Q103 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics

If $u=\sin \left(\frac{x}{y}\right), x=e^t$ and $y=t^2$, then$t^6\left(\frac{d u}{d t}\right)^2 \div e^{2 t}(t-2)^2=$

  • A
    $2 \mathrm{u}$
  • B
    $u^2$
  • Correct answer
    C
    $1-u^2$
  • D
    $\cos u$

Solution

Given $u=\sin \left(\frac{x}{y}\right), x=e^t, y=t^2 \Rightarrow u=\sin \left(\frac{e^t}{t^2}\right)$ Now, $\frac{d u}{d t}=\cos \left(\frac{e^t}{t^2}\right)\left\{\frac{t^2 e^t-e^t \cdot 2 t}{\left(t^2\right)^2}\right\}$ $\begin{aligned} & =\cos \left(\frac{e^t}{t^2}\right)\left(\frac{t-2}{t^3}\right) e^t \\ & \Rightarrow\left(\frac{d u}{d t}\right)^2=\cos ^2\left(\frac{e^t}{t^2}\right) \frac{(t-2)^2}{t^6} e^{2 t} \\ & \Rightarrow t^6\left(\frac{d u}{d t}\right)^2 \div(t-2)^2 e^{2 t} \\ & =\cos ^2\left(\frac{e^t}{t^2}\right)=1-\sin ^2\left(\frac{e^t}{t^2}\right)=1-u^2 \end{aligned}$

Hence, the correct answer is option: 'C'.

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