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Question

Let αRbe such that the function

f(x)=cos1(1{x}2)sin1(1{x}){x}{x}3αx0x=0

is continuous atx=0, where x=x[x],[x] is the greatest integer less than or equal to x. Then:


A

α=π4

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B

No such α exists

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C

α=0

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D

α=π2

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Solution

The correct option is B

No such α exists


Explanation for the correct option:

Step 1: Solving the Right-hand limit

limx0+fx

We know that

x+x=xx=x-xx0+x=x-0x=x

limx0+cos1(1x2)sin1(1x)x(1x2)π2limx0+cos1(1x2)xsin1(1)=π2ByL'HospitalRuleddxcos-1=-11-x2π2limx0+-11(1x2)2.-2x1-3x2πlimx0+x2x2x4.11-3x2πlimx0+12x2.11-3x2π2

Step 2: Solving the Left-hand limit

limx0-fx

We know that

x+x=xx=x-xx0-x=x+1

limx0cos1(1(1+x)2)sin1(1x+1)(1+x)(1+x)3limx0cos1(1(1+x)2)sin1(x)(1+x)(1+x)3limx0cos1(-x2-2x)sin1(-x)(1+x)((1+x)21)π2limx0sin1(-x)(1+x)(-x2-2x)π2limx0sin1(x)x(1+x)(x+2)Weknowthatlimx0sin1(x)x=1π2×12=π4

As LHL is not equal to RHL, therefore f(x) is not continuous at x=0.

Therefore, option (B) is the correct answer.


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