Q.
Let α∈R be such that the function f(x)=cos-11-x2sin-11-x{x}-{x}3,x≠0α,x=0 is continuous at x=0, where {x}=x-[x], [x] is the greatest integer less than or equal to x. Then :
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a
α=0
b
no such α exists
c
α=π4
d
α=π2
answer is B.
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Detailed Solution
RHL=limx→0+cos-11-x2sin-1(1-x)x1-x2 =π2limx→0+cos-11-x2x =π2limx→0+-11-1-x22(-2x) (L'Hospital Rule) =πlimx→0+x2x2−x4=πlimx→0+12−x2=π2LHL=limx→0+cos-11-(1+x)2sin-1(-x)(1+x)-(1+x)3 =π2limx→0+sin-1x(1+x)(1+x)2-1 =π2limx→0+sin-1xx2+2x =π212=π4 As LHL≠RHL so f(x) is not continuous at x=0. Therefore no such α exists
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