QA #1 Equations, Expressions and Inequalities

Questions and Answers Sheet 1 Equations, Expressions and Inequalities Question #1 Proof of triangle inequality I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that |𝑎 + 𝑏| ≤ |𝑎| + |𝑏|. Answer: From your definition of the absolute value, establish first |𝑥 | = max{𝑥 , −𝑥} 𝑎𝑛𝑑 ± 𝑥 ≤ |𝑥 |. Then you can use 𝑎 + 𝑏 − 𝑎 − 𝑏 ≤ |𝑎| + 𝑏 ≤ |𝑎| + |𝑏|,and ≤ |𝑎 | − 𝑏 ≤ |𝑎 | + |𝑏 |. Question #2 Solve, please: 2𝑥 = 𝑥 2 Answer: Collect terms to 1 side and equate to zero. Thus, 𝑥 2 − 2𝑥 = 0 There is a common factor 𝑥, so we have 𝑥 (𝑥 − 2) = 0 Hence, 𝑥 = 0 𝑜𝑟 𝑥 − 2 = 0 → 𝑥 = 2 The answer is 0 and 2. Question #3 a) The proportion to represent the number of girls in the school. Given: At a school, the school population is 2 5 boys and there are 450 students at the school who are boys. b) The steps to solve the equations. c) The number of girls in school. Given: The population, 2 5 𝑥 = 450 Answer: a) Concept used: Rules of Addition/Subtraction: - Two numbers with similar sign always get added and the resulting number will carry the similar sign. - Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number. Rules of Multiplication/Division: - The product/quotient of two similar sign numbers is always positive. - The product/quotient of two numbers with opposite signs is always negative. Calculation: In order to find proportion to represent the number of girls in the school, observe the that the proportion of boys is 2 2 5 , so subtract it from 1 and similify further as: 5 2 1−5= 5−5 = 5−2 5 3 =5 Thus, the proportion of girls in school is 3 5 . Question #4 Find a Cartesian equation for the curve and identify it. 𝑟 2 cos(2θ) = 1 Answer: Step 1 Given: The polar equation 𝑟 2 cos(2θ) = 1 Step 2 Consider, 𝑟 2 cos(2θ) = 1 𝑟 2 (cos 2 θ − sin2 θ) = 1 (∵ cos(2θ) = cos 2 θ − sin2 θ) Consider the parametric equation 𝑥 = 𝑟 cos 𝑦 = 𝑟 sin The implies, 𝑥 𝑟 𝑦 𝑟 = cos = sin Then, 𝑥 2 (𝑟 ) = cos 2 𝑦 2 (𝑟 ) = sin2 Step 3 𝑥 2 Substitute (𝑟 ) = cos 2 𝑥 2 𝑦 2 𝑟 2 (( ) − ( ) ) = 1 𝑟 𝑥2 𝑟 𝑦2 𝑟 2 (𝑟 2 − 𝑟 2 ) = 1 𝑦 2 𝜃, ( 𝑟 ) = sin2 𝜃 𝑖𝑛 𝑟 2 (cos 2 θ − sin2 θ) = 1 1 𝑟 2 (𝑟 2 ( 𝑥 2 − 𝑦 2 )) = 1 𝑥2 − 𝑦2 = 1 Therefore, the Cartesian equation of 𝑟 2 cos(2θ) = 1 𝑖𝑠 𝑥 2 − 𝑦 2 = 1 Question #5 If 𝑃(0 < 𝑧 < 𝑘) = 3212 find and solve k. Answer: Solving: 𝑃(0 < 𝑧 < 𝑘) = 0.3212 ⇒ 𝑃(𝑧 > 𝑘) − 𝑃(𝑧 < 0) = 0.3212 using z-table, we get ⇒ 𝑃(𝑧 < 𝑘) − 0.5000 = 0.3212 ⇒ 𝑃(𝑧 < 𝑘) = 0.3212 + 0.5000 ⇒ 𝑃(𝑧 < 𝑘) = 0.0.8212 using z-table, we get 𝑃(𝑧 < 0.92) = 0.8212 𝑘 = 0.92 Question #6 Write the first expression in terms of the second if the terminal point determined by t is in the given quadrant. tan 𝑡 , 𝑐𝑜𝑠; Quadrant III Answer: We are given 𝑡𝑎𝑛 and need to rewrite the expression in terms of 𝑐𝑜𝑠 . sin 𝑡 Using the reciprocal identity, we get tan 𝑡 = cos 𝑡 so we need to rewrite the sine in terms of cosine. Using the Pythagorean Identity, we get: sin2 sin 2 𝑡 + cos 2 𝑡 = 1 Pythagorean Identity. 