Question

Let $g^{\prime }\left(x\right)>0$ and $f^{\prime }\left(x\right)<0\forall xϵR$, then

• A. $f\left(f\right(x+1\left)\right)>f\left(f\right(x-1\left)\right)$
• B. $f\left(g\right(x-1\left)\right)>f\left(g\right(x+1\left)\right)$
• C. $g\left(f\right(x+1\left)\right)>g\left(f\right(x-1\left)\right)$
• D. $g\left(g\right(x+1\left)\right)>g\left(g\right(x-1\left)\right)$

Answer 1

Answer: (A), (B), (D)

$f\left(g\right(x-1\left)\right)>f\left(g\right(x+1\left)\right)$
$f\left(f\right(x+1\left)\right)>f\left(f\right(x-1\left)\right)$
$g\left(g\right(x+1\left)\right)>g\left(g\right(x-1\left)\right)$

Given $g^{\prime }\left(x\right)>0$ and $f^{\prime }\left(x\right)<0$
$\Rightarrow g\left(x\right)$ is increasing and $f\left(x\right)$ is decreasing.
Now option A,
Hence, $f(x+1)<f(x-1)$
$\Rightarrow f\left(f\right(x+1\left)\right)>f\left(f\right(x-1\left)\right)$
option B,
Hence, $g(x-1)<g(x+1)$
$\Rightarrow f\left(g\right(x-1\left)\right)>f\left(g\right(x+1\left)\right)$
option C,
Hence, $f(x+1)<f(x-1)$
$\Rightarrow g\left(f\right(x-1\left)\right)>g\left(f\right(x+1\left)\right)$
option D,
Hence, $g(x-1)<g(x+1)$
$\Rightarrow g\left(g\right(x+1\left)\right)>g\left(g\right(x-1\left)\right)$
Therefore option A,B,D are correct