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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $x = (1 + t^2)^{-1}$ and $y = \frac{1}{t(1 + t^2)}$.First,we find $\frac{dx}{dt} = -1(1 + t^2)^{-2} \cdot (2t) = -\frac{2t}{(1 + t^2)^2}$.Ne

Vedclass · Accepted Answer

increasing in $(0, 1)$

Vedclass · Answer

(B) Given $x = (1 + t^2)^{-1}$ and $y = \frac{1}{t(1 + t^2)}$.First,we find $\frac{dx}{dt} = -1(1 + t^2)^{-2} \cdot (2t) = -\frac{2t}{(1 + t^2)^2}$.Next,we find $\frac{dy}{dt}$ using the quotient rule or product rule: $y = t^{-1}(1 + t^2)^{-1}$.$\frac{dy}{dt} = -t^{-2}(1 + t^2)^{-1} + t^{-1}(-1)(1 + t^2)^{-2}(2t) = -\frac{1}{t^2(1 + t^2)} - \frac{2}{(1 + t^2)^2} = -\frac{1 + t^2 + 2t^2}{t^2(1 + t^2)^2} = -\frac{1 + 3t^2}{t^2(1 + t^2)^2}$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-(1 + 3t^2) / [t^2(1 + t^2)^2]}{-2t / (1 + t^2)^2} = \frac{1 + 3t^2}{2t^3}$.Since $t > 0$,$1 + 3t^2 > 0$ and $2t^3 > 0$,so $\frac{dy}{dx} > 0$ for all $t > 0$.As $t 	o \infty$,$x 	o 0$,and as $t 	o 0$,$x 	o 1$. Thus,for $t \in (0, \infty)$,$x \in (0, 1)$.Since $\frac{dy}{dx} > 0$,the function $f$ is increasing 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:

Similar Questions

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27\,^{\circ}C$. If the work done during the process is $3\,kJ$,then the final temperature of the gas is ......$K$ $(C_V = 20\,J/K)$.

The conversion of $O$-acylated phenol to $C$-acylated phenol in the presence of $AlCl_3$ is an example of which type of organic reaction?

Identify the plant shown in the given figure.

If the lines $4x + 3y - 1 = 0$,$x - y + 5 = 0$ and $kx + 5y - 3 = 0$ are concurrent,then $k$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module