Quadratic problem...

We can differentiate it, we get 2k-2 and equate it to 0. Then we have k=1

for the minimum value, we want k^2-2k to be negative, and the only case that can happen is k^2<2k solving this we get k<2.since 0 doesn't make k^2-2k negative and negative numbers convert k^2-2k to k^2+2k, 1 is the only possible integer solution

-D/4a gives us the minimum value which is 9 in this case. Putting the whole value of expression as 9 we get k^2-2k+1=0 which is (k-1)^2 so we get k as 1

f(k)=k^2 -2k + 10 by adding and subtracting 1, we can make a perfect square with constant. So, f(k)= k^2 -2k +1 -1+10 = k^2 -(2)(k)(1) +1 -1+10 = (k-1)^2 +9
on putting k = 1 in the above function, we get, f(k)= (1-1)^2 +9 = 0^2 +9 = 9 which is the minimum value of f(k). Hence, answer is 1