1. Functions f and g are such that g-1 (x) = x - 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 - = y
x = y +
g (y) = y +
g (x) = x +
(f ∘ g) (x) = 3x - 1
f (g ) (x) = 3x - 1
f (x + ) = 3x - 1
= 3 (x + ) - 2
Hence f (x ) = 3x - 2
(g -1 ∘ f ) (x) = g -1 ( f (x ) )
= g -1 ( 3x - 2)
= 3x -
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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