Let f(x) and g(x) be two continuous functions | vedclass

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(C) Given that $f(x_1) > f(x_2)$ for all $x_1 > x_2$,$f$ is a strictly increasing function.Given that $g(x_1)  x_2$,$g$ is a s

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$(1, 4)$

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(C) Given that $f(x_1) > f(x_2)$ for all $x_1 > x_2$,$f$ is a strictly increasing function.Given that $g(x_1)  x_2$,$g$ is a strictly decreasing function.Since $f$ is strictly increasing,the inequality $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ implies $g(\alpha^2 - 2\alpha) > g(3\alpha - 4)$.Since $g$ is strictly decreasing,the inequality $g(\alpha^2 - 2\alpha) > g(3\alpha - 4)$ implies $\alpha^2 - 2\alpha Rearranging the terms,we get $\alpha^2 - 5\alpha + 4 Factoring the quadratic expression,we have $(\alpha - 1)(\alpha - 4) This inequality holds when $\alpha$ lies between the roots $1$ and $4$.Thus,the solution set is $\alpha \in (1, 4)$.

Let $f(x)$ and $g(x)$ be two continuous functions defined from $R \rightarrow R$,such that $f(x_1) > f(x_2)$ and $g(x_1) < g(x_2)$ for all $x_1 > x_2$. Then the solution set of $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ is

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