• # question_answer If the function $f\left( x \right)=\frac{x-1}{c-{{x}^{2}}+1}$ does not take any value in the internal $\left[ -1,-\frac{1}{3}, \right]$ then the largest integral value that $c$can attain is equal to A) 2 B) 1 C) $-1$ D) 0

 Let $y=f\left( x \right)=\frac{x-1}{c-{{x}^{2}}+1}$ Take $y=-t,$ where $t\in \left[ \frac{1}{3},1 \right],$ $\therefore$      $-t=\frac{x-1}{c-{{x}^{2}}+1}$$\Rightarrow$ ${{x}^{2}}-c-1=\frac{x-1}{t}$$\Rightarrow$            ${{x}^{2}}-\frac{1}{t}x+\frac{1}{t}-c-1=0$
 As$-t\in \left[ -1,-\frac{1}{3} \right],$hence the above must not possess real solution $\therefore$      ${{\left( \frac{1}{t} \right)}^{2}}-4\left( \frac{1}{t}-c-1 \right)<0$$\Rightarrow$$\frac{1}{{{t}^{2}}}-\frac{4}{t}+4<-4c\Rightarrow c<-\frac{1}{4}{{\left( \frac{1}{t}-2 \right)}^{2}}$ Now, $\frac{1}{3}\le t\le l\Rightarrow 1\le \frac{1}{t}-2\le 1$$\Rightarrow$$-\frac{1}{4}\le \frac{1}{4}{{\left( \frac{1}{t}-2 \right)}^{2}}\le 0$$\Rightarrow$$0\le \frac{-1}{4}{{\left( \frac{1}{t}-2 \right)}^{2}}\le \frac{1}{4}$