Let $C$ be a code with a distance $d=2t+u+1$. Determine the maximum number of errors that $C$ can detect if used simultaneously to correct $t$ errors.
I claim that the maximum number of errors that $C$ can detect if used simultaneously to correct $t$ errors is $t+u$. This is because the spheres about codewords of $C$ with radius e, then any pattern of $t+u$ errors cannot take a codeword into a word that is contained in some spheres about another codeword $(t+u+t =2t+u \neq d$). This, means that the a received word obtained by introducing $t+u$ errors cannot lie in any codeword sphere and C will detect it.
Well, this depends on your definition of "simultaneously detect".
If you count those errors that are corrected as also being detected, the answer is as you give.
If you only count errors that are detected only but not corrected ( as is customary)the answer is $u.$
To see this consider the extreme points $t=0,$ which gives $d=u+1,$ consistent with my claim. Similarly $u=0$ gives $d=2t+1,$ also consistent with my claim.