26 February, 06:38

# Which of the following subsets of P2P2 are subspaces of P2P2? A. p′ (t) p (t) B. p (t) C. p (t) p′ (6) = p (7) D. ∫10p (t) dt=0 ∫01p (t) dt=0 E. p′ (t) + 2p (t) + 8=0p (t) F. p (t) p (t)

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1. 26 February, 07:00
0
The correct options are;

E. p (t) p (t)

F. p (5) = 6 p (5) = 6

Step-by-step explanation:

A. 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) is not a subspace of P2

B. p (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) for all tt is not a subspace of P2

C. 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) is not a subspace of P2

D. ∫10p (t) dt=0p (t)

∫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) ∫01p (t) dt=0 is not a subspace of P2

E. p′ (t) + 2p (t) + 8=0p (t)

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) + 2p (t) + 8=0 p′ (t) + 2p (t) + 8=0 is a subspace of P2

F. p (5) = 6p (t)

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 is a subspace of P2