JEE Main 2024 — Mathematics Calculus

Doubtful points : $-1,0,1, \sqrt{2}, \sqrt{3}, 2$−1,0,1,2​,3​,2 at $x=\sqrt{2}, \sqrt{3}$x=2​,3​ $\begin{array}{cc} f(x)=\left(2 x^2+x-[x]\right)+\left[x^2\right]= \text{Discount} \\ \downarrow \qquad\qquad\quad \downarrow \\ \text{Cont.} \qquad\qquad \text{Cont.} \end{array}$f(x)=(2x2+x−[x])+[x2]=Discount↓↓Cont.Cont.​

at $x=-1$x=−1 : $\left.\begin{array}{rl}\text { RHL } \Rightarrow & f(x)=(2-1-(-1))+0=2 \\ & f(-1)=2-1-(-1)+1=3\end{array}\right\}$ RHL ⇒​f(x)=(2−1−(−1))+0=2f(−1)=2−1−(−1)+1=3​} Dis. at $x=2$x=2 : $\left.\begin{array}{rl}
\text {LHL } \Rightarrow f(x) & =8+2-1+3=12 \
f(2) & =8+2-2+4=12
\end{array}\right} \text { Cont. }$at$at\mathrm{x}=0$:$:\left.\begin{array}{rl}
\text {LHL } \Rightarrow & 0+0-(-1)+0=1 \
& \mathrm{f}(0)=0
\end{array}\right} \text { Dis. }$at$at\mathrm{x}=1$$\left.\begin{array}{rl}
\text {LHL } \Rightarrow & 2+1-0+0=3 \
& f(1)=3-1+1=3 \
\text {RHL } \Rightarrow & 2+1-1+1=3
\end{array}\right} \text { Cont. }$