# Question Bank - Algebra - Hemant Malhotra

• g(-2) < 4
so 4a-2m+n < 4 ----(1)
g(2) > -4
so 4a+2m+n > -4 so -4a-2m-n < 4 ---- (2)
subtract these two so 8a+2n < 0 so 4a+n < 0
now g(3) < -11
so 9a+3m+n < -11
and 4a-2m+n < 4
solve these two equations and eliminate m
so 18a+6m+2n < -22
12a-6m+3n < 12
30a+5n < -10
and 4a+n < 0
so 20a+5n < 0
subtract
10a < -10
so a < -1

or try from graphs

• Q50) Find out area bound by |x - 3| + |y +2| = 12

• Direct method
Area of |x-a| + |y-b| = k wil be = 2 * k^2 =2 * 12^2 = 288

• Q51) • G(x+1)-G(x)=x^2+1
G(1)-G(0)=0^2+1
G(2)-G(1)=1^2+1
G(3)-G(2)=2^2+1.
.
.
.
G(20)-G(19)=19^2+1
G(20) - G(0) = 1^2 + 2^2 + ... + 19^2 + 20
= 19 * 20 * 39/6 + 20
=19 * 10 * 13 + 20
= 2490

• Q52) Which of the following is not a possible value for (x^2 + 4x + 5)/(x + 2) for real x
a. - 3
b. - 2
c. - 1
d. 2

• y = (x^2+4x+5)/(x+2)
So x^2+x(4-y)+5-2y=0
x is real so D > = 0
So (4-y)^2 -4(5-2y)>=0
So 16+y^2-8y-20+8y>=0
So y^2-4y >=0
So y>=2 and y < = -2
So -1 is not possible

• Q53) How many pairs (a, b) of positive integers are there such that 2(a^2) = 3(b^3), where b < 1000

• a^2 = 3/2(b^3)
let b=2t
a^2 = 4 * 3 * t^3
let t=3l
a^2 = 4 * 3^4 * l^3
so l^3 is a square and l is a perfect sq
pairs can be like (18,6) for l=1
l=4 -then t=12 b=24
a= 2 * 9 * 8= 144 so (144,24)
l=9 so b=54
l=16 sob=96
till l=144 so b = 144 * 6 =864
so 12 pairs

• Q54) The coefficient of x^9 in the expansion of (1 + x) (1 + x^2) (1 + x^3)... (1 + x^100) is
(a) 6
(b) 7
(c) 8
(d) 9

• a + b = 9
so a=0, b=9
a=1, b=8
a=2, b=7
a=3, b=6
a=4, b=5

a + b + c = 9
a = 1, b = 2, c = 6
a = 1, b = 3, c = 5
a = 2, b = 3, c = 4
a + b + c + d = 9
a = 1, b = 2, c = 3, d = 4 which is greater than 9 so not possible
so only 8 ways possible

• Q55) • log_(x+5) (2x-1) = +-1
so (2x-1)=(x+5)^(+-1)
so 2x-1=x+5 or 2x-1=1/(x+5)
so x=6 is one value
(2x-1)*(x+5)=1
2x^2+9x-6=0
x=-9+-sqrt(81+48)/2
so out of these 2 values one value will satisfy domain part
so OA=2

• Q56) • (1+y+y^2)^100
now (1+y)^2=1+y^2+2y= 3 terms=2 * 1
(1+y+y^2)^2=1+y^4+y^2+2y^3+2y+2y^2=y^4+2y^3+3y^2+2y+1 =5 terms so 2*2+1
so (1+x+x^2....x^m)^n main numbers of terms will be mn+1

• Q57) • OA=0
x1^2/(1-x1) = x1/(1-x1) - x1
x2^2/(1-x2)=x2/(1-x2) - x2
.
.
.
x1/(1-x1)+x2/(1-x2)+....x2014/(1-x2014) -(x1+x2+x3....x2014)
1-1=0

• Q58) Find no of solution of a^3 + 2^(a+1) = a^4

• f(a)=a^4-a^3-2^(a+1)
f(0)=-2
f(-1)=1+1-1=1
so one root will lie between 0 and -1
and for other negative value f(a)>0 so no real root there
f(x)=a^3(a-1)-2^(a+1)
f(2)=2^3-2^(3)=0 so a=2 another root
now f(3)=3^4 * 2-2^4 > 0
f(4)=4^3 * 3 - 2^5 > 0
here f(4) < f(3)
and f(5) < f(4) so value will decrease so graph will cut x axis one more time so n
.
.f(15)=15^3 * 14-2^16 < 0 (by approximation))
so one root will lie between 2 and 15 ... somewhere near 14 .. and after than value will be negative so no more real roots
so Total 3 real roots

• Q59) x, y and z are positive integers. Then how many triplets (x, y, z) exist (x + y + z)^2 + (x + y − z)^2 + (x − y + z)^2 = 40
a) 1
b) 3
c) 5
d) 6
e) 2

92

103

199

31

121

30

42