If $f\left( x \right) = \sqrt {{x^2} - 2x + 1} $, then

$f'\left( x \right) = 1$ for all $x$

$f'\left( x \right) = - 1$ for all $x \le 1$

$f'\left( x \right) = 1$ for all $x \ge 1$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

f\left( x \right) = \sqrt {{x^2} - 2x + 1}

, then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the derivative is 1 for x > 1 and -1 for x < 1, and it's not defined at x=1.

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