Let n be a positive integer and $0 < a < b < 1$. The total number of real roots of the equation $(x − a)^{2n+1} + (x − b)^{2n+1} = 0$ is?
Taking the derivative $f'(x)>0$ Therefore it will cross x axis only once , so it has only one root.
Is this solution enough or should there be more tests taking second order derivative etc?
If $f '(x) > 0$ then there is at most one solution. You need to use something like $f(a) < 0 < f(b)$ to conclude that a solution actually exists (and it is between $a$ and $b$).
The unique solution is $(a+b)/2$ - see the answer of @lab bhattacharjee.
Hint:
$((x-a)/(x-b))^{2n+1}=-1$
The only real root of $z^p=-1$ for odd $p$ is $-1$
If $f(x) = (x − a)^{2n+1} + (x − b)^{2n+1} $, then $f'(x) = (2n+1)((x − a)^{2n} + (x − b)^{2n}) $ so that, as OP wrote $f'(x) > 0$ if $a \ne b$.
Also, as Catalin Zara pointed out, $x_0=\frac{a+b}{2}$ is a root, since $x_0-a = -(x_0-b) =\frac{b-a}{2}$.
Note that this holds for any real $a$ and $b$, not just between $0$ and $1$.
Even if $a=b$, there is still only one real root since $f'(x) = 0$ only at $x=a$ and $f'(x) > 0$ elsewhere, and $x=a$ is where $f(x) = 0$.
