JEE Main 2021 — Mathematics Calculus

As $[.]$[.] is discontinuous at integers so we will check at $x=-1,0,1$x=−1,0,1 only

at $x=-1$x=−1

$[-{1}^{-}]=-2$[−1−]=−2 and $[-{1}^{+}]=-1$[−1+]=−1

we have

$$\mathrm{LHL}=(-2)\cdot (0)+\mathrm{sin}(\pi )+1=1$$LHL=(−2)⋅(0)+sin(π)+1=1

$$\mathrm{RHL}=(-1)\cdot (0)+\mathrm{sin}(\frac{\pi }{2})-0=1$$RHL=(−1)⋅(0)+sin(2π​)−0=1

$f(-1)=1\Rightarrow$f(−1)=1⇒ Continuous at $x=-1$x=−1

Again at $x=0$x=0

[{0}^{-}]=-1&[{0}^{+}]=0

$$\mathrm{LHL}=(-1)\cdot (1)+\mathrm{sin}(\frac{\pi }{2})-0=0$$LHL=(−1)⋅(1)+sin(2π​)−0=0

$$\mathrm{RHL}=0+\mathrm{sin}(\frac{\pi }{3})-1\neq \mathrm{LHL}$$RHL=0+sin(3π​)−1=LHL

Hence, discontinuous at $x=0$x=0

Again at $x=1$x=1,

[{1}^{-}]=0&[{1}^{+}]=1

We have

$$\mathrm{LHL}=0+\mathrm{sin}(\frac{\pi }{3})-1$$LHL=0+sin(3π​)−1

$$\mathrm{RHL}=0+\mathrm{sin}(\frac{\pi }{4})-2\neq \mathrm{LHL}$$RHL=0+sin(4π​)−2=LHL

Hence, discontinuous at $x=1$x=1

Hence, discontinuous at exactly $2$2 points