𝑡 = 1 − cos 2 task Subtract cos 2 task on both sides. sin 𝑡 = ±√1 − cos 2 𝑡 Square root both sides. Since t lies in Quadrant lll, then sin 𝑡 < 0 so we need to use the negative root. Therefore, sin 𝑡 tan 𝑡 = cos 𝑡 = −√1−cos2 Result: tan 𝑡 = − 𝑡 cos 𝑡 √1−cos2 𝑡 cos 𝑡 Question #7 is sin2 𝑥 equal to sin 𝑥 2 ? Answer: sin2 𝑥 is NOT equal to sin 𝑥 2 . The easiest way to show that they are not equal is to evaluate both expressions for a nonzero value of 𝑥. For example, we will evaluate both expressions at 𝑥 = 90∘ (similar for x-values in radians). sin2 90∘ = (sin 9 0∘ )2 = 12 = 1 sin(90∘ )2 = sin 8 100∘ = sin(45 × 180∘) = sin 1 80∘ = 0 We then note that the two expressions cannot be equal, because one expression becomes 1 and the other 0 at 𝑥 = 90∘ . Note: If you ever doubt if an equality is true, it is always useful to check for a few values of 𝑥 to be sure. Question #8 What quantum numbers specify these subshells: 1𝑠 4𝑝 5𝑑 ⋯ 𝑛 =? 𝑙 =? Answer: Step 1 The principal quantum number (𝑛) for 1𝑠 is 1 and the azimuthal quantum number (𝑙 ) for 1𝑠 is 0. Calculate the range of azimuthal quantum number (𝑙 )as follows: 𝑙=𝑛−1 Substitute 1 for n in the above equation for the calculation of l as follows: 𝑙 =1−1 =0 The value of 𝑛 is 1 and 𝑙 is 0 for the 1𝑠 orbital. The principal quantum number (𝑛) defines the shell of the orbital. For 1𝑠, the shell is 1. The azimuthal quantum number (𝑙 ) defines the subshell of the orbital. It is calculated as the difference between the 𝑛 and 1. For 1𝑠, the subshell is 0. Step 2 The principal quantum number (𝑛) for 4𝑝 is 4. Calculate the range of azimuthal quantum number (𝑙 ) as follows: 𝑙=𝑛−1 Substitute 4 for n in the above equation for the calculation of l as follows: 𝑙 =4−1 =3 The value of 𝑙 for 4𝑝 lies between 0, 1, 2, 3 . For 𝑝 orbital, the value of 𝑙 is 1. Therefore, the azimuthal quantum number (𝑙 ) for 4𝑝 is 1. The value of 𝑛 is 4 and 𝑙 is 1 for the 4𝑝 orbital. The principal quantum number (𝑛) defines the shell of the orbital. For 4𝑝, the shell is 4. The azimuthal quantum number (𝑙 ) defines the subshell of the orbital. It is calculated as the difference between the 𝑛 and 1. For 4𝑝 orbital, the value of l can be 0, 1, 2, 3. The orbital is 𝑝. For p orbital, the azimuthal quantum number is 1. Therefore, for 4𝑝 orbital, 𝑙 is 1. Step 3 The principal quantum number (𝑛) for 5𝑑 is 5. Calculate the range of azimuthal quantum number (𝑙 ) as follows: 𝑙=𝑛−1 Substitute 5 for n in the above equation for the calculation of l as follows: 𝑙 =5−1 =4 The value of 𝑙 for 5𝑑 lies between 0,1,2,3,4. For d orbital, the value of 𝑙 is 2. Therefore, the azimuthal quantum number (𝑙 ) for 5𝑑 is 2. The value of 𝑛 is 5 and 𝑙 is 2 for the 5𝑑 orbital. The principal quantum number (𝑛) defines the shell of the orbital. For 5𝑑, the shell is 5. The azimuthal quantum number (𝑙 ) defines the subshell of the orbital. It is calculated as the difference between the 𝑛 and 1. For 5𝑑 orbital, the value of 𝑙 can be 0, 1, 2, 3, 4. The orbital is 𝑑. For 𝑑 orbital, the azimuthal quantum number is 2. Therefore, for 5𝑑 orbital, 𝑙 is 2. Question #9 For Exercise, solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. 