Question

If $g^{\prime }\left(x\right)>0$ 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), (C), (D)

The correct options are
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 $g\left(g\right(x+1\left)\right)>g\left(g\right(x-1\left)\right)$

Point to be noted is that the Monotonically increasing (decreasing) function does not (does too) changes the sign of the inequality between two numbers.
From the data given the function $f$ is monotonically decreasing, whereas $g$ is monotonically increasing in Real domain. Why? because the first order slopes $f^{\prime }\left(x\right)$ and $g^{\prime }\left(x\right)$ of the functions $f$ and $g$ are, less than and grater than zero respectively.
So as $\forall x:x+1>x-1$
Now $f(x+1)<f(x-1)$ ... that is an inequality sign change, (because ${}^{\prime }{f}^{\prime }$ is monotonically decreasing). Similarly, $g(x+1)>g(x-1)$ i.e no sign change. Now if we apply these functions on both side of the inequality just obtained, one by one, we can see that all the option follows from the given data.