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Question

# Let gâ€²(x)>0 and fâ€²(x)<0âˆ€xÏµR, then

A
f(f(x+1))>f(f(x1))
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B
f(g(x1))>f(g(x+1))
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C
g(f(x+1))>g(f(x1))
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D
g(g(x+1))>g(g(x1))
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Solution

## The correct options are B f(g(x−1))>f(g(x+1)) C f(f(x+1))>f(f(x−1)) D g(g(x+1))>g(g(x−1))Given g′(x)>0 and f′(x)<0 ⇒g(x) is increasing and f(x) is decreasing.Now option A,Hence, f(x+1)<f(x−1)⇒f(f(x+1))>f(f(x−1))option B,Hence, g(x−1)<g(x+1)⇒f(g(x−1))>f(g(x+1))option C,Hence, f(x+1)<f(x−1)⇒g(f(x−1))>g(f(x+1))option D,Hence, g(x−1)<g(x+1)⇒g(g(x+1))>g(g(x−1))Therefore option A,B,D are correct

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