IIT JEE , AIEEE Forum - Mathematics , Algebra

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consider the function f(x) = {tan(inverse)x}^2 + 2/root(1+x^2)}

differentiate.. we get 2tan(inv)x/(1+x^2) -2x/(1+x^2)^3/2..

if we prove this is greater than 0 in [1/e,e] sum over as it will be an inc function..

so ve have to prove tan(inv)x/x>1/root(1+x^2) put x=tant

it reduces to (sint/t)<1 which is obvious in given domain of x(1/e to e)
hence as x inc f(x) inc..hence result follows

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