Problem 1 :
If f(x) = 2x - 1, g(x) = (x + 1)/2, show that
f o g = g o f = x
Solution :
f o g = f[g(x)] = f[(x + 1)/2] = 2((x + 1)/2) - 1 = x + 1 - 1 = x ----(1) |
g o f = g[f(x)] = g[2x - 1] = (2x - 1 + 1)/2 = 2x/2 = x ----(2) |
From (1) and (2),
f o g = g o f = x
Problem 2 :
If f(x) = x2 - 1, g(x) = x - 2 find a, if g o f(a) = 1.
Solution :
g o f(a) = 1
g[f(a)] = 1
g[a2 - 1] = 1
a2 - 1 - 2 = 1
a2 - 3 = 1
a2 = 4
Take square root on both sides.
a = ±2
Problem 3 :
Find k, if f(k) = 2k - 1 and f o f(k) = 5.
Solution :
f o f(k) = 5
f[f(k)] = 5
f[2k - 1] = 5
2(2k - 1) - 1 = 5
4k - 2 - 1 = 5
4k - 3 = 5
4k = 8
k = 2
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