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  • <img alt="\mathrm{\text { Let } f(x)=\left\{\begin{array}{ll} x+1, & x<0 \\ |x-1|, & x \geq 0 \end{array} \text { and } g(x)= \begin{cases}x+1, & x<0 \\ (x-1)^2, & x \geq 0\en

\mathrm{\text { Let } f(x)=\left\{\begin{array}{ll} x+1, & x<0 \\ |x-1|, & x \geq 0 \end{array} \text { and } g(x)= \begin{cases}x+1, & x<0 \\ (x-1)^2, & x \geq 0\end{cases}\right.}

Then the number of points at which the function g(f(x)) is not differentiable is

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

So, g(f(x)) is continuous for all x but not differentiable for
x = ±1.

Posted by

mansi

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