Question

Let $f\left(x\right)$ and $g\left(x\right)$ be two continuous functions defined from $R\to R$, such that $f\left(x_1\right)>f\left(x_2\right)$

and $g\left(x_1\right)<g\left(x_2\right)$ , $also$ $x_1<x_2$, then solution set of $f\left(g\right({\alpha }^2-2\alpha \left)\right)>f\left(g\right(3\alpha -4\left)\right)is$

• A. $R$
• B. $[1,4]$
• C. $(1,4)$
• D. $R-[1,4]$

Answer 1

The correct option is C $(1,4)$

$givenf\left(x1\right)>f\left(x2\right)wherex1<x2g\left(x1\right)<g\left(x2\right)wherex1<x2Now,\Rightarrow f\left(g\right(x1\left)\right)>f\left(g\right(x2\left)\right)sinceg\left(x1\right)<g\left(x2\right)forx1<x2\Rightarrow Forf\left(g\right({\alpha }^2-2\alpha \left)\right)>f\left(g\right(3\alpha -4\left)\right)\Rightarrow {\alpha }^2-2\alpha <3\alpha -4\Rightarrow {\alpha }^2-5\alpha +4<0\Rightarrow {\alpha }^2-4\alpha -\alpha +4<0\Rightarrow (\alpha -4)(\alpha -1)<0\alpha ϵ(1,4)$

Hence, the option is $"C"$