MCQ
If $\text{x}=2\text{ at},\text{y}=\text{at}^2,$ where a is a constant, then $\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{ at}\ \text{x}=\frac{1}{2}$ is :
  • $\frac{1}{2}\text{a}$
  • B
    $1$
  • C
    $2a$
  • D
    None of these

Answer

Correct option: A.
$\frac{1}{2}\text{a}$
Here,
$\text{x}=2\text{ at},\text{y}=\text{at}^2,$
Differentiating $\text{w.r.t. x},$ we get
$\frac{\text{dx}}{\text{dt}}=2\text{a}$ and $\frac{\text{dy}}{\text{dt}}=2\text{at}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2\text{at}}{2\text{a}}=\text{t}$
Differentiating $\text{w.r.t. x},$ we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=1\times\frac{\text{dt}}{\text{dx}}=\frac{1}{2\text{a}}$
Now, $\Big[\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big]_{\text{x}=\frac{1}{2}}=\frac{1}{2\text{a}}$

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