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+x-42 < 0

    solution :-

    (x+7)(x-6) < 0

    x € (-7,6)

    ( € = belongs to)

    Q. 7x-3/2x+5  = < 3.

    solution :-

    (7x-3)/(2x+5) - 3 = < 0

    7x-3-6x-15 / 2x+5  = < 0

    x-18 / 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+29x-20 / x^2 +7x+12 < 0.

    solution :-

    here for x = 1,numerator can give value 0.

    hence (x-1) is a factor of numerator.

    = > x^3-x^2-9x^2+9x+20x-20 / x^2+3x+4x+12  < 0.

    = > x^2(x-1)-9x(x-1)+20(x-1) / x(x+3)+4(x+3) < 0.

    = > (x-1) (x^2-9x+20) / (x+3)(x+4) < 0.

    = > (x-1)(x-4)(x-5)/(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 (2x-3)^3 / (x-5) (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 (x-8 )(x-10)(x-12)..........(x-100) < 0.

    solution :-

    x = 8,10,12..................100.

    total terms = (100-8 )/2 + 1 = 47.

    value of x for which total product should be negative.

    so for 8-10,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 = {(100-8 )/2}/2 = 23.

    Hence 23 values of x for which product < 0.

    Q. (x+1)(x-3)(x+5)(x-7).........(x+97)(x-99) < 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. (x-1)(x+5)(x-9).......(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)(x-2)(x+3).........(x-98 ) < 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

    2|x-3|+3|x+2|+5|x-2| = 40.

    solution :-

    Here values of x = 2,3,-2.

    Firstly we take extreme value of x = 3.

    At x = 3

    2|x-3|+3|x+2|+5|x-2| = 20.

    At x = 4,

    it will increase by 10,because 2+3+5 = 10.

    2|4-3|+3|4+2|+5|4-2| = 30.

    At x = 5,

    it will give value 40.

    2|5-3|+3|5+2|+5|5-2| = 40.

    So max value of x = 5.

    For min value of x = -2.

    at x = -2,

    2|x-3|+3|x+2|+5|x-2| = 30.

    at x = -3

    It will give 40.

    So sum of values of x = 5 - 3 = 2.

    Q. |x+2|+|x-3|+|x-5| < 15.

    Find the range of x.

    Solution :-

    First we take extreme value of x = 5.

    At x = 5,

    |x+2|+|x-3|+|x-5| = 9.

    at x = 6,

    |x+2|+|x-3|+|x-5| = 12.

    at x = 7,

    |x+2|+|x-3|+|x-5| = 15.

    For minimum take x = -2(min)

    at x = -2,

    |x+2|+|x-3|+|x-5| = 12.

    at x = -3,

    |x+2|+|x-3|+|x-5| = 15.

    x € (-3,7).

    Q. Find minimum value of |x-1|+|2x-1|+|3x-1|+|4x-1|+|5x-1|

    solution :-

    |x-1|+2|x-1/2|+3|x-1/3|+4|x-1/4|+5|x-1/5|

    Total terms = 1+2+3+4+5 = 15.

    median = 8th term.

    we will get 8th term at x-1/4 = 0

    = > x = 1/4.

    at x = 1/4

    |x-1|+|2x-1|+|3x-1|+|4x-1|+|5x-1|

    = 3/4 + 2/4 + 1/4 + 1/4

    = 7/4.

    Q. |x-1|+|x-5|+|x-9|+|x-13| < = 42.

    solution :-

    first we take extreme value at x-13 = 0

    = > x = 13

    at x = 13.

    |x-1|+|x-5|+|x-9|+|x-13| = 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

    |x-1|+|x-5|+|x-9|+|x-13|

    = 16.5 + 12.5 + 8.5 + 4.5

    = 42.

    max value of x = 17.5

    for minimum value of x-1 = 0

    = > x = 1.

    at x = 1

    |x-1|+|x-5|+|x-9|+|x-13|

    = 24.

    again to increase by 18,we need to reduce value of x by 4.5.

    x = 1 - 4.5 = -3.5

    |x-1|+|x-5|+|x-9|+|x-13|

    = 4.5 + 8.5 + 12.5 + 16.5

    = 42.

    x € [-3.5,17.5]

     

     


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