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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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 :