A) \[\frac{2}{\sqrt{5}}\]
B) \[\sqrt{5}\]
C) \[\frac{1}{\sqrt{5}}\]
D) \[5\sqrt{5}\]
Correct Answer: C
Solution :
[c] \[y={{x}^{4}}+3{{x}^{2}}+2x\therefore \frac{dy}{dx}=4{{x}^{3}}+6x+2\] Point on curve which is nearest to the line \[y=2x-1\] is the point where tangent to curve is parallel to given line. Therefore, \[4{{x}^{3}}+6x+2=2\] or \[2{{x}^{3}}+3x=0\] or \[x=0,\,\,y=0\]. Therefore, point on the curve at the least distance from the line \[y=2x-1\] is (0, 0). Distance of this point from line is \[\frac{1}{\sqrt{5}}\].
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