Q103 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

If $y=t^2+t^3$ and $x=t-t^4$ then $\frac{d^2 y}{d x^2}$ at $t=1$ is

  • A
    $-\frac{2}{3}$
  • Correct answer
    B
    $-\frac{4}{3}$
  • C
    $\frac{8}{3}$
  • D
    4

Solution

$y=t^2+t^3$ $\begin{aligned} & x=t-t^4 \Rightarrow \frac{d y}{d t}=2 t+3 t^2 \\ & \frac{d y}{d t}=2+6 t \text { and } \frac{d x}{d t}=1-4 t^3 \\ & \frac{d^2 x}{d t^2}=-12 t^2\end{aligned}$ So, $\frac{d y}{d x}=\frac{2 t+3 t^2}{1-4 t^3}$ Differentiate w.r.t. $x$ $\begin{aligned} & \frac{d^2 y}{d x^2}=\frac{\left(1-4 t^3\right)(2+6 t)-\left(2 t+3 t^2\right)\left(-12 t^2\right)}{\left(1-4 t^3\right)^2} \times \frac{d t}{d x} \\ & \frac{d^2 y}{d x^2}=\frac{\left(1-4 t^3\right)(2+6 t)+12 t^2\left(2 t+3 t^2\right)}{\left(1-4 t^3\right)^3} \\ & \left.\frac{d^2 y}{d x^2}\right|_{t=1}=\frac{(1-4)(2+6)+12(2+3)}{(1-4)^3} \\ & =\frac{(-3)(8)+12(5)}{(-3)^3}=\frac{36}{-27}=\frac{-4}{3} \end{aligned}$

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

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