Question -
Answer -
Given that f (x) = 2x+ 1
⇒ f= {(1, 2(1)+ 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) +1)}
={(1, 3), (2, 5), (3, 7), (4, 9)}
Also given that g(x) =x2−2
⇒ g= {(3, 32−2), (5, 52−2), (7, 72−2), (9, 92−2)}
={(3, 7), (5, 23), (7, 47), (9, 79)}
Clearly f and g are bijections and, hence, f−1: B→A and g−1: C→ B exist.
So, f−1= {(3, 1), (5, 2), (7, 3), (9, 4)}
And g−1= {(7, 3), (23, 5), (47, 7), (79, 9)}
Now, (f−1 o g−1): C→A
f−1 o g−1 ={(7, 1), (23, 2), (47, 3), (79, 4)}……….(1)
Also, f: A→B and g: B → C,
⇒ gof: A → C, (gof) −1 : C→A
So, f−1 o g−1and (gof)−1 have same domains.
(gof) (x) =g (f (x))
=g (2x + 1)
=(2x +1 )2−2
⇒ (gof) (x) = 4x2 + 4x +1 − 2
⇒ (gof) (x) = 4x2+ 4x −1
Then, (gof) (1) = g (f (1))
= 4 + 4 − 1
=7,
(gof) (2) =g (f (2))
= 4(2)2 +4(2) – 1 = 23,
(gof) (3) =g (f (3))
= 4(3)2 +4(3) – 1 = 47 and
(gof) (4) =g (f (4))
= 4(4)2 +4(4) − 1 = 79
So, gof ={(1, 7), (2, 23), (3, 47), (4, 79)}
⇒ (gof)– 1 ={(7, 1), (23, 2), (47, 3), (79, 4)}…… (2)
From (1) and (2), we get:
(gof)−1 = f−1 o g−1