1. Let the roots of $x^2-mx+4=0$ be $\alpha,\beta$.

By Vieta's formulas, $\alpha+\beta=m, \qquad \alpha\beta=4.$

We are given:

• both roots are real and distinct,
• both lie in the interval $[1,5]$.

2. Since both roots lie in $[1,5]$, we have $1\le \alpha,\beta \le 5.$ Also, $\alpha\beta=4.$

Now let one root be $t$. Then the other root is $\dfrac{4}{t}$. So both roots are in $[1,5]$ implies $t\in[1,5] \quad \text{and} \quad \frac{4}{t}\in[1,5].$

From $1\le \frac{4}{t}\le 5,$ we get $\frac{4}{5}\le t\le 4.$ Combining with $t\in[1,5]$ gives $t\in[1,4].$

Thus the roots are of the form $t \text{ and } \frac{4}{t}, \qquad t\in[1,4].$

3. Then $m=t+\frac{4}{t}, \qquad t\in[1,4].$ We must also use the fact that the roots are distinct, so $t\ne \frac{4}{t} \implies t^2\ne 4 \implies t\ne 2.$

Now consider $f(t)=t+\frac{4}{t}, \qquad t\in[1,4],\ t\ne 2.$

Differentiate: $f'(t)=1-\frac{4}{t^2}.$ So,

• $f'(t)<0$ for $1\le t<2$,
• $f'(t)>0$ for $2<t\le 4$.

Hence $f(t)$ has minimum at $t=2$: $f(2)=2+2=4.$ But $t=2$ is not allowed because roots must be distinct. Therefore $m>4.$

At the endpoints, $f(1)=1+4=5, \qquad f(4)=4+1=5.$ So the possible values of $m$ are $4<m\le 5.$

But check $m=5$: $x^2-5x+4=0 \implies (x-1)(x-4)=0,$ whose roots are $1$ and $4$, both real, distinct, and in $[1,5]$. So actually $m=5$ is allowed.

Thus, $m\in(4,5].$

4. Now compare with the options:

• A: $(-5,-4)$ — impossible since sum of roots is positive.
• B: $(4,5)$ — misses the allowed value $m=5$.
• C: $(5,6)$ — impossible.
• D: $(3,4)$ — impossible since distinct roots require $m>4$.

So the mathematically correct interval is $\boxed{(4,5]}.$ This is not exactly present among the options. The closest listed option is B, but it excludes $m=5$, which should be included.

5. Verification with discriminant: For real and distinct roots, $m^2-16>0 \implies |m|>4.$ Since roots lie in $[1,5]$, their sum is positive, so $m>4$. Also the maximum sum under product $4$ with roots in $[1,5]$ occurs at $1$ and $4$, giving $m\le 5.$ Hence again, $m\in(4,5].$