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If \mathrm{f(x)=|x-a|+|x+b|, x \in \mathbf{R}, b>a>0}. Then

Option: 1

\mathrm{f^{\prime}(a+)=1}


Option: 2

\mathrm{f^{\prime}(a+)=0}


Option: 3

\mathrm{f^{\prime}(-b+)=0}


Option: 4

\mathrm{f^{\prime}(-b+)=1}


Answers (1)

best_answer

\mathrm{ f^{\prime}(a+)=\lim _{h \rightarrow 0+} \frac{f(a+h)-f(a)}{h}}
           \mathrm{ =\lim _{h \rightarrow 0+} \frac{|h|+|a+b+h|-|a+b|}{h}}
           \mathrm{ =\lim _{h \rightarrow 0+} \frac{h+a+b+h-(a+b)}{h} \text { as } h>0, a+b>0}

          \mathrm{ \quad=2}

\mathrm{ f^{\prime}(-b+) =\lim _{h \rightarrow 0+} \frac{f(-b+h)-f(-b)}{h}}
                 \mathrm{ =\lim _{h \rightarrow 0+} \frac{|-b+h-a|+|h|-|-b-a|}{h}}
                \mathrm{ =\lim _{h \rightarrow 0+} \frac{|a+b-h|+|h|-|a+b|}{h} }
              \mathrm{ =\lim _{h \rightarrow 0+} \frac{a+b-h+h-(a+b)}{h}=0 }.

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Ritika Jonwal

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