The value of d-dx -x-1- -x-5- at x - 3 is

The correct option is B 0 ddx |x| = |x|x Applying the above formula , ddx ( |x 1|+|x 5|) = ddx|x 1|+ddx|x 5|= |x 1|x 1+|x 5|x 5 Substituting x=3 , ...

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Standard XII

Mathematics

Integration by Partial Fractions

The value of ...

Question

The value of $\frac{d}{d x} (| x - 1 | + | x - 5 |)$ at x = 3 is

-2

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Solution

The correct option is B 0
$\frac{d}{d x} | x | = \frac{| x |}{x}$

Applying the above formula ,
$\frac{d}{d x} (| x - 1 | + | x - 5 |) = \frac{d}{d x} | x - 1 | + \frac{d}{d x} | x - 5 | = \frac{| x - 1 |}{x - 1} + \frac{| x - 5 |}{x - 5}$

Substituting $x = 3$ ,
$\frac{| 3 - 1 |}{3 - 1} + \frac{| 3 - 5 |}{3 - 5} = \frac{| 2 |}{2} + \frac{| - 2 |}{- 2} = \frac{2}{2} + \frac{2}{- 2} = 1 - 1 = 0$

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The value of ddx(|x−1|+|x−5|) at x = 3 is

The correct option is B 0ddx|x|=|x|xApplying the above formula ,ddx(|x−1|+|x−5|)=ddx|x−1|+ddx|x−5|=|x−1|x−1+|x−5|x−5Substituting x=3,|3−1|3−1+|3−5|3−5=|2|2+|−2|−2=22+2−2=1−1=0

The value of $\frac{d}{d x} (| x - 1 | + | x - 5 |)$ at x = 3 is