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  • Need solution for R.D. Sharma maths class 12 chapter 17 Maxima and Minima Exercise 17.3 Question 1 Sub Question 9.

Need solution for R.D. Sharma maths class 12 chapter 17 Maxima and Minima Exercise 17.3 Question 1 Sub Question 9.

Answers (1)

Answer:

 Point of local maxima value is \frac{a}{3}and it’s local maximum value is \frac{4 a^{8}}{27}. Also point of local minima is a and it’s local minimum value is  0.

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If f''(c) >0 then c is point of local minima.

If f’’(c)<0 then c is point of local maxima .

where c &c are critical points.

Put c and c in f(x) to get minimum value & maximum value.

Given:

f(x)=x^{3}-2 a x^{2}+a^{2} x, a>0

Explanation:

We have,

f(x)=x^{3}-2 a x^{2}+a^{2} x

\begin{aligned} &f^{\prime}(x)=3 x^{2}-4 a x+a^{2} \\ &f^{\prime \prime}(x)=6 x-4 a \end{aligned}

For maxima and minima f’(x) =0

\begin{aligned} &3 x^{2}-4 a x+a^{2}=0 \\ &x=\frac{-(-4 a) \pm \sqrt{(-4 a)^{2}-4 \times 3 \times a^{2}}}{2 \times 3} \\ &\quad=\frac{4 a \pm \sqrt{16 a^{2}-12 a^{2}}}{6} \\ &\quad=\frac{4 a \pm 2 a}{6} \\ &x=a \text { or } x=\frac{a}{3} \end{aligned}

At x= a,

\begin{aligned} f^{\prime \prime}(a) &=6(a)-4 a \\ &=2 a>0 \end{aligned}

So, x=a is point of local minima

Now at x=a,

\begin{aligned} f(a) &=a^{3}-2 a(a)^{2}+a^{2}(a) \\ &=a^{3}-2 a^{3}+a^{3}=0 \end{aligned}

Also, at \mathrm{x}=\frac{a}{3}

\begin{aligned} f^{\prime \prime}\left(\frac{a}{3}\right) &=6\left(\frac{a}{3}\right)-4(a) \\ &=-2 a<0 \end{aligned}

So, \frac{a}{3} is a point of local maxima.

At x=\frac{a}{3},

\begin{aligned} f\left(\frac{a}{3}\right) &=\left(\frac{a}{3}\right)^{2}-2 a\left(\frac{a}{3}\right)+a^{2}\left(\frac{a}{3}\right) \\ &=\frac{4 a^{8}}{27} \end{aligned}

Thus, point of local maxima is \frac{a}{3} & its max. value is \frac{4 a^{8}}{27} & point of local minima is a & its value is 0.

Posted by

infoexpert21

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