The correct options are
(a) If he crosses the river in minimum time, \(x = \frac{du}v\)
(c) For x to be minimum, he has to swim in a direction making an angle of \(\frac{\pi}{2}+ sin^{-1} \left(\frac vu\right)\) with the direction of the flow of water.
(A) For minimum time the swimmer will have to start moving perpendicular to the river flow. (vertical component of the swimmer's speed with respect to ground should be v)
Then the vertical component of velocity will be v and distance to be covered will be d. Hence time to cross the river = \(\frac dv\)
The x component of the swimmer's speed in ground frame will u.
Hence drift = \(u\times\frac dv\)
The time required for crossing = t = (d/v)cosθ ... (i)
x = (u−vsinθ)t = ud/vsecθ −d tanθ (from equation (i))
For x to be minimum, \(\frac{dx}{d\theta} = 0\)
This gives sin θ = \(\frac vu\)