Limits & Continuity: Let f(x)= [x^2-x ]+|-x+[x]| , where xinmathbbR and [t] denotes the… | JEE Main 2023

JEE Main 2023 (Online) 11th April Morning ShiftQuestion 1

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

Limits & Continuity

not continuous at $x=0$ and $x=1$continuous at $x=0$ and $x=1$continuous at $x=0$, but not continuous at $x=1$continuous at $x=1$, but not continuous at $x=0$✓ Correct answer

Solution

$f(x)=[x^2-x]+\{x\}$ (since $|-x+[x]|=\{x\}$). $f(0)=0$, $f(1)=0$. At $0$: for $x\to0^-$, $x^2-x\in(0,1)$ so $[\cdot]=0$ and $\{x\}\to1$: $f\to1\ne f(0)$. Not continuous at $0$. At $1$: for $x\to1^-$, $x^2-x<0$ small so $[\cdot]=-1$, $\{x\}\to1$: $f\to0$; for $x\to1^+$, $[\cdot]=0$, $\{x\}\to0$: $f\to0$. Continuous at $1$. Option (D).

Limits & Continuity - JEE Mathematics question (JEE Main 2023 (Online) 11th April Morning Shift)

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