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**Hint:**First of all, draw the figure for this problem with the help of the given data to get a clear idea of what we have to find. Then use the distance-time formula and second derivative test to find the shortest time period. So, use this concept to reach the solution to the given problem.

**Complete step-by-step solution:**Given that the swimmer is at a point S which is at a distance ‘d’ km from point A.

Let the house of the swimmer be at point B.

Since the distance between point A and his house i.e., B is \[L{\text{ km}}\]. So, \[AB = L{\text{ km}}\].

Let the swimmer land at point C on the shore and let \[AC = x{\text{ km}}\]as shown in the below figure:

Therefore, from the figure \[SC = \sqrt {{x^2} + {d^2}} \] and \[CB = \left( {L - x} \right)\]

We know that \[{\text{time}} = \dfrac{{{\text{distance}}}}{{{\text{speed}}}}\]

Given that speed of the swimmer to swim from S to C \[ = u{\text{ km/hr}}\]

So, time taken to travel from S to C \[ = \dfrac{{\sqrt {{x^2} + {d^2}} }}{u}\]

Also given that speed of the swimmer to walk from C to B \[ = v{\text{ km/hr}}\]

So, time taken to travel from S to B \[ = \dfrac{{L - x}}{v}\]

Let \[T\] be the time taken to travel from S to B

time taken to travel from S to B = time taken to travel from S to C + time taken to travel from C to B

\[ \Rightarrow T = \dfrac{{\sqrt {{x^2} + {d^2}} }}{u} + \dfrac{{L - x}}{v}\]

Let \[f\left( x \right) = T = \dfrac{{\sqrt {{x^2} + {d^2}} }}{u} + \dfrac{L}{v} - \dfrac{x}{v}\]

Differentiating \[f\left( x \right)\] with respective to \[x\], we have

\[

\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{{\sqrt {{x^2} + {d^2}} }}{u}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{L}{v}} \right) - \dfrac{d}{{dx}}\left( {\dfrac{x}{v}} \right) \\

\Rightarrow f'\left( x \right) = \dfrac{1}{u}\dfrac{d}{{dx}}\left( {\sqrt {{x^2} + {d^2}} } \right) + \dfrac{d}{{dx}}\left( {\dfrac{L}{v}} \right) - \dfrac{1}{v}\dfrac{d}{{dx}}\left( x \right) \\

\Rightarrow f'\left( x \right) = \dfrac{1}{u}.\dfrac{{2x}}{{2\sqrt {{x^2} + {d^2}} }} + 0 - \dfrac{1}{v} \\

\]

We know that for the least and greatest value of any function is obtained by equating its first derivate to zero.

For maximum or minimum, put \[f'\left( x \right) = 0\]

\[

\Rightarrow \dfrac{1}{u}.\dfrac{{2x}}{{2\sqrt {{x^2} + {d^2}} }} + 0 - \dfrac{1}{v} = 0 \\

\Rightarrow \dfrac{x}{{u\sqrt {{x^2} + {d^2}} }} = \dfrac{1}{v} \\

\Rightarrow \dfrac{{\sqrt {{x^2} + {d^2}} }}{x} = \dfrac{v}{u} \\

\]

Squaring on both sides, we get

\[

\Rightarrow \dfrac{{{x^2} + {d^2}}}{{{x^2}}} = \dfrac{{{v^2}}}{{{u^2}}} \\

\Rightarrow \dfrac{{{x^2}}}{{{x^2}}} + \dfrac{{{d^2}}}{{{x^2}}} = \dfrac{{{v^2}}}{{{u^2}}} \\

\Rightarrow 1 + \dfrac{{{d^2}}}{{{x^2}}} = \dfrac{{{v^2}}}{{{u^2}}} \\

\Rightarrow \dfrac{{{d^2}}}{{{x^2}}} = \dfrac{{{v^2}}}{{{u^2}}} - 1 \\

\Rightarrow \dfrac{{{d^2}}}{{{x^2}}} = \dfrac{{{v^2} - {u^2}}}{{{u^2}}} \\

\Rightarrow {x^2} = \dfrac{{{u^2}{d^2}}}{{{v^2} - {u^2}}} \\

\]

Rooting on both sides, we have

\[

\Rightarrow x = \pm \dfrac{{ud}}{{\sqrt {{v^2} - {u^2}} }} \\

\therefore x = \dfrac{{ud}}{{\sqrt {{v^2} - {u^2}} }}{\text{ or }}\dfrac{{ - ud}}{{\sqrt {{v^2} - {u^2}} }} \\

\]

We know that, the least value is obtained for the value of \[x\] at which \[\dfrac{{{d^2}f\left( x \right)}}{{d{x^2}}} > 0\] and the greatest value is obtained for the value of \[x\] at which \[\dfrac{{{d^2}f\left( x \right)}}{{d{x^2}}} < 0\].

Now, consider the second derivative of \[f\left( x \right)\]

\[ \Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {f'\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {\dfrac{x}{{u\sqrt {{x^2} + {d^2}} }}} \right)\]

\[

\Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {f'\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {\dfrac{x}{{u\sqrt {{x^2} + {d^2}} }}} \right) \\

\therefore f''\left( x \right) = \dfrac{{{d^2}}}{{u\left( {{x^2} + {d^2}} \right)\sqrt {{x^2} + {d^2}} }} \\

\]

For \[x = \dfrac{{ud}}{{\sqrt {{v^2} - {u^2}} }}\] we have \[f''\left( {\dfrac{{ud}}{{\sqrt {{v^2} - {u^2}} }}} \right) > 0\].

So, the shortest possible time is \[\dfrac{{ud}}{{\sqrt {{v^2} - {u^2}} }}\].

**Thus, the correct option is A. \[\dfrac{{ud}}{{\sqrt {{v^2} - {u^2}} }}\]**

**Note:**For the least and greatest value of any function is obtained by equating its first derivate to zero. The least value is obtained for the value of \[x\] at which \[\dfrac{{{d^2}f\left( x \right)}}{{d{x^2}}} > 0\] and the greatest value is obtained for the value of \[x\] at which \[\dfrac{{{d^2}f\left( x \right)}}{{d{x^2}}} < 0\].

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