Question

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

• A. Increasing in $(0,\frac{3}{2})$ & decreasing in $(\frac{3}{2},\infty )$
• B. Increasing in $(0,1)$
• C. Increasing in $(0,\infty )$
• D. Decreasing in $(0,1)$

Answer 1

Answer: (B)

Increasing in $(0,1)$

$\frac{dy}{dx}$ can be calculated as $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
Now, $\frac{dy}{dt}$ $=-\frac{1+3t^2}{t^2(1+t^2{)}^2}$
Similarly $\frac{dx}{dt}$ $=-\frac{2t}{(1+t^2{)}^2}$
Hence, $=\frac{dy}{dx}$ $=\frac{1+3t^2}{2t^3}$
$>0$ for it to be an increasing function.
$1+3t^2>0$
No real value.
Thus $f\left(x\right)>0$ for $t\ge 0$.
This implies $xϵ(0,1)$.