Answer:
D. Since y₁ is not a constant multiple of y₂ on (0,1), the functions are linearly independent on (0,1).
Explanation:
y₁=2Cos²t-1
y₂=6Cos2t
We want to determine if y₁ and y₂ are linearly independent in the interval (0,1).
To do this, we show that there does not exist any c₁ and c₂ in (0,1) that makes the expression:
f(t)=c₁(2Cos²t-1)+c₂6Cos2t=0.
If c₁ and c₂=0
f(t)=0
Let c₁=1 and c₂=-⅓
f(t)=c₁(2Cos²t-1)+c₂6Cos2t
=2Cos²t-1-⅓*6(cos²t-sin²t)
=2Cos²t-1-2cos²t+2sin²t
=2sin²t-1
Since f(t)≠0, y₁ is not a constant multiple of y₂ and the functions are linearly independent on (0,1).