If exactly one root of the quadratic equation $\displaystyle x^2 - (a + 1) x + 2a = 0$ lies in the interval $(0, 3)$, then the set of values $a$ is given by

Let f(x) = x^2 - (a+1)x + 2a = (x-a)(x-2). Roots are a and 2. For exactly one root in (0, 3), we check the boundary conditions. If a=2, both roots are 2 (not 'exactly one'). If a <= 0, root 2 is in (0, 3) and root a is <= 0 (outside). If a > 6, root 2 is in (0, 3) and root a > 6 (outside). If a=6, roots are 6 and 2 (only 2 is in range).