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Question Number 65930 by gunawan last updated on 06/Aug/19

$$\mathrm{If}t=\frac{x^2-3}{3x+7}\mathrm{then}$$ $$\log (1-\mid t\mid )\mathrm{can}\mathrm{to}\mathrm{find}\mathrm{for}$$ $$\mathrm{a}.2<x<6$$ $$b.-2<x<5$$ $$c.-2⩽x⩽6$$ $$\mathrm{d}.x⩽-2\mathrm{or}x>6$$ $$e.x<-1orx>3$$

Answered by MJS last updated on 06/Aug/19

$$x<-\frac{7}{3}\vee -\sqrt{3}⩽x⩽\sqrt{3}:t⩽0\Rightarrow$$ $$\Rightarrow 1-\mid t\mid =1+t=\frac{x^2+3x+4}{3x+7}$$ $$x<-\frac{7}{3}:1+t<0;x>-\frac{7}{3}:1+t>0$$ $$\Rightarrow \log (1+t)\mathrm{defined}\mathrm{for}-\sqrt{3}⩽x⩽\sqrt{3}$$ $$-\frac{7}{3}<x<-\sqrt{3}\vee \sqrt{3}<xt>0\Rightarrow$$ $$\Rightarrow 1-\mid t\mid =1-t=\frac{-(x-5)(x+2)}{3x+7}$$ $$x<-\frac{7}{3}\vee -2<x<5:1-t>0;$$ $$-\frac{7}{3}<x⩽2\vee 5⩽x:1-t⩽0$$ $$\Rightarrow \log (1-t)\mathrm{defined}\mathrm{for}-2⩽x⩽-\sqrt{3}\vee \sqrt{3}⩽x⩽5$$ $$$$ $$\Rightarrow \log (1-\mid t\mid )\mathrm{defined}\mathrm{for}-2<x<5$$

Commented bygunawan last updated on 06/Aug/19

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

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