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Homework Problems (Ch 2&3) Module 1 – Week 1 Exercise 2.5: Function and exponents 1. Graph...

Homework Problems (Ch 2&3) Module 1 – Week 1

Exercise 2.5: Function and exponents 1. Graph the functions a) y = 16 + 2x b) y = 8 – 2x c) y = 2x +12 (In each case, consider the domain as consisting of nonnegative real numbers only) 5. Condense the following expressions: b) x^a\times x^b\times x^c c){\ \ x}^3\times y^3\times z^3 6. Find: b) \left(x^{1/2}\times x^{1/3}\right)/x^{2/3}

Exercise 3.2: Single Unknown 2. Let the demand and supply function be as follows: a) Q_d= 51 – 3P b) Q_d=30-2P Q_s=6P-10 Q_s=-6+5P Find P* and Q* by elimination of variables. (Use fractions rather than decimals) 4. If (b + d) =0 in the linear market model, can an equilibrium solution be found by using (3.4) and (3.5)? Why or why not?

Exercise 3.3: Nonlinear equations 1. Find the zeros of the following functions graphically: b) g\left(x\right)=2x^2-4x-16 4 c) For the following function determine if x = 1 is a root: 5. Find the rational roots, if any, of the following: b) 8x^3+6x^2-3x-1=0 6. Find the equilibrium solution for each of the following models: a) Q_d=Q_s Q_d=3-P^2 Q_s=6P-4

Exercise 3.5: More than one unknown 2. Let the national-income model be: Y=C+I_0+G

C=a + b (Y-T_0) (a > 0, 0 < b < 1)

G =gY (< g < 1)

a) Identify the endogenous variables.

b) Give the economic meaning of the parameter g.

c) Find the equilibrium national income.

d) What restriction on the parameters is needed for a solution to exist?

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2:5) 1) Graphing the functions: The best way to graph a linear function is simply finding the horizontal (x-axis) and vertical (y- axis) intercepts, and joining the two. Since we are considering domain as consisting of non negative real number only, we are concerned only with the first quadrant of XY-plane. ay 162x y-2x- 16 Dividing whole expression by 16, (-2x/16+(y/16) 1 Or: x (-8)+ y(16) -1 So, we have horizontal intercept as (-8) and vertical intercept as 16. Following is the required graph:40 У-16:+ 2x 36 32 28 24 20 12 8 64-2 2 4 68 102 14 6 18 20b)y=8.2x 2x + y = 8Dividing whole expression by 8: x/4+y/8-1 So, we have horizontal intercept as 4 and vertical intercept as 8. Following is the required graph:10 7 ど8.2x) >嚚c)y-2x + 12 -2x+y 12 Dividing whole expression by 12: x-6)+y/12 1 So, we have horizontal intercept as (-6) and vertical intercept as 12. Following is the required graph:y-2x + 12 36 32 28 24 20 16 129 8 6 42 0 2 468 10 12 14 16 18 202) Condensing the expressions: A term written as x* denotes X as the base and Y as the exponent or power. Basic rules for condensing of multiplication and division, especially required for this question involves: 1) For the expression with terms with same base carrying exponents, the expression can be shortened as base raised to the power of addition of the exponents of terms which are multiplied and subtraction of the exponents of terms which are divided 2) For the expression with terms with same exponents over say different bases, the expression can be shortened as taking the power common for ll. and writing the bases togetheras originally multiplied or divided This will soon become more clear: b) xa xb X 2.c-ra+0+c(using rule (1) above) c) .r3 уз ,3. (x x y x ~ )3(using rule (2) above) 6. Solving expressions: We have to find: 22/3Clearly, the expression has a common base with different powers. So, using rule number 1 from above x1/2 × x1/3 = X12 + 13-233.2) As we equate respective demand functions and supply functions we get froma) Q d- 51 - 3P and Q s 6P -10 as Qd-Q_s theref2x*(x-4)+4 (x-4)-0 (2x+4)*(x-4)-0 X-4 or -4/2--2 5) Given function 8x^3+6x 2-3x-1-0 If X-1 then above function has a value 8*1A3+6*1A2-3-1-10Since at X=1 we are getting non zero value so X=l is not root of above equation. Rearranging above equation 8xA3-1+6x12-3x-0 (2x-1)*(4x 2+1+2x)+3x*(2x-1)-0 (2x-1) (4x 2+2x+1+3x)-0 (2x-1)*(4x2+5x+1)0 (2x-1)*(4x 2+4x+x+1)-0 X=1/2 or-1/4 or-1 6) Given Qd-Qs3-P 2-6P-4 P2+6P-7-0 PA2+7P-P-7-0 P*(P+7)-1*(P+7)0 (P-1)*(P+7-0 P 1 or -7 Since price is always positive so P-1 And Qd-3-1A2-2 Qs-6*1-4-2a) Endogenous variables: Y, C and G b) on govemment expenditure c) At equilibrium, Y-C+I+G That is, d) In order for solution to exist, 1-g-b must be a positive number and not equal to 0.

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