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Question Number 66355 by gunawan last updated on 13/Aug/19

$$V\mathrm{alue}\mathrm{of}x\mathrm{satiesfied}y=\frac{{\log }_4(x^2-1)}{4x^2+2x+1}$$ $$negativevalue\mathrm{is}...$$ $$\mathrm{a}.-1<x<\sqrt{2}$$ $$b.-\sqrt{2}<x<1$$ $$c.-\sqrt{2}<x<\sqrt{2}$$ $$d.-\sqrt{2}<x<-1$$ $$e.x<-2$$

Commented bykaivan.ahmadi last updated on 13/Aug/19

$$x^2-1>0\Rightarrow x>1orx<-1$$ $$4x^2+2x+1>0\Rightarrow x\in \mathbb{R}$$ $$\Rightarrow D_y=x>1orx<0$$ $$y<0\Rightarrow {log}_4(x^2-1)<0\Rightarrow x^2-1<1\Rightarrow x^2<2\Rightarrow -\sqrt{2}<x<\sqrt{2}$$ $$sotheansweris-\sqrt{2}<x<-1and1<x<\sqrt{2}$$

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