Mathematics - Functions Question with Solution | TestHub

Domain of ${\sin }^{-1}(\sin \mathrm{x})$ is which of R $-{\log }_{\frac{x+4}{2}}{\log }_2\left(\frac{2x-1}{3+x}\right)>0$ i.e. ${\log }_{\frac{\mathrm{x}+4}{2}}\left({\log }_2\left(\frac{2\mathrm{x}-1}{3+\mathrm{x}}\right)\right)<0$ case-I $0<\frac{\mathrm{x}+4}{2}<1$ i.e. $-4<\mathrm{x}<-2$ then ${\log }_2\frac{2\mathrm{x}-1}{3+\mathrm{x}}>1$ i.e. $\frac{2\mathrm{x}-1}{3+\mathrm{x}}>2$ i.e. $\frac{2\mathrm{x}-1-6-2\mathrm{x}}{3+\mathrm{x}}>0$ i.e. $x+3<0$ i.e. $x<-3$ $\therefore -4<\mathrm{x}<-3$

case-II if $\frac{\mathrm{x}+4}{2}>1$ i.e. $\mathrm{x}>-2$ then $0<{\log }_2\frac{2\mathrm{x}-1}{3+\mathrm{x}}<1$ i.e. $1<\frac{2\mathrm{x}-1}{3+\mathrm{x}}<2$ i.e. $\frac{2\mathrm{x}-1-3-\mathrm{x}}{\mathrm{x}+3}>0$ and $\frac{2\mathrm{x}-1-6-2\mathrm{x}}{\mathrm{x}+3}<0$ 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. $\mathrm{x}>4$...(iii) from (i) and (ii) $\mathrm{x}\in (-4,-3)\cup (4,\infty )$ $\therefore a=-4,b=-3,c=4$ and so $a+b+3c=5$Ans.