Answer

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**Hint: -**We will simply use $speed = \dfrac{{dis\tan ce}}{{time}}$ along with some geometry. In order to find the maximum and minimum of the particular quantity we will put slope equal to zero.

**Complete step by step answer:**

Given: Velocity of man relative to water $ = v$

Width of river $ = d$

Velocity of river $ = u$

Now, time required for crossing the river is given by

$\Rightarrow$ $t = \dfrac{d}{{{v_{parallel}}}}$

From figure, ${v_{parallel}} = v\cos \theta $ and ${v_{perpendicular}} = v\sin \theta $

$\therefore t = \dfrac{d}{{v\cos \theta }} \cdots \left( 1 \right)$

For the time to be minimum, $\theta $ should be zero

$\Rightarrow$ $t = \dfrac{d}{{v\cos 0}} = \dfrac{d}{v}$

Now, $x = $ horizontal velocity $ \times $ time

$\Rightarrow$ $x = \left( {u - v\sin \theta } \right) \times \dfrac{d}{{\cos \theta }}$

For minimum time,

$

x = \left( {u - v\sin 0} \right)\dfrac{d}{v} \\

x = \dfrac{{du}}{v} \\

$

From above,

\[

x = (u - v\sin \theta ) \times \dfrac{d}{{v\cos \theta }} \\

x = \dfrac{{ud\sec \theta }}{v} - d\tan \theta \\

\]

For $x$ to be minimum,

$\dfrac{{dx}}{{d\theta }} = 0$

That is,

$

\dfrac{{d\left[ {u\dfrac{d}{v}\sec \theta - d\tan \theta } \right]}}{{d\theta }} = 0 \\

\Rightarrow \dfrac{{ud}}{v} \times \sec \theta \tan \theta - d{\sec ^2}\theta = 0 \\

\Rightarrow \dfrac{u}{v}\tan \theta - \sec \theta = 0 \\

\dfrac{{u\sin \theta }}{{v\cos \theta }} = \dfrac{1}{{\cos \theta }} \\

\sin \theta = \dfrac{v}{u} \\

\theta = {\sin ^{ - 1}}\left( {\dfrac{v}{u}} \right) \\

$

The swimmer has to swim in the direction making an angle of $\dfrac{\pi }{2} + {\sin ^{ - 1}}\left( {\dfrac{v}{u}} \right)$ with the direction of flow of water.

**Hence, option (A) and (C) are correct.**

**Note: -**Since, in the question, $(u > v)$ hence, we have to add $\dfrac{\pi }{2}$ . If $v > u$ then, $\dfrac{\pi }{2}$ will be subtracted. Hence, for $x$ to be minimum, the swimmer has to swim in a particular direction with the direction of flow of water.

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