A straight rod of length L extends from $$x = a$$ to $$x = L + a$$. The gravitational force it exerts on a point mass 'm' at $$x = 0$$, if the mass per unit length of the rod is $$A + Bx^2$$, is given by:

We have a point mass of magnitude $$m$$ placed at the origin $$x = 0$$. A thin straight rod lies along the positive $$x$$-axis, beginning at $$x = a$$ and ending at $$x = a + L$$. Its mass per unit length (linear mass density) varies with position according to

$$\lambda(x)=A + Bx^{2}.$$

Our aim is to find the total gravitational force (magnitude) exerted by every small element of the rod on the mass at the origin. Throughout we shall use Newton’s law of gravitation, which in differential form for two small masses separated by a distance $$r$$ is

$$dF=\frac{G\,dm_1\,dm_2}{r^{2}}.$$

Here $$dm_1 = m$$ (the point mass) and $$dm_2 = dm$$ (an infinitesimal element of the rod). The separation between the element located at position $$x$$ and the origin is simply $$r = x$$. Substituting these facts, we get

$$dF = \frac{G\,m\,dm}{x^{2}}.$$

Because each element lies on the positive $$x$$-axis while the point mass is at the origin, every elemental force acts along the negative $$x$$-direction. We shall calculate its magnitude first and later attach the direction if required.

The infinitesimal mass element of the rod is given by the product of its linear density and an infinitesimal length $$dx$$:

$$dm = \lambda(x)\,dx = (A + Bx^{2})\,dx.$$

Placing this expression for $$dm$$ into the formula for $$dF$$, we obtain

$$dF = \frac{G\,m\,(A + Bx^{2})}{x^{2}}\;dx.$$

Now we separate the fraction:

$$dF = Gm\left(\frac{A}{x^{2}} + B\right)\,dx.$$

To find the total force, we integrate $$dF$$ from the near end of the rod at $$x = a$$ to the far end at $$x = a + L$$:

$$F = Gm \int_{a}^{a+L} \left(\frac{A}{x^{2}} + B\right)\,dx.$$

We evaluate the integral term by term.

First term: for $$\displaystyle \int \frac{A}{x^{2}}\,dx$$, we recall the standard result

$$\int x^{-2}\,dx = -x^{-1} + C,$$

so

$$\int \frac{A}{x^{2}}\,dx = A\int x^{-2}\,dx = -\frac{A}{x}.$$

Second term: for $$\displaystyle \int B\,dx$$, $$B$$ is a constant, hence

$$\int B\,dx = Bx.$$

Combining these results, the definite integral becomes

$$F = Gm\left[\,-\frac{A}{x} + Bx\;\right]_{x=a}^{x=a+L}.$$

We substitute the limits. First, the upper limit $$x = a + L$$:

Upper contribution  $$= -\frac{A}{a + L} + B(a + L).$$

Next, the lower limit $$x = a$$:

Lower contribution  $$= -\frac{A}{a} + Ba.$$

Subtracting the lower contribution from the upper gives the net value of the bracketed expression:

$$\begin{aligned} F &= Gm\Bigg[\left(-\frac{A}{a + L} + B(a + L)\right) \;-\; \left(-\frac{A}{a} + Ba\right)\Bigg] \\ &= Gm\Bigg[-\frac{A}{a + L} + B(a + L) + \frac{A}{a} - Ba\Bigg]. \end{aligned}$$

The two $$Ba$$ terms cancel:

$$B(a + L) - Ba \;=\; BL.$$

Hence the expression inside the brackets simplifies to

$$-\frac{A}{a + L} + \frac{A}{a} + BL.$$

Reordering the terms more conventionally, we write

$$\frac{A}{a} - \frac{A}{a + L} + BL.$$

Therefore the magnitude of the gravitational force is

$$F = Gm\left[A\!\left(\frac{1}{a} - \frac{1}{a + L}\right) + BL\right].$$

This matches the expression given in Option D.

Hence, the correct answer is Option D.