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Let $f$ and $g$ be two differentiable functions on $R$ such that $f^{\prime}(x)>0$ and $g^{\prime}(x) < 0$ for all $x \in R$. Then for all $\mathrm{x}$ :
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Verified Answer
The correct answer is:
$f(g(x))>f(g(x+1))$
$f(g(x))>f(g(x+1))$
Since $f^{\prime}(x)>0$ and $g^{\prime}(x) < 0$, therefore $f(x)$ is increasing function and $g(x)$ is decreasing function.
$\Rightarrow f(x+1)>f(x)$ and $g(x+1) < g(x)$
$\Rightarrow g[f(x+1)] < g[f(x)]$ and $f[g(x+$
1) $] < f[g(x)]$
Hence option (b) is correct.
$\Rightarrow f(x+1)>f(x)$ and $g(x+1) < g(x)$
$\Rightarrow g[f(x+1)] < g[f(x)]$ and $f[g(x+$
1) $] < f[g(x)]$
Hence option (b) is correct.
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