Respuesta :
Answer:
The correct options are;
E. {p(t) | p′(t)+2p(t)+8=0}{p(t) | p′(t)+2p(t)+8=0}
F. {p(t) | p(5)=6}{p(t) | p(5)=6}
Step-by-step explanation:
A. {P(t)| P'(t)} is constant is given by
P(t)=a+bt+ct^2
P'(t) = b + 2ct which is not a constant, therefore
{P(t)| P'(t)} is not a subspace of P2
B. {p(t) | p(−t)=p(t){p(t) | p(−t)=p(t) for all t}t}
Here, we have
P(t)=a+bt+ct^2 and
P(-t)=a-bt+ct^2
P(t) is not equal to P(-t)
Therefore {p(t) | p(−t)=p(t){p(t) | p(−t)=p(t) for all t}t} is not a subspace of P2
C. {p(t) | p′(6)=p(7)}{p(t) | p′(6)=p(7)}
p'(t) = b+ct and p′(6) = b+6c
p(7) = a+7b+49c
Therefore
p′(6) is not equal to p(7) and
{p(t) | p′(6)=p(7)}{p(t) | p′(6)=p(7)} is not a subspace of P2
D. {p(t) | ∫10p(t)dt=0}{p(t) | ∫01p(t)dt=0}
∫10p(t)dt
= at+0.5×bt^2+(ct^3)/3
= a + 0.5b+cr/3 which is not equal to 0
Therefore
{p(t) | ∫10p(t)dt=0}{p(t) | ∫01p(t)dt=0} is not a subspace of P2
E. {p(t) | p′(t)+2p(t)+8=0}{p(t) | p′(t)+2p(t)+8=0}
Here, we have
p′(t)+2p(t)+8=0 given by
b+2c+2 (a+bt+ct^2)+8=0...(1)
When t=2, we have
a = -(16+5b+4c)/2
Substituting the value of a into the equation (1), and simplying we have
(2t-4)×b+(2t-4)×c+ 2t^2-8=0
Therefore when t=2 the above equation =0
Hence
{p(t) | p′(t)+2p(t)+8=0}{p(t) | p′(t)+2p(t)+8=0} is a subspace of P2
F. {p(t) | p(5)=6}{p(t) | p(5)=6}
Here we have
p(t) = a+bt+ct^2
p(5) = a+5b+25c = 6
a=6-5b-25c
Substitution gives
6-5b-25c+bt+ct^2
Which gives on factorization
(t-5)b+(t^2-25)c+6 which is equal to 6 when t=5
Therefore
{p(t) | p(5)=6}{p(t) | p(5)=6} is a subspace of P2
