JEE Main 2021 — Mathematics Calculus

$f:[0,\infty )\rightarrow [0,\infty )$f:[0,∞)→[0,∞), $\displaystyle f(x)={\int }_{0}^{x}[y]dy$f(x)=∫0x​[y]dy

Let $x=n+f,f\in (0,1)$x=n+f,f∈(0,1)

$\Rightarrow [x]=n,{x}=f$⇒[x]=n,x=f.

So

$$f(x)={\int }_{0}^{1}[y]dy+{\int }_{1}^{2}[y]dy+{\int }_{2}^{3}[y]dy+ \ldots +{\int }_{n-1}^{n}[y]dy+{\int }_{n}^{n+f}[y]dy$$f(x)=∫01​[y]dy+∫12​[y]dy+∫23​[y]dy+…+∫n−1n​[y]dy+∫nn+f​[y]dy

$$f(x)={\int }_{0}^{1}0dy+{\int }_{1}^{2}1dy+{\int }_{2}^{3}2dy+ \ldots +{\int }_{n-1}^{n}(n-1)dy+{\int }_{n}^{n+f}ndy$$f(x)=∫01​0dy+∫12​1dy+∫23​2dy+…+∫n−1n​(n−1)dy+∫nn+f​ndy

$$f(x)=0[1-0]+1[2-1]+2[3-2]+ \ldots +(n-1)[n-(n-1)]+n[n+f-n]$$f(x)=0[1−0]+1[2−1]+2[3−2]+…+(n−1)[n−(n−1)]+n[n+f−n]

$$f(x)=0+1+2+\ldots +(n-1)+nf$$f(x)=0+1+2+…+(n−1)+nf

$$\Rightarrow f(x)=\frac{n(n-1)}{2}+nf$$⇒f(x)=2n(n−1)​+nf

$$=\frac{[x]([x]-1)}{2}+[x]{x}$$=2[x]([x]−1)​+[x]x

Note $\underset{x\rightarrow {n}^{+}}{\mathrm{lim}}f(x)=\frac{n(n-1)}{2}$x→n+lim​f(x)=2n(n−1)​

$$\underset{x\rightarrow {n}^{-}}{\mathrm{lim}}f(x)=\frac{(n-1)(n-2)}{2}+(n-1)$$x→n−lim​f(x)=2(n−1)(n−2)​+(n−1)

$$=\frac{n(n-1)}{2}$$=2n(n−1)​

$$f(x)=\frac{n(n-1)}{2}(n\in {N}_{0})$$f(x)=2n(n−1)​(n∈N0​)

so $f(x)$f(x) is continuous $\forall x\geq 0$∀x≥0 and differentiable except at integer.'