Mathematics - Functions Question with Solution | TestHub

Domain of ${\sin }^{-1}(\sin x)$ is which of $R$

$$-{\log }_{\frac{x+4}{2}}{\log }_2\left(\frac{2x-1}{3+x}\right)>0$$

i.e. ${\log }_{\frac{x+4}{2}}\left({\log }_2\left(\frac{2x-1}{3+x}\right)\right)<0$

case-I $0<\frac{x+4}{2}<1$ i.e. $-4<x<-2$

then ${\log }_2\frac{2x-1}{3+x}>1$ i.e. $\frac{2x-1}{3+x}>2$ i.e. $\frac{2x-1-6-2x}{3+x}>0$ i.e. $x+3<0$ i.e. $x<-3$

$\therefore -4<\mathrm{x}<-3$ ...(i)

case-II if $\frac{x+4}{2}>1$ i.e. $x>-2$

then $0<{\log }_2\frac{2x-1}{3+x}<1$ i.e. $1<\frac{2x-1}{3+x}<2$

i.e. $\frac{2x-1-3-x}{x+3}>0$ and $\frac{2x-1-6-2x}{x+3}<0$

i.e. $\frac{x-4}{x+3}>0$ and $\frac{-7}{x+3}<0$

i.e. $\{x<-3$ or $x>4\}$ and $x>-3$ i.e. $x>4$ ...(iii)

from (i) and (ii) $\mathrm{x}\in (-4,-3)\cup (4,\infty )$

$\therefore \mathrm{a}=-4,\mathrm{b}=-3,\mathrm{c}=4$ and so $\mathrm{a}+\mathrm{b}+3\mathrm{c}=5$ Ans.