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If f(x)=1+2sinx−1−2sinxx is continuous on R, then f(0)=

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a

1

b

2

c

3

d

4

answer is B.

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Detailed Solution

f(0)=Ltx→0f(x) =Ltx→01+2sinx−1−2sinxx =Ltx→0(1+2sinx)-(1-2sinx)1+2sinx+1−2sinx.1x on rationalisation=Ltx→04sinxx[1+2sinx+1−2sinx]=2

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