Question Bank  Algebra  Hemant Malhotra

0^2=0
1^2=1
2^2=4
3^2=9
4^2=16
5^2=25
6^2=36
so 40 is even number
so even + even + even
or odd + odd + even
so only one possible 36 + 4 + 0 = 40
so 2 triplets possible
(1,2,3) and (1,3,2)

Q60) How many integer solutions exist for the equation 5x  y  120 = 0 such that values that x assumes has opposite signs compared to the corresponding values of y

5xy=120
So x=24+(y/5)
When y positive then x will be positive so we can no take y as positive
So we are looking for negative values of y such that x is positive
x=24+(y/5)
When y=5 then x is positive
Till y=115 , x will be positive
So 5 to 115 with common difference of 5

Q61) If f(x) is an even function defined on the interval (5,5) , find total number of real values of x satisfying
f(x) = f(x+1/(x+2)

even function means f(x)=f(x)
means if f(a)=f(b) then a=b and a=b
f(x)=f(x+1/(x+2)
so x=(x+1/(x+2) so 2 values of x by this equation
and x=(x+1/(x+2) and two values of x by this equations
so total 4 real values of x

Q62) If (a + b)/(1  ab), b, (b + c)/(1  bc) are in AP then a, 1/b, c are in
a) AP
b) GP
c) HP
d) None of the above

Method1 If numbers are in AP ., always start from HP in these kind of questions ...and take any random values of a,1/b and c which will make a,1/b and c in HP
let a=1 , b=2 and c=1/3
so (a+b)/(1ab)=3/12=3
b=2
(b+c)/(1bc)=2+1/3/(12/3)=7
so our assumption is correct so a,1/b, c are in HPMethod 2
2b=(a+b)/(1ab)+(b+c)/(1bc)
by solving
2b=1/a+1/c
so a,1/b and c are in HP

Q63) In a soccer tournament, if all teams were divided into groups of 7 each, 2 teams got a bye to the next round. If the group size was 9 and 11, 3 and 4 teams got byes respectively. If atleast 500 teams participated in the tournament, what is the minimum possible number of teams?

7x + 2 = 9y + 3 = 11z + 4
7k = 2y + 1k = 2y+1/7 = y = 3 , 10 ...
y = 7p  463p 33 = 11z + 4
63p  37/11 = z
8p  37/11 = z
p = 6 , 17 , 28 , 39 , 50p = 17 gives minimum so y = 115
number = 9 * 115 + 3 = 1038

Q64) The expression f(x) is cubic in x, in which the coefficient of x^3 is 1. If f(1)=5, f(2)=8. f(3)=11. find f(4)

Q65) Two equations have a common root which is positive. The other roots of the equation satisfy x^2  9x + 18 = 0. The product of the sums of roots of two equations is 40. Find the common root.

Q66) The number of pairs of integers (a, b) such that (a + 2b)^2 + (2a + 5b  1/2)^2 < = 2 is


Q67) If the real numbers x and y satisfy x^3  3x^2 + 5x  17 = 0 and y^3  3y^2 + 5y + 11 = 0, then the numerical value of x + y is

x^33x^2+5x17=0
y^33y^2+5y+11=0,
add both equations
x^3+y^33(x^2+y^2)+5(x+y)6=0
(x+y)((x+y)^23xy)3((x+y))^26xy+5(x+y)6=0
(x+y)^33xy(x+y)3((x+y)^26xy+5(x+y)6=0
((x+y2))((x+y)^2(x+y))+((33xy))=0
so x+y=2

Q68) If a, b, c, d, e and f are non negative real numbers such that a + b + c + d + e + f = 1, then the maximum value of (ab + bc + cd + de + ef) is

Q69) If x and y are positive numbers and 1/x + 8/y = 1, find the minimum value of (x + y) + √(x^2 + y^2)

x + y + sqrt(x^2+y^2)=a
a(x+y))=sqrt(x^2+y^2)
a^2+(x+y)^22a(x+y)=x^2+y^2
a^2+2xy2a(x+y)=0
1/x+8/y=1
y+8x=xy so so y=8x/(x1)
put
a^2+2x((8x/(x1))2a((x+8x/(x1))=0
a^2 +16x^2/(x1) 2a((x^2+7x/(x1)=0
so a^2(x1)+16x^22ax^214ax=0
so x^2(162a)+x(a^214a)a^2=0
x real so D>=0
(a^214a)^2 4 (162a)(a^2)>=0
so a^2((a14)^2 +4a^2(162a)>=0
so a^2((a^2+19628a+648a)>=0
a^2((a^236a+260)>=0
so a>=26 min value

Q70) Aishwarya used a calculator to compute (a + b)/c, where a, b and c are positive integers. She pressed a, +, b, /, c and = in that order, and got the answer 11. When she pressed b, +, a, /, c and = in that order, she was surprised to get a different answer 14. Then she realized that the calculator performed the division before the addition. So she pressed (, a, +, b, ), /, c and = in that order. She ﬁnally got the correct answer, which is

Q71) a^2(b^2 + 1) + b^2(a^2 + 16) = 448
Find the number of ordered pairs of integers (a , b)