A) \[\frac{3}{4}\]
B) \[\frac{6}{5}\]
C) 1
D) None of these
Correct Answer: C
Solution :
Let \[x+y=4\] or \[y=4-x\] \[\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\]or\[f(x)=\frac{4}{xy}\]\[=\frac{4}{x(4-x)}\]; \[f(x)=\frac{4}{4x-{{x}^{2}}}\] Now \[f'(x)=\frac{-4}{{{(4x-{{x}^{2}})}^{2}}}.(4-2x)\] Put \[f'(x)=0\], then \[4-2x=0\] \[\therefore \] \[x=2\] and \[y=2\]; \[\therefore \min \left( \frac{1}{x}+\frac{1}{y} \right)=\frac{1}{2}+\frac{1}{2}=1\].
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