Problem Study (algebra)

ဒီေန႔ group မွာ ဆရာ ကိုၿဖိဳး ဆရာ ေက်ာ္ေဇျမင့္ တို႔ရဲ့ မွတ္သားဖြယ္ရာ ပုစာၦေတြကို အေျဖႏွင့္တကြ တင္ေပးလိုက္ပါတယ္။ ကၽြန္ေတာ္စဥ္စားမိသလို တြက္ေပးထားတာ ျဖစ္တဲ့အတြက္ ဒီထက္ပိုမို အဆင္ေျပ ေကာင္းမြန္တဲ့ တြက္နည္းမ်ား ရွိမွာ အမွန္ပါ။ ေက်ာင္းသားမ်ားအတြက္၊
ေလ့လာသူအတြက္ တစ္စံုတရာ
အက်ိဳးရွိေစမယ္လို႔ ယံုၾကည္လ်က္....။

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