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Question

Let f:RR be a function such that f(f(x))=x2x+1 for all x and the value of tan1(f(1))+sec1(f(1))+cos1(f(0))+cot1(f(0)) is equal to aπb where a and b are co-prime. Then the value of a+b is

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Solution

f(f(x))=x2x+1
Replace x by f(x)
f(f(f(x)))=(f(x))2f(x)+1
f(x2x+1)=(f(x))2f(x)+1 (1)
Put x=0 in equation (1)
f(1)=(f(0))2f(0)+1 (2)
Put x=1 in equation (1)
f(1)=(f(1))2f(1)+1
(f(1)1)2=0
f(1)=1
Put f(1)=1 in equation (2), we get
f(0)=0 or 1
But f(0)=0 is rejected because it does not satisfy f(f(x))=x2x+1
f(0) = 1 and f(1)=1

Now, tan1(1)+sec1(1)+cos1(1)+cot1(1)
=π4+0+0+π4=π2a+b=1+2=3

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