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Solution


Q.

Let f:[-1,2]→R be given by f(x)=2x2+x+[x2]-[x], where [t]denotes the greatest integer less than or equal to t. The number of points, where f is not continuous, is                [2024]
1 4  
2 6  
3 5  
4 3  

Ans.

(1)

f(x)=2x2+x+[x2]-[x]

We check continuity at

x=0,1,2,-1,2,3

At x=0

LHL=limx→0-f(x)=0+0+0-(-1)=1

RHL=limx→0+f(x)=0+0+0-(0)=-0

∵   LHL≠RHL at x=0

∴   f(x) is not continuous at x=0

        At x=1

LHL=limx→1-f(x)=2+1+0-0=3

RHL=limx→1+f(x)=2+1+1-1=3

f(1)=2+1+1-1=3

∵  LHL=RHL=f(x) at x=1

∴  f(x) is continuous at x=1

At   x=2

LHL=limx→2-f(x)=8+2+3-1=12

RHL=limx→2+f(x)=8+2+4-2=12

f(2)=8+2+4-2=12

∵  LHL=RHL=f(x) at x=2

∴  f(x) is continuous at x=2

At  x=2

LHL=limx→2-f(x)=4+2+1-1=4+2

RHL=limx→2+f(x)=4+2+2-1=5+2

∵   LHL≠RHL at x=2

∴   f(x) is not continuous at x=2

Similarly, LHL≠RHL at x=-1 and 3

So there are 4 points of discontinuity.