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Number of points of discontinuity of f(x...

Number of points of discontinuity of `f(x)=[sin^(-1)x]-[x]` in its domain is equal to (where [.] denotes the greatest integer function) a. 0 b. 1`` c. 2 d. 3

A

0

B

1

C

2

D

3

Text Solution

Verified by Experts

The correct Answer is:
D
Domain of the function is `[-1,1]`
`[sin^(-1)x]` is discontinuous at `x=-sin1,0, sin 1`
[x] is discontinuous at x = 0,1
Now `f(0^(+))=[0^(+)]-[0^(+)]=0-0=0`
`f(0^(-))=[0^(-)]-[0^(-)]=-1-(-1)=0`
Hence f(x) is continuous at x = 0.
Therefore points of discontinuity are `-sin 1, sin 1 and 1.`
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