If $f (x) = | (x - 4) (x - 5) |$ then $f^{'} (x)$ is equal to-

- 2 x + 9

, for all

x ϵ R

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2 x - 9

, if

4 < x < 5

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- 2 x + 9

, if

4 < x < 5

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None of these

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Solution

The correct option is B $- 2 x + 9$ , if $4 < x < 5$
$f (x) = | (x - 4) (x - 5) |$
$= ⎧ ⎨ ⎩ \begin{matrix} (x - 4) (x - 5) x < 4 - (x - 4) (x - 5) 4 \leq x < 5 (x - 4) (x - 5) x \geq 5 \end{matrix} = ⎧ ⎨ ⎩ \begin{matrix} x^{2} - 9 x + 20 x < 4 - x^{2} + 9 x - 204 \leq x < 5 x^{2} - 9 x + 20 x \geq 5 \end{matrix} ∴ f^{'} (x) = ⎧ ⎨ ⎩ \begin{matrix} 2 x - 9 x < 0 - 2 x + 94 \leq x < 5 2 x - 9 x \geq 5 \end{matrix}$

Suggest Corrections

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