If y -3 t 2-4 t- then minima of y will be at A- t -3 - 2B- t -3 - 4C- t -2 - 3D- t -4 - 3

The correct option is C t=2/3y=3t^2 4t For y to be minimum: dydt=0 and d^2 ydt^2 gt;0 dydt=6t 4=0 ⇒ t=23 d^2 ydt^2=6 gt;0 Hence, there will be minima at ...

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Standard XII

Physics

Basic Differentiation Rule

If y =3 t 2-4...

Question

If $y = 3 t^{2} - 4 t$ , then minima of $y$ will be at

t = 3 / 2

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t = 3 / 4

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t = 2 / 3

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t = 4 / 3

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Solution

The correct option is C $t = 2 / 3$
$y = 3 t^{2} - 4 t$
For y to be minimum:
$\frac{d y}{d t} = 0 and \frac{d^{2} y}{d t^{2}} > 0$
$\frac{d y}{d t} = 6 t - 4 = 0 \Rightarrow t = \frac{2}{3}$
$\frac{d^{2} y}{d t^{2}} = 6 > 0$
Hence, there will be minima at $t = \frac{2}{3}$

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If y=3t2−4t, then minima of y will be at

The correct option is C t=2/3y=3t2−4t For y to be minimum: dydt=0 and d2ydt2>0 dydt=6t−4=0⇒t=23 d2ydt2=6>0 Hence, there will be minima at t=23

If $y = 3 t^{2} - 4 t$ , then minima of $y$ will be at

The correct option is C $t = 2 / 3$
$y = 3 t^{2} - 4 t$
For y to be minimum:
$\frac{d y}{d t} = 0 and \frac{d^{2} y}{d t^{2}} > 0$
$\frac{d y}{d t} = 6 t - 4 = 0 \Rightarrow t = \frac{2}{3}$
$\frac{d^{2} y}{d t^{2}} = 6 > 0$
Hence, there will be minima at $t = \frac{2}{3}$