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If f(x)=xTan−1x then Ltx→1f(x)−f(1)x−1=

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a

π4

b

π+14

c

π+24

d

π+34

answer is C.

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Detailed Solution

limx→1 f(x)-f(1)x-1=0/0 use L hospital rule limx→1 f1(x)-01-0=f1(1) f(x)=xtan-1x⇒f1(x)=x1+x2+tan-1x f1(1)=12+π4

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