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A function y=f(x) is given by x=1/(1+t^2...

A function `y=f(x)` is given by `x=1/(1+t^2)` and `y=1/(t(1+t^2))` for all `tgt0` then `f` is

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The correct Answer is:
Increasing
`x=(1)/(1+t^(2)) and y=(1)/(t)(1+t^(2))`
`therefore (dx)/(dt)=-(2t)/(1+t^(2)(2)),(dy)/(dt) = (1+3t^(2))/(t^(2))(1+t^(2)2)`
`therefore (dy)/(dx)=(1+3t^(2))/(2t^(3))`
`(dy)/(dx)=(1+3t^(2))/(2t^(3))`
`(dy)/(dx)gt if tgt0 rarr x =(1)/(1+t^(2))in (0,1)`
Hence f(x) is an increasing funtion.
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