Question

plz answer 15,16 and 17 with explanation

Answer 1

ques 15 gonna b lngthy

we have
 
|2x-3| < |x+2|
critical points :  x = 3/2 and x = -2
case 1: 

x < -2

then |2x-3| = -(2x-3) = -2x + 3  nd  |x+2| = - (x+2) = -x - 2 

⇒ -2x + 3 <  -x-2  ⇒  5 <  x

now x  > 5  does not satisfy x ∈ (-∞ , -2 ) therefore results  in Ф hence case 1 fails.


case2:

-2 ≤ x <3/2

|2x-3| = - (2x-3) = -2x + 3  nd    |x+2| = x+2

⇒ -2x+3 <  x + 2  ⇒  1 < 3x ⇒ 1/3 < x

now x > 1/3 lies interval  x ∈  [-2 , 3/2)

therefore case 2 is true and intersection gives x ∈ (1/3, 3/2)

case3:  3/2 ≤ x

|2x-3|  = 2x-3    nd    |x+2|  = x+2

⇒ 2x-3 <  x+2  ⇒  x < 5 
since x < 5  lies interval  x  ∈ [3/2, ∞)

above case too true and  x ∈  [3/2, 5 )


conclusion : union of all cases gives x ∈ (1/3, 5)
option b hope so right 

ques 16 in same manner

ques 17
 |x + 1/x| =  |x² + 1 / x | > 2 
case 1 x < 0
solving gives  (x+1)² / x < 0  ⇒  x< 0

case 2 x ≥0
 solving gives (x-1)² / x > 0 ⇒  x> 0

please solve on your own if any queries ask me