Let $f$ and $g$ be two continuous function defined on $[0,1]$ such that $\sup f(x)= \sup g(x)$ for $x$ belongs to $[0,1]$ Then:
a) There exists $x$ in $[0,1]$ s.t. $f(x)=g(x)$.
b) There exists $x$ in $[0,1]$ s.t. $f(x)= g(x)-2$.
c) There exists $x$ in $[0,1]$ s.t. $f(x)=g(x)= \sup f(t)$ for $t$ in $[0,1]$.
d) There exists $x$ in $[0,1]$ s.t. $f(x)^2 +2f(x)=g(x)^2 +2g(x)$.
I tried this with the functions $f(x)=-x$ and $f(x)=x-1$. That proves choice b) false, and if choice a) is true, then choice d) true, as $f(x)-g(x)=0$. For c), take $f(x)= \dfrac{x}{x-1}$ and $g(x)=1$.
For option 'a'.
On a closed interval a continuous function reaches its supremum: there exists $x_f \in [0,1]: f(x_f) =\sup f(x)$.
If $f$ and $g$ reach their common supremum at the same $x$, we're done. So suppose they do not: $x_f\ne x_g$. And WLOG suppose $x_f<x_g$.
Then the difference $d(x) = f(x)-g(x)$ is continuous on $[0,1]$ hence on $[x_f,x_g]\subseteq[0,1]$, and it's positive at $x_f$ and negative at $x_g$.
Then the IVT implies there exists such $x_0\in[x_f,x_g]\subset[0,1]$ which zeroes $d$: $$d(x_0)=0$$ so $$f(x_0)=g(x_0)$$
....and of course a) implies d).