JEE 2023 — Mathematics PYQ

• Value of function: $f(0)=[0^2-0]+\{0\}=0+0=0$

• Left Hand Limit (LHL): As $x\to 0^{-}$, $x^2-x\to 0^{+}$ (kyunki small negative number ka square positive hota hai aur $-x$ bhi positive ho jayega).

  Isliye, $[x^2-x]=0$.

  Aur $\{x\}\to 1$ as $x\to 0^{-}$.

  $$\underset{x\to 0^{-}}{\lim }f(x)=0+1=1$$

• Nishkarsh: Kyunki $LHL\ne f(0)$, function $x=0$ par continuous nahi hai.

2. Continuity at $x=1$ ki jaanch:

• Value of function: $f(1)=[1^2-1]+\{1\}=0+0=0$

• Left Hand Limit (LHL): As $x\to 1^{-}$, let $x=0.9$.

  $x^2-x=(0.9{)}^2-0.9=0.81-0.9=-0.09$.

  Iska greatest integer $[-0.09]=-1$.

  Aur $\{x\}\to 1$ as $x\to 1^{-}$.

  $$\underset{x\to 1^{-}}{\lim }f(x)=-1+1=0$$

• Right Hand Limit (RHL): As $x\to 1^{+}$, let $x=1.1$.

  $x^2-x=(1.1{)}^2-1.1=1.21-1.1=0.11$.

  Iska greatest integer $[0.11]=0$.

  Aur $\{x\}\to 0$ as $x\to 1^{+}$.

  $$\underset{x\to 1^{+}}{\lim }f(x)=0+0=0$$

• Nishkarsh: Kyunki $LHL=RHL=f(1)$, function $x=1$ par continuous hai.

Sahi Option:

(3) hai: continuous at $x=1$, but not continuous at $x=0$.