A) for\[x+1>0,f\]and\[f'\]are continuous
B) for\[x+1>0,f\]is continuous but f is not continuous
C) at\[x=0,f\]and f are not continuous
D) at\[x=0,f\]is continuous but\[f'\]is not continuous
Correct Answer: A
Solution :
Given, \[f(x)=\int_{-1}^{x}{|t|}\,dt,x\ge -1\] If\[-1\le x\le 0,\] then\[f(x)=\int_{-1}^{x}{(-t)}\,dt\] \[=\left[ -\frac{{{t}^{2}}}{2} \right]_{-1}^{x}\] \[=\left[ -\frac{{{x}^{2}}}{2}+\frac{1}{2} \right]\] If\[x>0,\]then \[f(x)=\int_{-1}^{0}{(-t)dt}+\int_{0}^{x}{t\,dt}\] \[=\left[ -\frac{{{t}^{2}}}{2} \right]_{-1}^{0}+\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{x}\] \[=\frac{1}{2}+\frac{{{x}^{2}}}{2}\] \[=\frac{({{x}^{2}}+1)}{2}\] \[\therefore \]\[f(x)=\left\{ \begin{matrix} -\frac{1}{2}({{x}^{2}}-1), & -1\le x<0 \\ \frac{1}{2}({{x}^{2}}+1), & x\ge 0 \\ \end{matrix} \right.\] Here, function\[f(x)\] is continuous at\[x=0\] \[\therefore \]It is continuous for \[x>-1\] \[\Rightarrow \]It is continuous for \[x+1>0\] Now, \[Rf'(0)=0,Lf'(0)=0\] \[\therefore \] \[f(0)=0\] \[\therefore \] \[f'(x)=\left\{ \begin{matrix} -x, & -1<x<0 \\ 0, & x=0 \\ x, & x>0 \\ \end{matrix} \right.\] Here, function\[f'(x)\]is continuous at\[x=0.\] \[\therefore \]It is continuous for \[x>-1\] \[\Rightarrow \]It is continuous for \[x+1>0\] Hence, for\[x+1>0,\text{ }f\]and\[f'\]both are continuous.
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