Question Bank  Algebra  Hemant Malhotra

g(2) < 4
so 4a2m+n < 4 (1)
g(2) > 4
so 4a+2m+n > 4 so 4a2mn < 4  (2)
subtract these two so 8a+2n < 0 so 4a+n < 0
now g(3) < 11
so 9a+3m+n < 11
and 4a2m+n < 4
solve these two equations and eliminate m
so 18a+6m+2n < 22
12a6m+3n < 12
add both
30a+5n < 10
and 4a+n < 0
so 20a+5n < 0
subtract
10a < 10
so a < 1or try from graphs

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

Direct method
Area of xa + yb = 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
add all these so
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(4y)+52y=0
x is real so D > = 0
So (4y)^2 4(52y)>=0
So 16+y^28y20+8y>=0
So y^24y >=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=5a + 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) (2x1) = +1
so (2x1)=(x+5)^(+1)
so 2x1=x+5 or 2x1=1/(x+5)
so x=6 is one value
(2x1)*(x+5)=1
2x^2+9x6=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/(1x1) = x1/(1x1)  x1
x2^2/(1x2)=x2/(1x2)  x2
.
.
.
add all terms
x1/(1x1)+x2/(1x2)+....x2014/(1x2014) (x1+x2+x3....x2014)
11=0

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

f(a)=a^4a^32^(a+1)
f(0)=2
f(1)=1+11=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(a1)2^(a+1)
f(2)=2^32^(3)=0 so a=2 another root
now f(3)=3^4 * 22^4 > 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 * 142^16 < 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