A) \[y=\frac{1}{4}\]
B) \[y=4\]
C) \[y=\frac{1}{2}\]
D) \[y=0\]
E) \[y=2\]
Correct Answer: A
Solution :
Curve, \[y=\frac{1}{{{x}^{2}}+2x+5}\] ?.(i) Let the equation of line which is parallel to\[x-\]axis is, \[y=c\] ...(ii) The line (ii) is a tangent to curve (i), then slope of curve = slope of line \[\frac{-(2x+2)}{{{({{x}^{2}}+2x+5)}^{2}}}=0\] \[\left( \because \frac{dy}{dx}=\frac{-(2x+2)}{{{({{x}^{2}}+2x+5)}^{2}}} \right)\] \[\Rightarrow \] \[x=-1\] From Eq. (i), \[y=\frac{1}{1-2+5}=\frac{1}{4}\] From Eq. (ii), \[c=\frac{1}{4}\] Hence, the required equation of line is,\[y=\frac{1}{4}.\]
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