A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
Given, \[xy=1\] \[\Rightarrow \] \[y=\frac{1}{x}\] Let \[A=x+y\] \[\Rightarrow \] \[A=x+\frac{1}{x}\] \[\frac{dA}{dx}=1-\frac{1}{{{x}^{2}}}\] Put \[\frac{dA}{dx}=0,\] \[\Rightarrow \] \[1-\frac{1}{{{x}^{2}}}=0\] \[\Rightarrow \] \[{{x}^{2}}=1\] \[\Rightarrow \] \[x=\pm 1\] \[\frac{{{d}^{2}}A}{d{{x}^{2}}}=\frac{2}{{{x}^{3}}}\] At \[x=1,\frac{{{d}^{2}}A}{d{{x}^{2}}}\ge 0\] Hence, at\[x=1,\]A will be minimum. \[\therefore \]Minimum value\[=x+y=1+1=2\]
You need to login to perform this action.
You will be redirected in
3 sec
