MATH165 CO1 Exam Set G PDF

DEPARTMENT OF MATHEMATICS MATH165: CO1 Examination NAME Student Number: SECTION/PROF. Curricular Program...

DEPARTMENT OF MATHEMATICS MATH165: CO1 Examination NAME Student Number: SECTION/PROF. Curricular Program Test Set G Directions Calculators are NOT allowed. Cheat Sheets (e.g., index cards with formulas) are not allowed. In the space provided, present your complete solutions and answers. Use back pages as scratch papers. Use black or blue ink only. Do not use pencils or erasable pens (e.g. Frixion pens). Warning!!! Cheating in examinations is a major offense and has a sanction of either suspension or non-readmission from the University. (Reference: Student Discipline Handbook, 2016, pp. 10-11). I. Knowledge/Understanding Write T if the statement is TRUE, otherwise, write F. (2 pts @) Answer 2 1. 5 − 4[3 − 6(2 ∙ 3 − 12 ÷ 4)] = 353 T 2. |a - b| = |b - a| for any real numbers a and b. T 3. All integers are rational numbers. T 4. The expression 𝑥 2 − 𝑦 2 can be factored as (𝑥 − 𝑦)(𝑥 − 𝑦). F 4 4 5. The binomial 𝑎 + 4𝑏 is factorable. F 2 6. The complete solution of the inequality 𝑥 > 4 is 𝑥 > 2. F 7. A function is a relation where each input has exactly one output. T 8. 3 1 3𝑎 (7𝑎 + 2) − (9𝑎 − 7) = +5 T 4 2 4 3 9. √8𝑥 3 = 2𝑥 T 10. The absolute value of a number is always positive. T Provide concise and clear solutions to each question. Write legibly II. Skills/Applications and avoid erasures. Box your final answer. (10 pts each number) A. Find the following products using special products: 2 1. (√2 + √6) 2. (𝑐 3 + 4𝑏 2 )3 2 + 2√12 + 6 (𝑐 3 )3 + 3 ⋅ 𝑐 6 (4𝑏 2 ) + 3 ⋅ 𝑐 3 (4𝑏 2 )2 + (4𝑏 2 )3 𝒄𝟗 + 𝟏𝟐𝒄𝟔 𝒃𝟐 + 𝟒𝟖𝒄𝟑 𝒃𝟒 + 𝟔𝟒𝒃𝟔 Solution 8 + 2√12 𝟖 + 𝟒√𝟑 B. Factor the following completely: (write NOT FACTORABLE if the expression cannot be factored) 3. 10𝑧 3 − 15𝑧 2 − 4𝑧 + 6 4. 