Preliminary Concept - II

1. Concept of Brackets

{ }--- value is the solution, e.g. {3,4} => x =3 and 4

( )----- open interval , end points excluded

e.g. (3 4) => 3 < x < 4

[ ] ---- close interval, end points included

e.g. [3 4] => 

2. Number System:

3. Every number is a complex number

0 is both real as well as imaginary ( 0 + i0 )

4. Rational No. - p/q form =>   and p and q both are integers

                           and p/q =    recurring decimal --- rational

                                              non-recurring decimal --- irrational

5.  and e both are irrational

6. Standard Theorem :- Between two consecutive rational number there is only one irrational number and vice versa.

7. Real Line Method :

i) Algebraic equality & inequality

ii) Modulus equality & inequality

a. 

how to solve this inequality ,

the product of (x-1) and (x-2) is negative

=> either (x-1) is negative or (x-2) is negative

both cant be same sign when the product is negative

take (x-1), if x < 1 , x - 1 is negative , e.g. x = 0 < 1 => 0-1=-1 < 0 (-ve) 

and for x > 1 , (x - 1) is positive, e.g. take x = 2 => 2-1 = 1 >0 (+ve)

means here behaviour changing point is '1'.

same is for (x-2) , behaviour changing point is '2'.

clearly the product is negative when  ends point i.e x= 1 and x = 2 is not included because at x = 1 and x = 2 the product is zero and in question it is the inequality is strictly less than zero not equal to zero.

b. 

here, 1, -2 and -4 are behaviour changing points of (x-1), (x+2) and (x+4) respectively.

since denominator cannot be zero
therefore,  => 
and x = 1 and x= -2 because for these value expression is zero

 from above figure the solution is 
for infinity we always use open bracket .
since   , therefore open bracket is used in the above solution and x = -2 and x = 1 can be the solution of the above inequality , therefore closed bracket is used.

8. The more simple method to solve this type of inequality is wavy curvy method.

in this method first we find the behaviour changing points.
As we observed from above figures the behaviour of any expression is changing alternatively with respect to behaviour changing points.

for above inequality, 

 

there are three behaviour changing points -4 , -2 and 1. In wavy curvy method, first we observe the value for x > 1, certainly it is positive and then we draw the curve alternate positive and negative with behaviour changing points as shown above.

immediately we got the solution 

c.  

here 2 is not behaviour changing point, because   is always positive for any value of x.
therefore it cant change the sign of equality.

*Always remember if any term is always positive, it will not be the behaviour changing point.

but x = 2 is right , because on 2 inequality is satisfied i.e. whole expression gives the value 0 and ' is true

and also denominator cant be equal to zero.
=> 

here behaviour changing points are -1 and -5, 
because term (x + 1) is negative when x < -1 and positive when x > -1

 and similarly the term (x + 5) is negative when x < -5 and positive when x > -5.

clearly the solution of above inequality is 
we can also solve it by wavy curvy method.
since , therefore bracket is open on -5 but closed on -1 and dont forget x =2 was the solution of inequality because it will give the value 0.

d. 

here  is always positive , therefore it will not be a behaviour changing point.
and since denominator cant be zero
therefore 

-3 and -2 are behaviour changing points,
by wavy curvy method, right hand side is positive, then follow the alternate signs

therefore the solution is  , both the end points included because inequality is less than or equal to zero.