Problems Study

 

1. Functions f and g are such that  g-1 (x) = x  -   1 3  and (f ∘ g ) (x) = 3x - 1.
    Find (g -1 ∘ f ) (x) where x ∈ R.

   Solution 

   Let g -1(x ) = y then g (y) = x.
   Hence x - 1 3 = y
            x = y + 1 3
            g (y) = y + 1 3
            g (x) = x + 1 3
           (f ∘ g) (x) = 3x - 1
           f (g ) (x)   = 3x - 1
           f (x + 1 3)  = 3x - 1
                          = 3 (x + 1 3) - 2
   Hence f (x ) = 3x - 2
           (g -1 ∘ f ) (x) = g -1 ( f   (x ) )
                             = g -1 ( 3x - 2)
                             = 3x - 7 3

2. If x, y and z are any three consecutive even numbers, Show that x2 + y2 + z2 = 3y2 + 8.

   Solution 

   Let x = 2a where a is an integer.
   Since x, y and z are any three consecutive even numbers,
   y = 2a + 2 and z = 2a + 4
   x2 + y2 + z2 = (2a)2 + (2a + 2)2 + (2a + 4)2 
                     = 4a2 + 4a2 + 8a + 4 + 4a2 + 16a + 16
                     = 12a2 + 24a + 20
                     = 12a2 + 24a + 12 + 8
                     = 3 (4a2 + 8a + 4) + 8 
                     = 3 (2a + 2)2 + 8 
                     = 3 y2 + 8