Introduction:
We are given two linearly independent solutions, y1 and y2, of the differential equation y''(sin x)y = 0, where 0 ≤ x ≤ 1. We need to determine the properties of the Wronskian, g(x) = W(y1, y2)(x), and check which of the given options are correct.

Wronskian:
The Wronskian of two functions y1 and y2 is given by the formula:
W(y1, y2)(x) = y1(x)y2'(x) - y1'(x)y2(x)

Properties of the Wronskian:
1. If the Wronskian is not identically zero on an interval, then the two functions are linearly independent on that interval.
2. If the Wronskian is identically zero on an interval, then the two functions may or may not be linearly dependent on that interval.

Analysis of the given options:
(A) g′ > 0 on [0, 1]
(B) g′ < 0="" on="" [0,="" />
(C) g′ vanishes at only one point of [0, 1]
(D) g′ vanishes at all points of [0, 1]

Analysis of option (A):
To determine whether g′ > 0 on [0, 1], we need to analyze the sign of g′(x). Let's calculate g′(x) using the formula for the Wronskian and the product rule of differentiation:

g′(x) = (y1(x)y2''(x) - y1''(x)y2(x)) + (y1'(x)y2'(x) - y1'(x)y2'(x))
= y1(x)y2''(x) - y1''(x)y2(x)

The differential equation y''(sin x)y = 0 implies that y2''(x) = 0 for all x in [0, 1]. Substituting this in the above expression, we get:

g′(x) = y1(x)(0) - y1''(x)y2(x)
= -y1''(x)y2(x)

Since y1 and y2 are linearly independent solutions, y1''(x) ≠ 0 and y2(x) ≠ 0 for all x in [0, 1]. Therefore, g′(x) = -y1''(x)y2(x) < 0="" for="" all="" x="" in="" [0,="" />

Conclusion for option (A):
Option (A) is incorrect because g′ < 0="" on="" [0,="" />

Analysis of option (B):
To determine whether g′ < 0="" on="" [0,="" 1],="" we="" need="" to="" analyze="" the="" sign="" of="" g′(x).="" from="" the="" previous="" analysis,="" we="" know="" that="" g′(x)="-y1''(x)y2(x)" />< 0="" for="" all="" x="" in="" [0,="" />

Conclusion for option (B):
Option (B) is correct because g′ < 0="" on="" [0,="" />

Analysis of option (C