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Q.

Given a real valued function f such that f(x)={tan2{x}(x2−[x]2)for x >0 1for x = 0 {x}cot{x}for x <0 , where [x] is the integral part  and {x} is the fractional part of x then

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a

limx→0+f(x)=1

b

limx→0−f(x)=cot1

c

cot−1(limx→0−f(x))2=1

d

tan−1(limx→0−f(x))=π4

answer is .

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Detailed Solution

We have limk→0+f(x)=limk→0+tan2{x}(x2−[x]2)=limx→0+tan2xx2=1     ……..(1) [∵x→0+;[x]=0={x}=x]  Also limk→0−f(x)=limx→0−{x}cot{x}=cot1.....(2) [∵x→0−;[x]=−1⇒{x}=x+1⇒{x}→1] Also, cot−1(limx→0−f(x))2=cot−1(cot 1)=1
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