Question

Let $f^{\prime }\left(x\right)>0$ and $g^{\prime }\left(x\right)<0$ for all $xϵR.$ Then

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

Answer 1

Answer: (A), (C)

The correct options are
A $f\left\{g\left(x\right)\right\}>f\left\{g(x+1)\right\}$
C $g\left\{f\left(x\right)\right\}>g\left\{f(x+1)\right\}$

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