If f(x)=sincos-11-22x1+22x and its first derivative with respect to x is -baloge2 when x=1, where a and b are integers, then the minimum value of a2-b2

f(x)=sincos-11-22x1+22x2x=tanθ ∴f(x)=sincos-1(cos2θ)=sin2θf1(x)=2cos2θ·ddx(θ) =21-tan2θ1+tan2θ·ddxtan-12x=21-22x1+22x·2x·log21+22x Given f1(1)=ba·log2⇒2-352·log25=ba·log2 ba=-1225a=25k,b=-12ka2-b2=481k2 Minimum value of a2-b2=481