JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 10)
Let $$f(x)=\left[x^{2}-x\right]+|-x+[x]|$$, where $$x \in \mathbb{R}$$ and $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then, $$f$$ is :
continuous at $$x=0$$, but not continuous at $$x=1$$
continuous at $$x=0$$ and $$x=1$$
continuous at $$x=1$$, but not continuous at $$x=0$$
not continuous at $$x=0$$ and $$x=1$$
Explanation
We have,
$$\begin{aligned} f(x) & =\left[x^2-x\right]+|-x+[x]| \\\\ & =[x(x-1)]+\{x\} \end{aligned} $$
$$ f(x)=\left\{\begin{array}{ccc} x+1 & ; & -0.5 < x < 0 \\ 0 & ; & x=0 \\ -1+x & ; & 0 < x <1 \\ 0 & ; & x=1 \\ x-1 & ; & 1 < x < 1.5 \end{array}\right. $$
At $x=0$,
$$ \begin{array}{r} \text { LHL }=\lim\limits_{x \rightarrow 0^{-}} f(x)=1 \\\\ \text { RHL } \lim\limits_{x \rightarrow 0^{+}} f(x)=-1 \\\\ f(0)=0 \end{array} $$
$\therefore f(x)$ is not continuous at $x=0$
At $x=1$
$$ \begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0 \\\\ \mathrm{RHL} & =\lim _{x \rightarrow 1^{+}} f(x)=1-1=0 \\\\ f(1) & =0 \end{aligned} $$
$\therefore f(x)$ is continuous at $x=1$
Hence, $f(x)$ is continuous at $x=1$, but not continuous at $x=0$.
$$\begin{aligned} f(x) & =\left[x^2-x\right]+|-x+[x]| \\\\ & =[x(x-1)]+\{x\} \end{aligned} $$
$$ f(x)=\left\{\begin{array}{ccc} x+1 & ; & -0.5 < x < 0 \\ 0 & ; & x=0 \\ -1+x & ; & 0 < x <1 \\ 0 & ; & x=1 \\ x-1 & ; & 1 < x < 1.5 \end{array}\right. $$
At $x=0$,
$$ \begin{array}{r} \text { LHL }=\lim\limits_{x \rightarrow 0^{-}} f(x)=1 \\\\ \text { RHL } \lim\limits_{x \rightarrow 0^{+}} f(x)=-1 \\\\ f(0)=0 \end{array} $$
$\therefore f(x)$ is not continuous at $x=0$
At $x=1$
$$ \begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0 \\\\ \mathrm{RHL} & =\lim _{x \rightarrow 1^{+}} f(x)=1-1=0 \\\\ f(1) & =0 \end{aligned} $$
$\therefore f(x)$ is continuous at $x=1$
Hence, $f(x)$ is continuous at $x=1$, but not continuous at $x=0$.
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