A) -1
B) 3
C) -3
D) 1
Correct Answer: C
Solution :
[c] \[f(x)={{x}^{2}}+bx-b;f'(x)=2x+b\] |
\[\Rightarrow f'(1)=b+2\] |
Equation of tangent: \[y-1=(b+2)(x-1)\] |
Putting \[x=0\Rightarrow y=1-b-2=-b-1>0\] |
\[\Rightarrow b<-1\] |
Putting \[y=0\Rightarrow x-1=-\frac{1}{b+2}\Rightarrow x=\frac{-1}{b+2}+1\] |
\[=\frac{b+1}{b+2}>0\Rightarrow b<-2\] or \[b>-1\] |
Combining, the two conditions \[=b<-2\] |
Now, \[\frac{1}{2}\left| -b-1 \right|\left| \frac{b+1}{b+2} \right|=2;{{(b+1)}^{2}}=4\left| b+2 \right|\] |
\[=-4b-8\] |
\[\Rightarrow {{(b+3)}^{2}}=0\Rightarrow b=-3\] follows the condition \[b<-2\] |
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