If both the roots of the quadratic equation $$x^2 - mx + 4 = 0$$ are real and distinct and they lie in the interval (1, 5), then $$m$$ lies in the interval:
Note: In the actual JEE paper interval was $$[ 1, 5 ]$$

Let the two roots of the quadratic equation

$$x^{2}-mx+4=0$$

be denoted by $$\alpha$$ and $$\beta$$. Because the coefficient of $$x^{2}$$ is $$1$$, we can immediately apply Vieta’s relations. We have

$$\alpha+\beta = m \qquad\text{and}\qquad \alpha\beta = 4.$$

According to the question,

1. the roots are real and distinct,
2. each root lies strictly between $$1$$ and $$5$$, i.e. $$1\lt \alpha\lt 5$$ and $$1\lt \beta\lt 5$$.

We now translate every one of these verbal conditions into explicit inequalities in $$m$$ and then intersect them.

Condition for real and distinct roots. For a quadratic $$ax^{2}+bx+c=0$$, the discriminant formula is $$\Delta=b^{2}-4ac$$. Here $$a=1,\; b=-m,\; c=4,$$ so

$$\Delta = (-m)^{2}-4(1)(4)=m^{2}-16.$$

For two distinct real roots we must have $$\Delta\gt 0$$, therefore

$$m^{2}-16\gt 0 \;\Longrightarrow\; m^{2}\gt 16 \;\Longrightarrow\; |m|\gt 4.$$

Thus either $$m\gt 4$$ or $$m\lt -4$$.

Rough bounds from the interval (1,5). Because both $$\alpha$$ and $$\beta$$ lie inside $$(1,5)$$, we immediately get

$$1\lt \alpha\lt 5,\qquad 1\lt \beta\lt 5.$$

Add the two double inequalities term-wise to obtain a bound on their sum:

$$1+1\lt \alpha+\beta\lt 5+5 \;\Longrightarrow\; 2\lt \alpha+\beta\lt 10.$$

But $$\alpha+\beta=m,$$ so

$$2\lt m\lt 10.$$

Combine this with the previous real-root requirement $$(m\gt 4 \text{ or } m\lt -4)$$ to see that the only possible overlap is

$$4\lt m\lt 10.$$

Sharper upper bound using the sign of the quadratic at the end-points. Because the parabola opens upward $$(a=1\gt 0)$$, its values are positive outside the interval determined by the two roots and negative between them. Since both roots are to the left of $$x=5$$, the value of the quadratic at $$x=5$$ must be positive:

$$\bigl(5\bigr)^{2}-m\bigl(5\bigr)+4 \gt 0.$$

Simplify:

$$25-5m+4 \gt 0 \;\Longrightarrow\; 29-5m\gt 0 \;\Longrightarrow\; m\lt \frac{29}{5}=5.8.$$

Likewise, because both roots are to the right of $$x=1$$, the value at $$x=1$$ must also be positive:

$$\bigl(1\bigr)^{2}-m\bigl(1\bigr)+4 \gt 0.$$

This gives

$$1-m+4 \gt 0 \;\Longrightarrow\; 5-m\gt 0 \;\Longrightarrow\; m\lt 5.$$

The inequality $$m\lt 5$$ is stronger than $$m\lt 5.8$$, so we retain the tighter condition. Appending it to our previous lower bound $$m\gt 4$$, we arrive at

$$4\lt m\lt 5.$$

Finally, the discriminant is strictly positive for every $$m$$ in this open interval (because $$m\gt 4\Rightarrow m^{2}-16\gt 0$$), ensuring that the roots remain distinct. All conditions are now simultaneously satisfied exactly when

$$m \in (4,5).$$

The option that matches this interval is Option D.

Hence, the correct answer is Option D.