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Solved Examples

Illustration 1 :

If 2 ≤ 8 x - 1 ≤ 6, then x lies between:
1. 2 ≤ x ≤ 15             2. - 2 ≤ x ≤ - 15        3. 3/8 ≤ x ≤ 7/8                 4. 2/8 ≤ x ≤ 15/8

Solution

Transpose the function to get:
3 ≤ 8x ≤ 7, divide by 8, hence 3/8 ≤ x ≤ 7/8 (dividing by positive quantities)


 

Illustration 2:

If a > 0, b > 0, c < 0, then which of the following is necessarily true?
1. a-3 . b- 3 . c-3 > 0   2. c2/ab < 0      3. a c b > 0       4. cc . a3 < 0

Solution

Best solved by eliminating choices.
If c < 0, then the first choice is clearly wrong, the second choice c2 /ab must be positive as both quantities are positive; the fourth choice may be positive even if c is negative. Hence (3) is the correct answer: a raised to the power of a negative quantity is positive.

 
 

Illustration 3 : 

x2 + 4x > 2x + 8 is true when:
1. x > 4 or x < - 2            2. x > 4 > or x < 2             3. x > 2 or x < - 4              4. x > - 2 or x < 4        

Solution

x2 + 4x > 2x + 8, so x2 + 2x - 8 > 0
Factorise to get: (x + 4) (x - 2) > 0.
Either both quantities must be positive or both quantities must be negative, which is true if x < - 4 or x > 2


 

Illustration 4:

Solve this inequality |y + 9| < 15 for y.

Solution

Take both the sides y + 9 < 15 and y + 9 > - 15
y < 15 – 9 and y >  – 15 – 9 ⇒ y < 6 and y > - 24. - 24 < y < 6.

 
 

Illustration 5:

Solve this inequalities x2 – 7x + 16 < 4 for x.

Solution

Take all the terms given on one side.
x2 – 7x + 16 – 4 < 0 ⇒ x2 – 7x + 12 < 0.
x2 – 4x – 3x + 12 < 0 ⇒ (x – 4) (x – 3) < 0.
Now because their product is less than zero, implies one of these is negative and the other is positive. Now out of the factors, it can be seen that x – 3 is greater than x – 4. So it can be concluded that x – 3 is positive and x – 4 is negative.   
x – 4 < 0 and x – 3 > 0 ⇒ x < 4 and x > 3.
The solution set of the inequality is 3 < x < 4.


 

Illustration 6:

x2 – 11x + 28 >  0

Solution

x2 – 11x + 28 > 0. ⇒  (x – 7)(x – 4) > 0.
Since their product is positive means, either both are positive or both are negative.
Taking both to be positive x – 7 > 0 and x – 4 > 0 ⇒ x > 7 and x > 4.
Now choose that one out of the two, which includes the other also. Anything that is greater than 7, will definitely be greater than 4 also. So x > 7.
Now taking both to be negative x – 7 < 0 and x – 4 < 0 ⇒ x < 7 and x < 4.
Now choose that one out of these two, which includes the other also. Anything that is lesser than 4, will definitely be smaller than 7 also. So x < 4.
So the solution set is x < 4 or x > 7.





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