CATAlgebra > MediumEntered answer:✅ Correct Answer: 40Related 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 2020 Slot 1The number of real-valued solutions of the equation 2x+2−x=2−(x−2)22^x + 2^{-x} = 2 - (x - 2)^22x+2−x=2−(x−2)2 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) is