A) 2
B) 1
C) \[-1\]
D) 0
Correct Answer: D
Solution :
| 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, \[\frac{1}{3}\le t\le l\Rightarrow 1\le \frac{1}{t}-2\le 1\]\[\Rightarrow \]\[-\frac{1}{4}\le \frac{1}{4}{{\left( \frac{1}{t}-2 \right)}^{2}}\le 0\]\[\Rightarrow \]\[0\le \frac{-1}{4}{{\left( \frac{1}{t}-2 \right)}^{2}}\le \frac{1}{4}\] |
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