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Let f : R to R :f (x) = x ^(2) g : R to...

Let `f : R to R :f (x) = x ^(2) g : R to R : g (x) = x + 5,` then gof is:

A

`(x + 5)`

B

`(x + 5 ^(2))`

C

`(x ^(2) + 5 )`

D

`(x + 5) ^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
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