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Final Solution

# Final Solution - Economics 2P30 Foundations of Economic...

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Economics 2P30 Foundations of Economic Analysis Lester M.K. Kwong Department of Economics Brock University Winter 2013 Final Examination - Suggested Solutions Time: 3 Hours This examination will not be deposited in the library reserve. Section A: Definitions ∗ ∗ ∗ ∗ ∗ ∗ ∗ Define 6 of the following 8 terms in two sentences or less. ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ This Section of the exam is worth 10%. ∗ ∗ ∗ ∗ ∗∗ 1. Continuous function at some point x 2. Strictly concave function 3. Homogeneous function of degree k 4. Mean Value Theorem 5. Complement of a set X 6. Contrapositive of P R 7. Compact set 8. Diameter of a set X Solution : 1. A function f is said to be continuous at some point x if δ > 0, > 0 so that if d ( x,y ) < then d ( f ( x ) , f ( y )) < δ . 2. A function f is said to be strictly concave if for all x y , f ( x + ( 1 ) y ) > f ( x ) + ( 1 ) f ( y ) for all ( 0 , 1 ) . 3. A function f is said to be homogeneous of degree k if for all λ > 0 and for all x > 0, f ( λ x ) = λ k f ( x ) . 4. If f C 1 over [ a,b ] then there exists some c ( a,b ) so that: f ( c )( b a ) = f ( b ) f ( a ) 5. The complement of a set X is X c = { x x X } . 1

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6. The contrapositive of P R is R P . 7. A set is said to be compact if it is both closed and bounded. 8. The diameter of a set X is diam ( X ) = sup { d ( x,y ) x,y X } . Section B: Proofs ∗ ∗ ∗ ∗ ∗ ∗ ∗ Choose 6 of the following 8 questions. ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗ This section of the exam is worth 40%. ∗ ∗ ∗ ∗ ∗∗ For each of the following, answer true or false. If true, prove. If false, derive a counterexample 1. If f C 1 is homogeneous of degree k then f is homogeneous of degree k 1. Solution : True. If f is homogeneous of degree k then: f ( λ x ) = λ k f ( x ) Taking the derivative of the above identity with respect to x yields: f ( λ x ) λ = λ k f ( x ) f ( λ x ) = λ k 1 f ( x ) and hence f is homogeneous of degree k 1. 2. The function f ( x ) = 20 + x is weakly concave and weakly convex over R . Solution : True. f ′′ ( x ) = 0 and so f ′′ 0 and f ′′ 0 for all x R . Hence, f is weakly concave and weakly convex over R .
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