PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# F(X) = X 3 − 2 a X 2 + a 2 X , a > 0 , X ∈ R - PUC Karnataka Science Class 12 - Mathematics

#### Question

f(x) = $x^3 - 2a x^2 + a^2 x, a > 0, x \in R$ .

#### Solution

$f\left( x \right) = x^3 - 2a x^2 + a^2 x$

$\Rightarrow f'\left( x \right) = 3 x^2 - 4ax + a^2$

$\text { For the local maxima or minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow 3 x^2 - 4ax + a^2 = 0$

$\Rightarrow 3 x^2 - 3ax - ax + a^2 = 0$

$\Rightarrow 3x\left( x - a \right) - a\left( x - a \right) = 0$

$\Rightarrow \left( 3x - a \right)\left( x - a \right) = 0$

$\Rightarrow x = a \text { and } \frac{a}{3}$

$\text { Thus, x = a and x } = \frac{a}{3}\text { are the possible points of local maxima or local minima }.$

$\text { Now,}$

$f''\left( x \right) = 6x - 4a$

$\text { At } x = a:$

$f''\left( a \right) = 6\left( a \right) - 4a = 2a > 0$

$\text { So, x = a is the point of local minimum }.$

$\text { The local minimum value is given by }$

$f\left( a \right) = a^3 - 2a \left( a \right)^2 + a^2 \left( a \right) = 0$

$\text { At }x = \frac{a}{3}:$

$f''\left( \frac{a}{3} \right) = 6\left( \frac{a}{3} \right) - 4a = - 2a < 0$

$\text { So }, x = \frac{a}{3} \text { is the point of local maximum }.$

$\text { The local maximum value is given by }$

$f\left( \frac{a}{3} \right) = \left( \frac{a}{3} \right)^3 - 2a \left( \frac{a}{3} \right)^2 + a^2 \left( \frac{a}{3} \right) = \frac{a^3}{27} - \frac{2 a^3}{9} + \frac{a^3}{3} = \frac{4 a^3}{27}$

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Solution F(X) = X 3 − 2 a X 2 + a 2 X , a > 0 , X ∈ R Concept: Graph of Maxima and Minima.
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