) 19 (d) 5 2. If p is a prime number and p divides a2(a is a positive integer),then which of the following is true: (a) p does not divide a (b) p divides a (c) p2 divides a (d) p divides a 3. If ax + by = a2 – b2 and bx + ay = 0,then value of (x +y) is : (a) a2 – b2 (b) b -a (c) a-b (d) a2 + b2 4. The pair of equation y = 0 and y = -5 has (a) one solution (b) two solution (c) infinitely many solution (d) No solution 5. The length of the diagonal of rhombus are 24cm and 32 cm . The length of the altitude of the rhombus in cm is : (a) 12 (b) 12.8 (c) 19 (d) 19.2 6. Sin 200 Cos 700 + Cos 200 Sin 700 is (a) 2 (b) 1 (c) 0 (d) 2 7. If tan + cot = 2 then tan2 + cot2 is (a) 4 (b) 6 (c) 2 (d) 1 8. If “less than” type and “more than type” of Ogive intersect each other at (20.5,15.5), than the median of the given data is : (a) 36.0 (b) 20.0 (c) 15.5 (d) 5.5
Can (x - 2) be the remainder on division of a polynomial p(x) by(2x +3) ? Justify
Answer: No , (x - 2) cannot be the remainder on division of a polynomial p(x) by(2x +3) Because degree of remainder must be less than degree of Divisor. Here degree of x-2 and 2x-3 is same that is 1 10th Math's Chapter 14 -Statistics – Mean, Mode and Median of Grouped Data and Graphical Representation of Cumulative Frequency Distribution
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