For particle A,

Initial velocity = o

Final velocity = v

Acceleration = f

Time taken by A = t + m sec

distance travel by A = x + n

For particle B,

Initial velocity = o

Final velocity = v

Acceleration = f'

Time taken by B = t sec

Distance travel by B = x

For particle A,

v = f(t + m)

For particle B,

v = f' t

$$\therefore$$ f'(t) = f(t + m)

$$\Rightarrow$$ f'(t) $$-$$ ft = fm

$$ \Rightarrow {t^2} = {{{f^2}{m^2}} \over {{{(f' - f)}^2}}}$$

Also, for particle A,

$$n + x = {1 \over 2}f{(t + m)^2}$$

For particle B,

$$x = {1 \over 2}f'{t^2}$$

$$\therefore$$ $$x + {1 \over 2}f'{t^2} = {1 \over 2}f{(t + m)^2}$$

$$ \Rightarrow n + {1 \over 2}f'{t^2} = {1 \over 2}f{t^2} + ftm + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {1 \over 2}(f' - f){t^2} = fm \times {{fm} \over {f' - f}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {1 \over 2}(f' - f) \times {{{f^2}{m^2}} \over {{{(f' - f)}^2}}} = {{{f^2}{m^2}} \over {(f' - f)}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {{{f^2}{m^2}} \over {2(f' - f)}} = {{{f^2}{m^2}} \over {f' - f}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n - {{{f^2}{m^2}} \over {2(f' - f)}} = {1 \over 2}f{m^2}$$

$$ \Rightarrow n(f' - f) - {{{f^2}{m^2}} \over 2} = {1 \over 2}f(f' - f){m^2}$$

$$ \Rightarrow n(f' - f) - {{{f^2}{m^2}} \over 2} = {1 \over 2}ff'{m^2} - {1 \over 2}{f^2}{m^2}$$

$$ \Rightarrow n(f' - f) = {1 \over 2}ff'{m^2}$$

$$y = b{e^{ - {x \over a}}}$$

বা, $${e^{{x \over a}}}y = b$$

বা, $${x \over a} + \log y = \log b$$

বা, $${1 \over a} + {1 \over y}.{{dy} \over {dx}} = 0$$

বা, $${{dy} \over {dx}} = - {y \over a}$$

(0, 0), (a, 0) ও (2a, 0) বিন্দুতে $${{dy} \over {dx}} = 0$$

$$\therefore$$ (A), (C) ও (D) সিদ্ধ হয় না।

(B) এর ক্ষেত্রে $${{dy} \over {dx}} = - {b \over a}$$

$$\therefore$$ $$y = b$$