A function y = f(x) is given by x = frac{1}{1 | vedclass

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(B) Given $x = \frac{1}{1+t^2}$ and $y = \frac{1}{t(1+t^2)}$.Since $t > 0$,$1+t^2 > 1$,so $0 < x < 1$.We have $1+t^2 = \frac{1}{x}$,which implies $t^2

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increasing in $(0, 1)$

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(B) Given $x = \frac{1}{1+t^2}$ and $y = \frac{1}{t(1+t^2)}$.Since $t > 0$,$1+t^2 > 1$,so $0 We have $1+t^2 = \frac{1}{x}$,which implies $t^2 = \frac{1-x}{x}$.Also,$y = \frac{1}{t(1+t^2)} = \frac{1}{t} \cdot x$.Squaring both sides: $y^2 = \frac{x^2}{t^2} = \frac{x^2}{(1-x)/x} = \frac{x^3}{1-x}$.Since $t > 0$,$y > 0$,so $y = \sqrt{\frac{x^3}{1-x}}$.Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{x^3}{1-x}}} \cdot \frac{(1-x)(3x^2) - x^3(-1)}{(1-x)^2} = \frac{1}{2y} \cdot \frac{3x^2 - 3x^3 + x^3}{(1-x)^2} = \frac{3x^2 - 2x^3}{2y(1-x)^2}$.For $f$ to be increasing,$\frac{dy}{dx} > 0$.Since $y > 0$ and $(1-x)^2 > 0$,we need $3x^2 - 2x^3 > 0$.$x^2(3 - 2x) > 0$.Since $x^2 > 0$ for $x \in (0, 1)$,we need $3 - 2x > 0$,which means $x Since $x$ is restricted to $(0, 1)$,the function is increasing for all $x \in (0, 1)$.

$A$ function $y = f(x)$ is given by $x = \frac{1}{1 + t^2}$ and $y = \frac{1}{t(1 + t^2)}$ for all $t > 0$. Then $f$ is:

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