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Let g prime(x)> 0 and f prime(x)< 0 AA x...

Let `g prime(x)> 0 and f prime(x)< 0 AA x in R,` then

A

`g(f(x+1))gt g(f(x-1))`

B

`f(g(x-1)) gt f(g(x + 1))`

C

`g(f(x+1)) lt g(f(x-1))`

D

`g(g(x+1))lt g(g(x-1))`

Text Solution

Verified by Experts

The correct Answer is:
C
`because g(x)` is increasing and f(x) is decreasing
`g(x+1)gt g(x-1)` and `f(x+1)lt f(x-1)`
or `f(g(x+1))lt f(g(x-1))`
and `g(f(x+1))lt g(f(x-1))`
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