Question

If $x=\frac{\left(1+t\right)}{t^3},y=\left(\frac{3}{2t^2}\right)+\left(\frac{2}{t}\right)$ satisfying $f\left(x\right)f{\left(y\text{'}\right)}^3=1+y\text{'}$, then $f\left(x\right)$ is

• A. $\frac{x^2}{\left[1+x^2\right]}$
• B. $\frac{\left[x^2+1\right]}{x^2}$
• C. $x+\left(\frac{1}{x}\right)$
• D. $x$

Answer 1

The correct option is D

$x$

Explanation for the correct option.

Given, $x=\frac{\left(1+t\right)}{t^3},y=\left(\frac{3}{2t^2}\right)+\left(\frac{2}{t}\right)$, satisfying $f\left(x\right)f{\left(y\text{'}\right)}^3=1+y\text{'}$.

Step 1: Find the $\frac{dx}{dt}$

Differentiate $\begin{array}{rcl}x& =& \frac{\left(1+t\right)}{t^3}\end{array}$ with respect to $t$.

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{\left[{\displaystyle \frac{d}{dt}\left(1+t\right)\times t^3-\left(1+t\right)\times \frac{d}{dt}{t}^3}\right]}{{\left(t^3\right)}^2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}\frac{\mathbf{u}}{\mathbf{v}}\mathbf{\right)}\mathbf{=}\frac{\left(v{\displaystyle \frac{d}{\mathrm{dx}}}u-u{\displaystyle \frac{d}{\mathrm{dx}}}v\right)}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{]}\\ \frac{dx}{dt}& =& \frac{t^3-\left(1+t\right)\times 3t^2}{t^6}\\ \frac{dx}{dt}& =& -\frac{\left(3+2t\right)}{t^4}\end{array}$

Step 2: Find the $\frac{dy}{dt}$

Differentiate $y=\left(\frac{3}{2t^2}\right)+\left(\frac{2}{t}\right)$ with respect to $t$.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}\left[\left(\frac{3}{2t^2}\right)+\left(\frac{2}{t}\right)\right] \\ \frac{dy}{dt}=\frac{d}{dt}\left(\frac{3}{2t^2}\right)+\frac{d}{dt}\left(\frac{2}{t}\right) \\ \frac{dy}{dt}=\left(\frac{0\times 2t^2-3\times 4t}{{\left(2t^2\right)}^2}\right)+\left(\frac{0\times t-2\times 1}{t^2}\right)\begin{array}{rcl}& & \mathbf{}\end{array}\begin{array}{rcl}& & \mathbf{}\end{array}\begin{array}{rcl}& & \mathbf{}\end{array}\begin{array}{rcl}& & \mathbf{}\end{array}\begin{array}{rcl}& & \mathbf{}\end{array}\begin{array}{rcl}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}\frac{\mathbf{u}}{\mathbf{v}}\mathbf{\right)}& \mathbf{=}& \frac{\left(v{\displaystyle \frac{d}{\mathrm{dx}}}u-u{\displaystyle \frac{d}{\mathrm{dx}}}v\right)}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{]}\end{array} \\ \frac{dy}{dt}=-\left(\frac{12t}{4t^4}\right)-\left(\frac{2}{t^2}\right) \\ \frac{dy}{dt}=-\left(\frac{3}{t^3}+\frac{2}{t^2}\right) \\ \frac{dy}{dt}=-\frac{\left(3+2t\right)}{t^3} \end{gathered}$$

Step 3: Find the $\frac{dy}{dx}$

Now divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$.

$$\begin{gathered} \frac{dy}{dx}=t.\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ x=\frac{1+t}{t^3} \\ x{t}^3=1+t \\ x{\left(\frac{dy}{dx}\right)}^3=1+\frac{dy}{dx}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{F}\mathbf{r}\mathbf{o}\mathbf{m}\mathbf{}\mathbf{e}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)\mathbf{]} \end{gathered}$$

Compare this value with $f\left(x\right)f{\left(y\text{'}\right)}^3=1+y\text{'}$.

$f\left(x\right)=x$

Hence, the correct option is D.