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  • Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 46.

Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 46.

Answers (1)

t = 2

Hint: For maxima and minima value of V, we must have \frac{dV}{dt}=0

Given:

\begin{aligned} &S=\frac{t^{4}}{4}-2 t^{3}+4 t^{2}-7 \\ &V=\frac{d S}{d t}=t^{3}-6 t^{2}+8 t \\ &a=\frac{d V}{d t}=3 t^{2}-12 t+8 \end{aligned}

Solution:

\frac{dV}{dt}=0

3t^2-12t+8 =0

On solving the equatin we get

t=2\pm \frac{2}{\sqrt{3}}

Now,

\begin{aligned} &\frac{d^{2} V}{d t^{2}}=6 t-12 \\ &\text { At } t=2 \pm \frac{2}{\sqrt{3}} \\ &\frac{d^{2} V}{d t^{2}}=6\left(2 \pm \frac{2}{\sqrt{3}}\right)-12,=\frac{-12}{3}<0 \end{aligned}

So, the velocity is maximum at t = 2-\frac{2}{\sqrt{3}}

Again,

\frac{da}{dt}=6t-12

For maximum or minimum value of a

\frac{da}{dt}=0

6t-12=0

t=2

Now,  \frac{d^2a}{dt^2}=6>0

So acceleration is minimum at t = 2

Posted by

infoexpert24

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