A) 1
B) -2
C) 4
D) None of these
Correct Answer: C
Solution :
[c] The equation of the line is \[y-3=\frac{3+2}{0-5}(x-0),i.e.,x+y-3=0\] \[y=\frac{c}{x+1}\] or \[\frac{dy}{dx}=\frac{-c}{{{(x+1)}^{2}}}\] Let the line touch the curve at \[(\alpha ,\beta )\]. Then \[\alpha +\beta -3=0,{{\left( \frac{dy}{dx} \right)}_{\alpha ,\beta }}=\frac{-c}{{{(\alpha +1)}^{2}}}=-1,\] and \[\beta =\frac{c}{\alpha +1}\] \[\therefore \frac{c}{{{(c/\beta )}^{2}}}=1\] or \[{{\beta }^{2}}=c\] or \[{{(3-\alpha )}^{2}}=c={{(\alpha +1)}^{2}}\] or \[3-\alpha =\pm (\alpha +1)\] or \[3-\alpha =\alpha +1\] or \[\alpha =1\]. So, \[c={{(1+1)}^{2}}=4\].
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