The function is given by: (1)² + a(1) + b = 2(1) +1 1 + a + b = 3 a + b = 2 Now, for differentiability at x=1, the left-hand derivative must be equal to the right-hand derivative. The derivative of x² + ax + b is 2x + a. The derivative of 2x+1 is 2. Left-hand derivative at x=1: For differentiability, f′(1^-) = f′(1⁺): 2+a=2 a=0 Now substitute the value of a into the continuity equation a + b = 2 0+ b=2 b=2 So, the values of a and b are a=0 and b=2. Therefore, (a,b)=(0,2).