# Question Bank - Algebra - Hemant Malhotra

• 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)^2-2a(x+y)=x^2+y^2
a^2+2xy-2a(x+y)=0
1/x+8/y=1
y+8x=xy so so y=8x/(x-1)
put
a^2+2x((8x/(x-1))-2a((x+8x/(x-1))=0
a^2 +16x^2/(x-1) -2a((x^2+7x/(x-1)=0
so a^2(x-1)+16x^2-2ax^2-14ax=0
so x^2(16-2a)+x(a^2-14a)-a^2=0
x real so D>=0
(a^2-14a)^2 -4 (16-2a)(-a^2)>=0
so a^2((a-14)^2 +4a^2(16-2a)>=0
so a^2((a^2+196-28a+64-8a)>=0
a^2((a^2-36a+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)

• a^2 * b^2 + a^2 + a^2 b^2 + 16 * b^2 = 448
so 2a^2 * b^2 + a^2 + 16b^2 = 448
a^2((2b^2+1) + 8(2b^2+8)=448+16
(a^2+ 8)(2b^2 + 1) = 456 = 8 * 3 * 19
now 2b^2+1 will be odd so it could be 3 or 19
when 2b^2+1=3 so 2b^2=2 so b=1 , 2b^2+1=19 so b^2=9 so b=3
when b=1 then a=12
when b=3 then a=4
so total will be 4 * 2=8

• Q72) Number of positive integral pairs (x, y) such that x ≤ y, that satisfies the equation √x + √y = √2009

• sqrtx = a * sqrt41
sqrty = b * sqrt41
sqrt2009= 7 * sqrt41
asqrt41+bsqrt41 = 7sqrt41
a+b=7
since x < = y
the possiblities are
so a=3 b=4
a=2, b=5
a=1 and b=6
3 cases

• Q73) If p, q, r and s are roots of x^4 + 4x^3 - 6x^2 + 7x - 9 = 0 then find value of (1 + p^2)(1 + q^2)(1 + r^2)(1 + s^2)

• Q74) Find all possible values of a for which x^2 + (a - 3)x + a = 0 has two distinct positive real roots.
a) a > 0
b) a < 1
c) 0 < a < 3
d) 0 < a < 1 or a > 9
e) 0 < a < 1

• Q75) a, b, c are distinct roots of x^3 - x + 1 = 0. Find a^16 + b^16 + c^16

• Q76) The polynomial x^3 + px^2 + qx + r has three roots in the interval (0,2). find range of p + q + r

• Q77) Let the roots of the cubic x^3 + 2x - 2 = 0 be a, b and c. Find sum of roots of the cubic equation whose roots are (a – b)^2, (b – c)^2 and (c – a)^2 .

• Q78) If a and b are positive integers, find the maximum possible value of the product of a and b, if 7a + 4b = 1860.

• Q79) Find sum of coefficient of (x + 3y + 2)^2

• Q80) f(x^2 +1) = x^4 - 4x^2 + p1
f(x^3 +4) = x^6 + 2x^3 + p2
Find p1 - p2
a) 1
b) 2
c) 3
d) 8
e) nota

• f(x^2+1)=x^4-4x^2+p1
find p1 so x^4=4x^2 so put x=2
so f(5)=p1
f(x^3+4)=x^6+2x^3+p2
now x^3+4=5 so x=1
so f(5)=3+p2
so p1=3+p2
so p1-p2 = 3

• Q81) Find the sum of the series given below.
2 + 1 + 4 + 2 + 8 + 3 + 16 + 4 + 32 + 5 + … + 20th term.
(a) 2047
(b) 2092
(c) 2147
(d) 2101

• 2, 4, 8, 16, 32 ... in GP, sum of 10 terms of GP
1, 2, 3, 4, 5 ..... in AP, sum of 10 terms of AP
SO 2101

• Q82) The sum of all the possible values of integral a such that (a^2 - 2a)^(a^2 + 47) = (a^2 - 2a)^(16a - 16) is
(a) 16
(b) 17
(c) 18
(d) 19
(e) none

47

71

142

121

34

102