PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# F(X) = (X − 1) (X − 2)2. - PUC Karnataka Science Class 12 - Mathematics

#### Question

f(x) = (x $-$ 1) (x $-$ 2)2.

#### Solution

$\text { Given: } f\left( x \right) = \left( x - 1 \right) \left( x - 2 \right)^2$

$= \left( x - 1 \right)\left( x^2 - 4x + 4 \right)$

$= x^3 - 4 x^2 + 4x - x^2 + 4x - 4$

$= x^3 - 5 x^2 + 8x - 4$

$\Rightarrow f'\left( x \right) = 3 x^2 - 10x + 8$

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

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

$\Rightarrow 3 x^2 - 10x + 8 = 0$

$\Rightarrow 3 x^2 - 6x - 4x + 8 = 0$

$\Rightarrow \left( x - 2 \right)\left( 3x - 4 \right) = 0$

$\Rightarrow x = 2 \text { and }\frac{4}{3}$

$\text { Thus, x = 2 and } x = \frac{4}{3} \text { are the possible points of local maxima or local minima } .$

$\text { Now },$

$f''\left( x \right) = 6x - 10$

$At x = 2:$

$f''\left( 2 \right) = 6\left( 2 \right) - 10 = 2 > 0$

$\text { So, x = 2 is the point of local minimum }.$

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

$f\left( 2 \right) = \left( 2 - 1 \right) \left( 2 - 2 \right)^2 = 0$

$\text { At }x = \frac{4}{3}:$

$f''\left( \frac{4}{3} \right) = 6\left( \frac{4}{3} \right) - 10 = - 2 < 0$

$\text { So, x} = \frac{4}{3}\text { is the point of local maximum } .$

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

$f\left( \frac{4}{3} \right) = \left( \frac{4}{3} - 1 \right) \left( \frac{4}{3} - 2 \right)^2 = \frac{1}{3} \times \frac{4}{9} = \frac{4}{27}$

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Solution F(X) = (X − 1) (X − 2)2. Concept: Graph of Maxima and Minima.
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