lf $f (x) = | x^{2} - 4 x + 3 |$ , then $f^{'} (x)$ is

2 x - 4

for

1 < x < 3

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4 - 2 x

for

1 < x < 3

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2 x - 4

for

1 \leq x \leq 3

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4 - 2 x

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Solution

The correct option is B $4 - 2 x$ for $1 < x < 3$
$f (x) = | (x - 1) (x - 3) |$
$f (x) = {\begin{matrix} \begin{matrix} (x - 1) (x - 3) & x < 1, x > 3 - (x - 1) (x - 3) & 1 < x < 3 \end{matrix} \end{matrix}$
$f (x) = {\begin{matrix} \begin{matrix} x^{2} - 4 x + 3 & x < 1, x > 3 - (x^{2} - 4 x + 3) & 1 < x < 3 \end{matrix} \end{matrix}$
$∴ f^{'} (x) = {\begin{matrix} \begin{matrix} 2 x - 4 & x < 1, x > 3 4 - 2 x & 1 < x < 3 \end{matrix} \end{matrix}$
Hence, option $B$ is correct

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