Problems Study

 

1. Functions f and g are such that  g^-1 (x) = x  -${\displaystyle \frac{1}{3 }}$ 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 - ${\displaystyle \frac{1}{3}}$ = y
            x = y + ${\displaystyle \frac{1}{3}}$
            g (y) = y + ${\displaystyle \frac{1}{3}}$
            g (x) = x + ${\displaystyle \frac{1}{3}}$
           (f ∘ g) (x) = 3x - 1
           f (g ) (x)   = 3x - 1
           f (x + ${\displaystyle \frac{1}{3}}$)  = 3x - 1
                          = 3 (x + ${\displaystyle \frac{1}{3}}$) - 2
   Hence f (x ) = 3x - 2
           (g ^-1 ∘ f ) (x) = g ^-1 ( f   (x ) )
                             = g ^-1 ( 3x - 2)
                             = 3x - ${\displaystyle \frac{7}{3}}$

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

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