Which of the following subsets of P2P2 are subspaces of P2P2? A. {p(t) | p′(t){p(t) | p′(t) is constant }} B. {p(t) | p(−t)=p(t){p(t) | p(−t)=p(t) for all t}t} C. {p(t) | p′(6)=p(7)}{p(t) | p′(6)=p(7)} D. {p(t) | ∫10p(t)dt=0}{p(t) | ∫01p(t)dt=0} 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}

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

the y will be in still but your answer will be y x 3

step-by-step explanation:

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the answer is d on edge

step-by-step explanation: