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  • Please solve rd sharma class 12 Chapter 17 Maxima and Minima excercise 17.3 part 1question 10 maths textbook solution

Please solve rd  sharma class 12 Chapter 17  Maxima and Minima excercise 17.3  part 1 question 9  maths textbook solution

Answers (1)

Answer:

 Point of local maxima value is \frac{a}{3} and it’s local maximum value is \frac{4 a^{3}}{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^{\prime \prime}\left(\mathrm{c}_{1}\right)>0 then c_1 is point of local minima.

If f^{\prime \prime}\left(\mathrm{c}_{2}\right)<0 then c_2 is point of local maxima .

where c_1 & c_2 are critical points.

Put c_1 and c_2 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,

\begin{aligned} &f(x)=x^{3}-2 a x^{2}+a^{2} x \\ &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} \\ &=\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\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^{3}}{27} \end{aligned}

Thus, point of local maxima is \frac{a}{3}& its max. value is \frac{4 a^{3}}{27}& point of local minima is 9 & its value is 0.

 

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