2𝑎6 − 128 (10𝑧 3 − 15𝑧 2 ) − (4𝑧 − 6) 2(𝑎6 − 64) 2[(𝑎3 )2 − (23 )2 ] Solution 5𝑧 2 (2𝑧 − 3) − 2(2𝑧 − 3) 2(𝑎3 − 23 )(𝑎3 + 23 ) (𝟐𝒛 − 𝟑)(𝟓𝒛𝟐 − 𝟐) 2(𝑎 − 2)(𝑎2 + 2𝑎 + 4)(𝑎 + 2)(𝑎2 − 2𝑎 + 4) 𝟐(𝒂 − 𝟐))(𝒂 + 𝟐)(𝒂𝟐 + 𝟐𝒂 + 𝟒(𝒂𝟐 − 𝟐𝒂 + 𝟒) 5 & 6) Simplify and determine the domain: C. 2 7 𝑦2 − 1 − ∙ 𝑦 𝑦+1 𝑦+8 5) Simplification: 2 7 𝑦 2 − 1 2 7(𝑦 − 1)(𝑦 + 1) 2 7(𝑦 − 1) 2(𝑦 + 8) − 7𝑦 2𝑦 + 16 − 7𝑦 𝟏𝟔 − 𝟓𝒚 − ∙ = − = − = = = 𝑦 𝑦+1 𝑦+8 𝑦 (𝑦 + 1)(𝑦 + 8) 𝑦 (𝑦 + 8) 𝑦(𝑦 + 8) 𝑦(𝑦 + 8) 𝒚(𝒚 + 𝟖) Solution 6) Domain: 𝐷 = {𝑦 ∈ ℝ: 𝑦 ≠ −8, −1, 0}  D. 7 & 8) Simplify and then determine the domain: 1 1+ 1 1+ 2+𝑥 7) Simplification: 1 2+𝑥 2+𝑥 2+𝑥 3+𝑥+2+𝑥 𝒙+𝟓 1+ ⋅ =1+ =1+ = = 1 1+2+𝑥 2+𝑥 2+𝑥+1 3+𝑥 3+𝑥 𝒙+𝟑 Solution 8) Domain: 𝐷 = {𝑥 ∈ ℝ: 𝑥 ≠ −2, −3} E. Define 𝑓(𝑥) = 3𝑥 2 − 𝑥 − 5 and 𝑔(𝑥) = 2 − 3𝑥. Evaluate the following: 𝑓+𝑔 𝑓 2 9) ( ) (1) 10) (𝑓𝑔)(−1) + ( ) (2) 11) (𝑓 ∘ 𝑔) ( ) 𝑓−𝑔 𝑔 3 𝑓+𝑔 𝑓(2) 2 =( ) (1) = 𝑓(−1)𝑔(−1) + = 𝑓 (𝑔 ( )) 𝑓−𝑔 𝑔(2) 3 2 𝑓(1) + 𝑔(1) = [3(−1) − (−1) − 5][2 − 3(−1)] 2 = (3(2)2 − 2 − 5) = 𝑓 (2 − 3 ( )) Solution 𝑓(1) − 𝑔(1) 3 + 2 − 3(2) = 𝑓(0) (3(1)2 − 1 − 5) + (2 − 3(1)) (12 − 7) = [3 = 3(0)2 − 0 − 5 2 (3(1) − 1 − 5) − (2 − 3(1)) = + 1 − 5][2 + 3] + 2−6 = −𝟓 (−3) + (−1) −4 5 −20 + 5 = = = (−1)(5) + = (−3) − (−1) −2 −4 −4 −15 𝟏𝟓 =𝟐 = = −4 𝟒 Provide concise and clear solutions to each question. Write legibly and avoid III. Analysis erasures. (15 pts @) F. 𝑥 12. Solve the inequality: ≥2 2𝑥+3 12) Critical values: 13) Create the Table of Signs 𝑥 −2≥0 2𝑥 + 3 −2 −3/2 −1 𝑥 − 2(2𝑥 + 3) 3 ≥0 Interval (−∞, −2) (−2, − ) 3 2𝑥 + 3 2 (− , −∞ ) 2 −3(𝑥 + 2) ≥0 /Factors 𝑥 = −3 𝑥 = −1.7 𝑥=0 2𝑥 + 3 Solution 𝑥+2 − + + 𝑥+2 ≤0 2𝑥 + 3 − − + 2𝑥 + 3 3 𝑥+2 CV: 𝑥 = −2; − 2 + − + 2𝑥 + 3 3 3 14) Set Notation: {𝑥 ∈ ℝ: −2 ≤ 𝑥 < − 2} 15) Interval notation: [−2, − 2) 16) Graph: −2 −3/2 −1 G. Find the domain and/or range of the given function. 1 17 & 18) Find the domain and range of 𝑓(𝑥) = + 2. 17) Domain: {𝑥 ∈ ℝ: 𝑥 ≠ 3} 3−𝑥 1−6−2𝑥 Range: 𝑦 = 3−𝑥 (3 − 𝑥)𝑦 = −5 − 2𝑥 18) Range: {𝑦 ∈ ℝ: 𝑦 ≠ 2} 3𝑦 − 𝑥𝑦 + 2𝑥 = −5 −𝑥(𝑦 − 2) = −5 − 3𝑦 Solution 5+3𝑦 𝑥 = 𝑦−2 𝑓+𝑔 (𝑥) if 𝑓(𝑥) = 𝑥 2 − 𝑥 − 5 and 𝑔(𝑥) = 2 − 5𝑥 20) Domain: {𝑥 ∈ ℝ: 𝑥 ≠ 3, −7} 𝑓−𝑔 𝑓+𝑔 19) Find 𝑓−𝑔 (𝑥) 𝑓+𝑔 𝑥 2 −𝑥−5+2−5𝑥 𝑥 2 −6𝑥−3 (𝑥) = = 𝑥 2 +4𝑥−7 𝑓−𝑔 𝑥 2 −𝑥−5−2+5𝑥 𝑓+𝑔 𝑥 2 −6𝑥−3 (𝑥) = 𝑓−𝑔 (𝑥−3)(𝑥+7)