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# Solution for F(X) = (X − 2) √ X − 1 in [ 1 , 9 ] . - CBSE (Science) Class 12 - Mathematics

#### Question

f(x) = (x $-$ 2) $\sqrt{x - 1} \text { in }[1, 9]$ .

#### Solution

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

$\Rightarrow f'\left( x \right) = \sqrt{x - 1} + \frac{\left( x - 2 \right)}{2\sqrt{x - 1}}$

$\text { For a local maximum or a local minimum, we must have }$

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

$\Rightarrow \sqrt{x - 1} + \frac{\left( x - 2 \right)}{2\sqrt{x - 1}} = 0$

$\Rightarrow 2\left( x - 1 \right) + \left( x - 2 \right) = 0$

$\Rightarrow 2x - 2 + x - 2 = 0$

$\Rightarrow 3x - 4 = 0$

$\Rightarrow 3x = 4$

$\Rightarrow x = \frac{4}{3}$

$\text { Thus, the critical points of f are } 1, \frac{4}{3} \text { and } 9 .$

$\text { Now },$

$f\left( 1 \right) = \left( 1 - 2 \right)\sqrt{1 - 1} = 0$

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

$f\left( 9 \right) = \left( 9 - 2 \right)\sqrt{9 - 1} = 14\sqrt{2}$

$\text { Hence, the absolute maximum value when x } = 9 is 14\sqrt{2}\text { and the absolute minimum value when x } = \frac{4}{3} is - \frac{2}{3\sqrt{3}} .$

$\text { Disclaimer: The solution given in the book is incorrect . The solution here is created according to the question given in the book }.$

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Solution for question: F(X) = (X − 2) √ X − 1 in [ 1 , 9 ] . concept: Graph of Maxima and Minima. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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