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Q.

Let f(x)=[x]+|1x| for -1x3 where [x] is the integral part of x. Then the number of values of x in [-1, 3] at which f is not continuous is.

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a

2

b

1

c

0

d

4

answer is D.

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Detailed Solution

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f(x)={x       1x<01x    0x<1x            1x<21+x    2x<3

Clearly f is discatinuous at x=0,1,2 no.of points = 3

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