Question

If ${\log }_{(x+3)}(x^2-x)<1$, then the value of x lies in

• A. $\left(-3,-2\right)$
• B. $\left(-1,0\right)\cup (1,3)$
• C. $\left(-3,-1\right)$
• D. $\left(-1,3\right)$

Answer 1

Answer: (A), (B)

$\left(-1,0\right)\cup (1,3)$

$\mathrm{Given}\mathrm{that},{\log }_{(x+3)}\left(x^2-x\right)<1\mathrm{For}{\log }_{(x+3)}\left(x^2-x\right)\mathrm{to}\mathrm{be}\mathrm{defined},x^2-x>0\Rightarrow x(x-1)>0\Rightarrow x<0\mathrm{or}x>1...\left[1\right]\mathrm{If}x+3>1\Rightarrow x>-2\mathrm{then},x^2-x<x+3\Rightarrow x^2-2x-3<0\Rightarrow \left(x-3\right)\left(x+1\right)<0\Rightarrow -1<x<3....\left[2\right]\mathrm{From}\left[1\right]\mathrm{and}\left[2\right],\mathrm{we}\mathrm{get}x\in \left(-1,0\right)\cup \left(1,3\right)\mathrm{If}0<x+3<1\Rightarrow -3<x<-2...\left[3\right]\mathrm{then},x^2-x>x+3\Rightarrow x^2-2x-3>0\Rightarrow \left(x-3\right)\left(x+1\right)>0\Rightarrow x<-1\mathrm{or}x>3....\left[4\right]\mathrm{From}\left[3\right]\mathrm{and}\left[4\right],\mathrm{we}\mathrm{get}x\in \left(-3,-2\right)$