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Q.

Let f:R→R be defined by f(x)=α+sin[x]x,ifx>02,ifx=0β+sinx−xx3,ifx<0 where [x] denotes the integral part of x. If f  continuous at x = 0, then β-α=

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a

-1

b

1

c

0

d

2

answer is B.

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Detailed Solution

LHLlimx→0  β+sinx-xx3⇒use l hospital rule =β+limx→0 cosx-13x2=β+limx→0  -sinx6x=β-1RHL=α+0x=αα+0=2=β−1α=β−1⇒β−α=1
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