JEE Main 2022 — Mathematics Calculus

Given,

$$f(x)=a\mathrm{sin}(\frac{\pi [x]}{2})+[2-x],a\in \mathbb{R}$$f(x)=asin(2π[x]​)+[2−x],a∈R

Now given $\underset{x\rightarrow -1}{\mathrm{lim}}f(x)$x→−1lim​f(x) exists,

So $\underset{x\rightarrow -{1}^{+}}{\mathrm{lim}}a\mathrm{sin}(\pi \frac{[x]}{2})+[2-x]=-a+2$x→−1+lim​asin(π2[x]​)+[2−x]=−a+2

And $\underset{x\rightarrow -{1}^{-}}{\mathrm{lim}}asin(\pi \frac{[x]}{2})+[2-x]=0+3=3$x→−1−lim​asin(π2[x]​)+[2−x]=0+3=3

So, $\underset{x\rightarrow -1}{\mathrm{lim}}f(x)$x→−1lim​f(x) exist when $-a+2=3\Rightarrow a=-1$−a+2=3⇒a=−1

Now,

$${\int }_{0}^{4}f(x)dx={\int }_{0}^{1}f(x)dx+{\int }_{1}^{2}f(x)dx+{\int }_{2}^{3}f(x)dx+{\int }_{3}^{4}f(x)dx$$∫04​f(x)dx=∫01​f(x)dx+∫12​f(x)dx+∫23​f(x)dx+∫34​f(x)dx

$$\Rightarrow {\int }_{0}^{4}f(x)dx={\int }_{0}^{1}-\mathrm{sin}(\frac{\pi [x]}{2})+[2-x]dx+{\int }_{1}^{2}-\mathrm{sin}(\frac{\pi [x]}{2})+[2-x]dx+{\int }_{2}^{3}-\mathrm{sin}(\frac{\pi [x]}{2})+[2-x]dx+{\int }_{3}^{4}-\mathrm{sin}(\frac{\pi [x]}{2})+[2-x]dx$$⇒∫04​f(x)dx=∫01​−sin(2π[x]​)+[2−x]dx+∫12​−sin(2π[x]​)+[2−x]dx+∫23​−sin(2π[x]​)+[2−x]dx+∫34​−sin(2π[x]​)+[2−x]dx

$$={\int }_{0}^{1}(0+1)dx+{\int }_{1}^{2}(-1+0)dx+{\int }_{2}^{3}(0-1)dx+{\int }_{3}^{4}(1-2)dx$$=∫01​(0+1)dx+∫12​(−1+0)dx+∫23​(0−1)dx+∫34​(1−2)dx

$$=1-1-1-1=-2$$=1−1−1−1=−2