Practicing Success
Let $f'(x)>0$ and $g'(x)<0$ for all $x \in R$. Then, (a) $f\{g(x)\}>f\{g(x+1)\}$ |
(a), (b) (a), (c) (b), (c) (b), (d) |
(a), (c) |
We have, $f^{\prime}(x)>0$ and $g^{\prime}(x)<0$ for all $x \in R$. $\Rightarrow f(x)$ is an increasing function and $g(x)$ is a decreasing function on $R$. $\Rightarrow f(x-1)<f(x)<f(x+1)$ and $g(x-1)>g(x)>g(x+1)$ for all $x \in R$. $\Rightarrow g\{f(x-1)\}>g\{f(x)\}>g\{f(x+1)\}$ and, $f\{g(x)\}>f\{g(x)\}>f\{g(x+1)\}$ for all $x \in R$ $\Rightarrow g\{f(x)\}>g\{f(x+1)\}$ and, $f\{g(x)\}>f\{g(x+1)\}$ for all $x \in R$. Hence, options (a) and (c) are correct. |