This question has Statement - I and Statement - II. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement - I: A point particle of mass $$m$$ moving with speed $$v$$ collides with stationary point particle of mass $$M$$. If the maximum energy loss possible is given as $$f\left(\frac{1}{2}mv^2\right)$$ then $$f = \left(\frac{m}{M+m}\right)$$.
Statement - II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

We have a one-dimensional situation in which a point particle of mass $$m$$ is moving with speed $$v$$ and strikes another point particle of mass $$M$$ that is initially at rest. The kinetic energy before collision is therefore

$$K_{\text{initial}}=\dfrac12\,m\,v^{2}.$$

Linear momentum is conserved in every type of collision, so the total initial momentum is

$$p_{\text{initial}} = m\,v.$$

Statement II tells us that the greatest possible loss of kinetic energy occurs when the two particles stick together. Such a collision is called a perfectly inelastic collision. We now analyse this perfectly inelastic case in detail.

After sticking, the two particles move as a single composite body of mass $$m+M$$ with some common speed $$V.$$ By conservation of linear momentum,

$$p_{\text{initial}} = p_{\text{final}}, \qquad\text{so}\qquad m\,v = (m+M)\,V.$$

Hence,

$$V = \dfrac{m\,v}{m+M}.$$

The kinetic energy after the collision is

$$K_{\text{final}} = \dfrac12\,(m+M)\,V^{2}.$$

Substituting the value of $$V$$ obtained above, we get

$$\begin{aligned} K_{\text{final}} &= \dfrac12\,(m+M)\left(\dfrac{m\,v}{m+M}\right)^{2} \\ &= \dfrac12\,(m+M)\,\dfrac{m^{2}\,v^{2}}{(m+M)^{2}} \\ &= \dfrac12\,\dfrac{m^{2}\,v^{2}}{m+M}. \end{aligned}$$

The loss of kinetic energy, $$\Delta K,$$ is the difference between the initial and final kinetic energies:

$$\begin{aligned} \Delta K &= K_{\text{initial}} - K_{\text{final}} \\ &= \dfrac12\,m\,v^{2} \;-\; \dfrac12\,\dfrac{m^{2}\,v^{2}}{m+M}. \end{aligned}$$

We now factor out the common factor $$\dfrac12\,m\,v^{2}:$$

$$\begin{aligned} \Delta K &= \dfrac12\,m\,v^{2}\left[1 - \dfrac{m}{m+M}\right]. \end{aligned}$$

Simplifying the bracket,

$$\begin{aligned} 1 - \dfrac{m}{m+M} &= \dfrac{m+M}{m+M} - \dfrac{m}{m+M} \\ &= \dfrac{m+M - m}{m+M} \\ &= \dfrac{M}{m+M}. \end{aligned}$$

Therefore, the maximum possible loss of kinetic energy is

$$\Delta K_{\max} = \dfrac12\,m\,v^{2}\left(\dfrac{M}{m+M}\right).$$

Comparing this result with the form stated in Statement I, namely $$f\left(\dfrac12\,m\,v^{2}\right),$$ we see that

$$f = \dfrac{M}{m+M},$$

whereas Statement I claims $$f = \dfrac{m}{M+m}.$$ Hence Statement I is false.

Statement II, asserting that maximum energy loss occurs when the particles stick together, is exactly what we have used above; it is therefore true.

Combining these conclusions, we find that Statement I is false while Statement II is true. This matches Option B in the list provided.

Hence, the correct answer is Option B.