CATAlgebra > Medium202020232323454545484848✅ Correct Option: 2Related questions:CAT 2020 Slot 2If xxx and yyy are positive real numbers satisfying x+y=102x + y = 102x+y=102, then the minimum possible value of 2601(1+1x)(1+1y)2601(1+\frac{1}{x})(1+\frac{1}{y})2601(1+x1)(1+y1) isCAT 2018 Slot 1Let f(x)=f(x) =f(x)= min{2x2,52−5x2x², 52-5x2x2,52−5x}, where x is any positive real number. Then the maximum possible value of f(x)f(x)f(x) isCAT 2018 Slot 2Let f(x)=f(x)=f(x)= max{5x5x5x, 52−2x252-2x^252−2x2}, where xxx is any positive real number. Then the minimum possible value of f(x)f(x)f(x) is