Free Online Preparation for CAT 2009 with Minglebox e-CAT Prep. Cover basic concepts of Inequalities under Quantitative Aptitude for MBA Entrance Exam Preparation with Study material, solved examples and tests prepared by CAT coaching experts.
A comparison relationship between two algebraic expression or quantities is known as an Inequalities.
In any inequality problem variable x is assumed to be real number unless otherwise . Any inequality is generally either one of the following
'greater then' (>)
'greater than or equal to '( >)
'less than' (<)
' less than or equal to' ( < )
Solution of an inequalities
By solution of an inequalities we seek to find a set of value for the variable involve in the problem so that inequalities holds true for all the values lying in the solution set.
3x+7 > x+1
We find that any value of x > -3 is a solution of the above inequalities. For instance
let us take x = -2, -1, 0, .5 etc. We find
3(-2)+7 > -2+1 1 > -1 hence true
3(-1)+7 > -1+1 4 > 0 hence true
3(0)+7 > 0+1 7 > 0 hence true
if x > y then for all c x+c > y+c and x-c > y-c
If x > y and c > 0 then cx > cy and
If x > y then for c < 0 , xc < yc and that is sign of inequalities gets reversed when both sides of
the inequalities are multiplied or divided by the same negative quantity. If x > y >0 then
8 > 2 > 0
also if 0 > x > y then
but x> 0 > y then
Relationship Between Power of a Number
If x > 1 then …. ............. …........
but if 0< 1 then …. .....
In general if x > 1 then if m> n and if 0 < x < 1 then if m < n
Solving quadratic inequalities
Let us study how the sign of a quadratic expression changes as we very the value of x.
let us consider that is
(x-2) (x-1) for x > 2 we find x-2 > 0 and x-1>0 therefore for all x >2 (x-2) (x-1) > 0
for 1 < x < 2 x-2 < 0 but x-1 > 0 (x-2) (x-1) < 0
where for x < 1 then x-2 < 0 and x-1 < 0 (x-2) (x-1) < 0
thus if we have to solve > 0 we proceed as follows.
While Factorizing the expression in linear factor make sure to make the sign of coefficient of as positive.
Observe for all and for , > 0. Therefore is the positive factor of the
above expression hence we can divided both sides by we get also
we again observe for all and for
Hence we just need to consider (x-1) (x-3) for solving inequalities now let's equal each linear term equal to zero we get x = 1 and x = 3 therefore for all x >3 (x-1) > 0 and (x-3) > 0 hence (x-1) (x-3) > 0
also for x < 1 , (x-1) < 0 and (x-3) < 0 hence again (x-1) (x-3) > 0 therefore solution set is given by
Solve - |x| + 2 < 0
We know that for all x > 0
Hence become factorizing into linear factor we get
Equality each linear factor equal to zero , we get
Hence if then
but for x > 0 and
= -x for -x < 0
above solution is 1 < x < 2 or 1 < -x < -2 or -1 > x > -2 hence following we get solution is
Adding -(4 – x) to both side we get
now for all therefore are multiply both side by we get
Discriminant of is given by
= 4 – 48 = -44 < 0
for all real value of 2.
dividing both sides by as the solution . Hence is given by
If x > 3, y > -2 then which of the following holds good
(1) xy > -6
(2) xy < -6
(4)none of these
We observe that y > -2 y can be positive as well as negative that is if y > -2 y can value living between 0 > y > -2 as well as y > 0
Hence we solve this using method of options let x = 4 and y = -1
then xy = -4 > -6 therefore 'a' is true but 'b' is not also 4 > 3 hence c also holds let us take x = 5 and
y = - ½ we get
hence '3' is not trueFinally if we take x = 18 & y = -1/2 we get
hence 'a' also does not s good
Hence answer is '4'
Solve If x > 5 and y < -1 , hence which of the following statements is true ?
(1)(x + 4y) > 1,
(x > -4y)
(3)-4x < 5y
none of these
Let us go by each option and take specified values of x and y
If x = 6 and y = -10
Then x+4y = 6 + 4(-10) = -34 < 1 hence (1) is not true
also if y < 1
-4y > 4
but x > 5 therefore taking x =6 , y = -10 we get 6 < 40 hence (2) is also not true
From (3) we find that
x > 5
-4x < -20
as y < -1
again taking x = 6 and y = - 10
we find that - 24 > -50 hence (3) is also not true
Hence answer is (4) .