JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 3)
If $$\mathrm{t}=\sqrt{x}+4$$, then $$\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)_{\mathrm{t}=4}$$ is :
4
zero
8
16
Explanation
Given,
$t=\sqrt{x}+4$, Squaring on both
$$ \begin{aligned} & x=(t-4)^2=t^2-8 t+16 \\\\ & \frac{d x}{d t}=2 t-8\\\\ & \text { at } t=4 \\\\ & \frac{d x}{d t}=8-8=0 \end{aligned} $$
$t=\sqrt{x}+4$, Squaring on both
$$ \begin{aligned} & x=(t-4)^2=t^2-8 t+16 \\\\ & \frac{d x}{d t}=2 t-8\\\\ & \text { at } t=4 \\\\ & \frac{d x}{d t}=8-8=0 \end{aligned} $$
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