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# mid1sol - Math 115a Lecture 1 Fall 2009 Midterm 1 Name...

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Math 115a Midterm 1 Lecture 1 Fall 2009 Name: Signature: Instructions: There are 5 problems. Make sure you are not missing any pages. Unless stated otherwise, you may use without proof anything proven in the sections of the book covered by this test. You may only cite an exercise from the book if it was assigned as homework. Question Points Score 1 10 2 10 3 10 4 15 5 15 Total: 60

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1. (10 points) Recall that P 5 ( R ) is the vector space of polynomials of degree less than or equal to 5. Let A = { f P 5 ( R ): f (0) = 0 } and B = { f P 5 ( R ): f (1) = 0 } . Prove that A B is a subspace of P 5 ( R ) . Solution: Let p ( x ) = 0 for all x R . Certainly p (0) = 0, so p A . Also, p (1) = 0, so p B . Hence p A B . This proves that the zero polynomial is in A B . Now let f, g A B and λ R . We know ( f + g )( x ) = f ( x ) + g ( x ) for any x R , by the definition of addition in P 5 ( R ). In particular, ( f + g )(0) = f (0) + g (0) = 0 + 0 = 0 so f + g A , and ( f + g )(1) = f (1) + g (1) = 0 + 0 = 0 so f + g B . Hence f + g A B . Similarly, ( λf )( x ) = λf ( x ) for any x R , by the definition of scalar multiplication in P 5 ( R ). In particular,
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mid1sol - Math 115a Lecture 1 Fall 2009 Midterm 1 Name...

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