QJEEM21Mab430: JEE Main 2021 Mathematics - Class 12 JEEM21Mab430 Solution: Let [t] denote the greatest integer t. The number of points…

$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$

$$f\left(x\right)=\{\begin{array}{cc}-2\left|x^2-1\right|+1,& -2<x<-1\\ -\left|x^2-1\right|+1,& -1\le x<0\\ \sin \frac{\pi }{3}+1,& 0\le x<1\\ \left|x^2-1\right|+\frac{1}{\sqrt{2}}-2,& 1\le x<2\end{array}$$

$\therefore$ at x = -1, $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 1$ and $\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 1$

Hence continuous at x = –1

Similarly check at x = 0,

$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$ and $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$

So, f(x) discontinuous and at x = 0

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$ and $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {1 \over {\sqrt 2 }} - 2$

So, f(x) discontinuous and at x = 1

Hence 2 points of discontinuity.