The least value of k for which f(x) = x2 + kx +1 is increasing on (1, 2), is

To find the least value of k for which f(x) = x^2 + kx + 1 is increasing on (1, 2), we need to find the derivative of the function and set it greater than 0, because a function is increasing where its derivative is positive.

Step 1: Find the derivative of f(x) f'(x) = 2x + k

Step 2: Set the derivative greater than 0 and solve for k 2x + k > 0 k > -2x

Since we want the function to be increasing on the interval (1, 2), we substitute these values into the inequality.

For x = 1, k > -2(1) => k > -2 For x = 2, k > -2(2) => k > -4

Since we want the least value of k for which the function is increasing on the entire interval (1, 2), we take the maximum of these two values, which is -2.

Therefore, the least value of k for which f(x) = x^2 + kx + 1 is increasing on (1, 2) is -2.