Q.

Let a function fx,x≠0 be such that fx+f1x=fx.f1x then f(x) can be

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a

1-x2013

b

x+1

c

π2tan-1x

d

21+klnx

answer is A.

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Detailed Solution

f1x-1=1fx-1i.e.,   f1x-1is  reciprocal  of   fx-1.Now,   for fx=π2tan-1x                            fx=-1=cos-1xtan-1x,f1x-1=tan-1xcot-1xAlso   for   fx=21+klnx      fx-1=1-klnx1+klnx,f1x-1=1+klnx1-klnx
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