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Question Number 66355 by gunawan last updated on 13/Aug/19
Value of x satiesfied y=((log_4 (x^2 −1))/(4x^2 +2x+1))  negative value is...  a. −1<x<(√2)  b. −(√2)<x<1  c. −(√2)<x<(√2)  d. −(√2)<x<−1  e. x<−2
$${V}\mathrm{alue}\:\mathrm{of}\:{x}\:\mathrm{satiesfied}\:{y}=\frac{\mathrm{log}_{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} \\ $$$${negative}\:{value}\:\mathrm{is}… \\ $$$$\mathrm{a}.\:−\mathrm{1}<{x}<\sqrt{\mathrm{2}} \\ $$$${b}.\:−\sqrt{\mathrm{2}}<{x}<\mathrm{1} \\ $$$${c}.\:−\sqrt{\mathrm{2}}<{x}<\sqrt{\mathrm{2}} \\ $$$${d}.\:−\sqrt{\mathrm{2}}<{x}<−\mathrm{1} \\ $$$${e}.\:{x}<−\mathrm{2} \\ $$
Commented by kaivan.ahmadi last updated on 13/Aug/19
x^2 −1>0⇒x>1 or x<−1  4x^2 +2x+1>0⇒x∈R  ⇒D_y =x>1 or x<0  y<0⇒log_4 (x^2 −1)<0⇒x^2 −1<1⇒x^2 <2⇒−(√2)<x<(√2)  so the answer is  −(√2)<x<−1 and 1<x<(√2)
$${x}^{\mathrm{2}} −\mathrm{1}>\mathrm{0}\Rightarrow{x}>\mathrm{1}\:{or}\:{x}<−\mathrm{1} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}>\mathrm{0}\Rightarrow{x}\in\mathbb{R} \\ $$$$\Rightarrow{D}_{{y}} ={x}>\mathrm{1}\:{or}\:{x}<\mathrm{0} \\ $$$${y}<\mathrm{0}\Rightarrow{log}_{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)<\mathrm{0}\Rightarrow{x}^{\mathrm{2}} −\mathrm{1}<\mathrm{1}\Rightarrow{x}^{\mathrm{2}} <\mathrm{2}\Rightarrow−\sqrt{\mathrm{2}}<{x}<\sqrt{\mathrm{2}} \\ $$$${so}\:{the}\:{answer}\:{is}\:\:−\sqrt{\mathrm{2}}<{x}<−\mathrm{1}\:{and}\:\mathrm{1}<{x}<\sqrt{\mathrm{2}} \\ $$

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