A) ? 2
B) 2
C) 1
D) ? 1
Correct Answer: C
Solution :
Let the positive number \[\left( x+\frac{1}{x} \right)\]will be minimum, when \[\frac{dy}{dx}=0\] and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}>0\]. Differentiate with respect to x, we have \[1-\frac{1}{{{x}^{2}}}=0\] Þ \[x=-1,\,1\]and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=+\frac{2}{{{x}^{3}}}\] Þ \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=1}}>0\] So, at \[x=1,\,\left( x+\frac{1}{x} \right)\] will be minimum.
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