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Quantitative Aptitude > Inequalities
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A comparison relationship between two algebraic expression or quantities is known as an Inequalities. For example
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.
An Example Question 3x+7 > x+1 Solution 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 3(-1)+7 > -1+1 3(0)+7 > 0+1
Law of inequalities
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 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 In general if x > 1 then
Solving quadratic inequalities
Let us study how the sign of a quadratic expression changes as we very the value of x. let us consider (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 where for x < 1 then x-2 < 0 and x-1 < 0 thus if we have to solve
While Factorizing the expression in linear factor make sure to make the sign of coefficient of
Solved Examples Question
Solve
Solution
Observe for all above expression hence we can divided both sides by
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
Question
Solve
Solution
We know that
Hence
Equality each linear factor equal to zero , we get
Hence if but = -x for -x < 0 above solution is 1 < x < 2 or 1 < -x < -2 or -1 > x > -2 hence following we get solution is
Question
Solve
Solution
Adding -(4 – x) to both side we get
now
Discriminant of
dividing both sides by
Question
If x > 3, y > -2 then which of the following holds good (1) xy > -6 (2) xy < -6 (3) (4)none of these
Solution
We observe that y > -2 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'
Question
Solve If x > 5 and y < -1 , hence which of the following statements is true ? (1)(x + 4y) > 1,
Solution
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 5y c-5 again taking x = 6 and y = - 10 we find that - 24 > -50 hence (3) is also not true Hence answer is (4) .
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7 > 0 hence true
