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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
Question
The differential coefficient of the function $|x - 1| + |x - 3|$ at the point $x = 2$ is
✓

Answer

b
(b) $f(x) = |x - 1| + |x - 3|$

$f(x) = \left\{ {\begin{array}{*{20}{c}}{ - (x - 1) - (x - 3),}&{x < 1}\\{(x - 1) - (x - 3),}&{x > 1}\\{(x - 1) - (x - 3),}&{x < 3}\\{(x - 1) + (x - 3),}&{x > 3}\end{array}} \right.$

$ = \left\{ {\begin{array}{*{20}{c}}{4 - 2x,}&{x < 1}\\{2\,\,\,\,\,\,\,\,\,,}&{1 < x < 3}\\{2x - 4,}&{x > 3}\end{array}} \right.$

At $x = 2$, $f(x) = $ $2$ .

Hence $\,f'(x) = 0$.

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