CATAlgebra > HardEntered answer:✅ Correct Answer: 34Related 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 2023 Slot 2Let k be the largest integer such that the equation (x−1)2+2kx+11=0(x - 1)^2 + 2kx + 11 = 0(x−1)2+2kx+11=0 has no real roots. If y is a positive real number, then the least possible value of k/4y+9yk/4y + 9yk/4y+9y 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