Tackling Inequalities  Vikas Saini

Basics of inequality is based on line system.In such problem it is measured that value of x lies in which range.
Let's take some problems of it.
Type 1
Q. x^2+x42 < 0
solution :
(x+7)(x6) < 0
x € (7,6)
( € = belongs to)
Q. 7x3/2x+5 = < 3.
solution :
(7x3)/(2x+5)  3 = < 0
7x36x15 / 2x+5 = < 0
x18 / 2x+5 < = 0.
x = 18, x = 2/5.
on line 2/5 will lie on negative side while 18 will lie on positive side.
 inf to 5/2 ~ positive value
5/2 to 18 ~ negative
18 to inf ~ positive.
So here on 5/2 value can be positive which will not satisfy to the question.
While on 18 it will give perfect 0.
Hence x € ( 5/2, 18]
Q. x^3  10x^2+29x20 / x^2 +7x+12 < 0.
solution :
here for x = 1,numerator can give value 0.
hence (x1) is a factor of numerator.
= > x^3x^29x^2+9x+20x20 / x^2+3x+4x+12 < 0.
= > x^2(x1)9x(x1)+20(x1) / x(x+3)+4(x+3) < 0.
= > (x1) (x^29x+20) / (x+3)(x+4) < 0.
= > (x1)(x4)(x5)/(x+3)(x+4) < 0.
= > x = 1,4,5,3,4.
on line 3 & 4 will lie on left side while 1,4,5 will lie on right side.
inf to 4 ~ negative
4 to 3 ~ positive
3 to 1 ~ negative
1 to 4 ~ positive
4 to 5 ~ negative
5 to inf ~ positive.
x € (inf,4) U (3,1) U (4,5).
Q. (x+7)^2 (2x3)^3 / (x5) (x+1) < 0.
solution :
x = 7.
x = 3/2.
x = 5.
x = 1.
7 & 1 will lie left side of line while 3/2 & 5 will lie right side of line.
inf to 7 ~ positive
7 to 1 ~ positive
1 to 3/2 ~ positive
3/2 to 5 ~ negative
5 to inf ~ positive
x € (inf,7)U(7,1)U(1,3/2)U(5,inf).
Type 2
Q. How many positive integer values can x take that satisfy the inequality (x8 )(x10)(x12)..........(x100) < 0.
solution :
x = 8,10,12..................100.
total terms = (1008 )/2 + 1 = 47.
value of x for which total product should be negative.
so for 810,if we put x = 9 then one value will be positive,rest 46 values will be negative. Product of negative nos(even no of negative terms) give a positive value.
For x = 11, 2 positive & 45 negative
hence product negative.
so negative value of x = {(1008 )/2}/2 = 23.
Hence 23 values of x for which product < 0.
Q. (x+1)(x3)(x+5)(x7).........(x+97)(x99) < 0.
Here x = 1,5,9............97.
x = 3,7.........99
for x € (1,99) product will come negative.
total terms = (1+97)/4 + 1
= 25.
but in between two terms there will be few values for which product will become negative.
Let's take any two term 1 & 5.
in between 1 & 5, 3 values x=2,3,4 product will become negative.It means only 3 values in between any two terms.
Hence total values = 3 x 25 = 75.
Q. (x1)(x+5)(x9).......(x+61) < 0.
solution :
x = 1,9....57.
x = 5,13.....61.
here terms of x which will give product in negative.
total term = (5+61)/8 + 1
= 7 + 1
= 8.
total values in between 5 & 13 which will give negative values are :
6,7,8,9,10,11,12
total 7 values.
Total = 8 x 7 = 56.
Q. (x+1)(x2)(x+3).........(x98 ) < 0.
solution :
x = 1,3................97.
x = 2,4......................98.
so total terms which will give negative value = (1+97)/2 + 1
= 49.
The value in between two 1 & 3 which will give negative value is only one 2.
so total values = 49 x 1 = 49.
Type 3
Q. Find the sum of all values of x which satisfy the equation
2x3+3x+2+5x2 = 40.
solution :
Here values of x = 2,3,2.
Firstly we take extreme value of x = 3.
At x = 3
2x3+3x+2+5x2 = 20.
At x = 4,
it will increase by 10,because 2+3+5 = 10.
243+34+2+542 = 30.
At x = 5,
it will give value 40.
253+35+2+552 = 40.
So max value of x = 5.
For min value of x = 2.
at x = 2,
2x3+3x+2+5x2 = 30.
at x = 3
It will give 40.
So sum of values of x = 5  3 = 2.
Q. x+2+x3+x5 < 15.
Find the range of x.
Solution :
First we take extreme value of x = 5.
At x = 5,
x+2+x3+x5 = 9.
at x = 6,
x+2+x3+x5 = 12.
at x = 7,
x+2+x3+x5 = 15.
For minimum take x = 2(min)
at x = 2,
x+2+x3+x5 = 12.
at x = 3,
x+2+x3+x5 = 15.
x € (3,7).
Q. Find minimum value of x1+2x1+3x1+4x1+5x1
solution :
x1+2x1/2+3x1/3+4x1/4+5x1/5
Total terms = 1+2+3+4+5 = 15.
median = 8th term.
we will get 8th term at x1/4 = 0
= > x = 1/4.
at x = 1/4
x1+2x1+3x1+4x1+5x1
= 3/4 + 2/4 + 1/4 + 1/4
= 7/4.
Q. x1+x5+x9+x13 < = 42.
solution :
first we take extreme value at x13 = 0
= > x = 13
at x = 13.
x1+x5+x9+x13 = 24.
if we inverse value of x by 1 then total value will increase by 4.
we need to increase value of 18,we will have to increase value by 4.5.
To increase by 4 = +1.
To increase by 1 = +1/4.
To increase by 18 = 18/4 = 4.5.
at x = 13+4.5 = 17.5
x1+x5+x9+x13
= 16.5 + 12.5 + 8.5 + 4.5
= 42.
max value of x = 17.5
for minimum value of x1 = 0
= > x = 1.
at x = 1
x1+x5+x9+x13
= 24.
again to increase by 18,we need to reduce value of x by 4.5.
x = 1  4.5 = 3.5
x1+x5+x9+x13
= 4.5 + 8.5 + 12.5 + 16.5
= 42.
x € [3.5,17.5]