3𝑥(𝑥 − 1) = 𝑥 + 6 Answer: Concept used: Quadratic Formula 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 𝑥= −𝑏±√𝑏2 −4𝑎𝑐 2𝑎 Now, Solving the equation 3𝑥(𝑥 − 1) = 𝑥 + 6 3𝑥 2 − 3𝑥 = 𝑥 + 6 3𝑥 2 − 3𝑥 − 𝑥 − 6 = 0 3𝑥 2 − 4𝑥 − 6 = 0 Now 𝑥 = −𝑏±√𝑏2 −4𝑎𝑐 2𝑎 Here, 𝑎 = 3, 𝑏 = −4, 𝑐 = −6 −(−4)±√(−4)2 −4×3×(−6) Then, 𝑥 = 2×3 𝑥= 4±√16+72 𝑥= 4±√88 𝑥= 4±2√22 6 𝑥= 2±√22 6 6 3 Answer: Hence, the values of 𝑥 are 𝑥= 2+√22 3 𝑎𝑛𝑑 𝑥 = 2−√22 3 Question #10 For Exercise, solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. log 2(3𝑥 − 1) = log2 (𝑥 + 1) + 3 Answer: Property used: 𝑚 Property 1: log 2(𝑚) − log 2(𝑛) = log2 ( ) 𝑛 Now to simplifying the given equation: log 2(3𝑥 − 1) = log2 (𝑥 + 1) + 3 log 2(3𝑥 − 1) − log2 (𝑥 + 1) = 3 3𝑥−1 log 2 ( 𝑥+1 ) = 3 [Using Property 1.] Now taking antilog 2 and solving: 3𝑥−1 ( 𝑥+1 ) = 23 (3𝑥 − 1) = 8(𝑥 + 1) (3𝑥 − 1) = 8𝑥 + 8 3𝑥 − 8𝑥 = 8𝑥 + 1 −5𝑥 = 9 9 𝑥 = −5 Since, The solution does not satisfy the given equation. Hence there is no solution for 𝑥 ∈ 𝑅 . Question #11 For Exercise, solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. (𝑥 2 − 9)2 − 2(𝑥 2 − 9) − 35 = 0 Answer: Now, Solving the equation (𝑥 2 − 9)2 − 2(𝑥 2 − 9) − 35 = 0{∵ (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏2 } 𝑥 4 + 81 − 18𝑥 2 − 2𝑥 2 + 18 − 35 = 0 𝑥 4 − 20𝑥 2 + 64 = 0 Let 𝑡 = 𝑥 2 Then, 𝑡 2 − 20𝑡 + 64 = 0 𝑡 2 − (16 + 4)𝑡 + 64 = 0 𝑡 2 − 16𝑡 − 4𝑡 + 64 = 0 𝑡(𝑡 − 16) − 4(𝑡 − 16) = 0 (𝑡 − 16)(𝑡 − 4) = 0 𝑡 = 16 and 𝑡 = 4 Now, Substitute 𝑡 = 𝑥 2 So, 𝑥 2 = 16 and 𝑥 2 = 4 𝑥 = ±4 and 𝑥 = ±2 Answer: Hence, the values of 𝑥 are 𝑥 = 4, 𝑥 = −4, 𝑥 = 2 and 𝑥 = −2 Question #12 Statement showing a relationship between numbers that are not necessarily equal using the symbols > , <, ≤, ≥, 𝑜𝑟 ≠ are called _. 1) inferences 2) inconsistencies 3) unequals 4) inequalities Answer: Step 1 Given: Statements showing a relationship between numbers that are not necessarily equal using the symbols > , <, ≥, ≤ 𝑜𝑟 ≠ Step 2 Explanation: Equation: An equation is a mathematical statement that two things are equal. It consists of two expressions, one on each side of an 'equals' sign. For example: 15 = 9 + 6 The most basic and common algebraic equations in math consist of one or more variables. For instance, 2𝑥 + 6 = 14 is an equation Inequalites: An inequality is a mathematical relationship between two expressions that are not equal. It can be represented using one of the following: ≤: ''less than or equal to'' <: ''less than'' ≠ : ''not equal to'' >: ''greater than'' ≥: ''greater than or equal to'' Step 3 Answer: The Statements showing a relationship between numbers that are not necessarily equal using the symbols >, <, ≥, ≤ 𝑜𝑟 ≠ are called Inequalities Question #13 For Exercise, solve the equations and inequalities. Write the solution sets to the inequalities in intervalnotation. 5 ≤ 3 + |2𝑥 − 7| Answer: Step 1 Simplify the given expression as follows. 3 + |2𝑥 − 7| ≥ 5 3 + |2𝑥 − 7| − 3 ≥ 5 − 3 (subtract 3 frpm both sides) |2𝑥 − 7| ≥ 2 Step 2 By the absolute rule, if |𝑢| ≥ 𝑎, 𝑎 > 0 then |𝑢| ≤ −𝑎 𝑛𝑑 |𝑢| ≥ 𝑎. 2𝑥 − 7 ≤ −2 and 2𝑥 − 7 ≥ 2 (by absolute rule) 2𝑥 − 7 + 7 ≤ −2 + 7 and 2𝑥 − 7 + 7 ≥ 2 + 7 (add 7 on both sides) 2𝑥 ≤ 5 and 2𝑥 ≥ 9 2𝑥 2 5 ≥ 2 and 2𝑥 2 5 9 ≤ 2 (divide by 2 on both sides) 9 𝑥 ≤ 2 and 𝑥 ≥ 2 5 9 Therefore, the solution set of the given inequality is (−∞, 2] ⋃ [2 , ∞) . Question #14 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation if possible. 1 1 −5 ≤ − 4 𝑥 + 3 < 2 Answer: Given inequality i 1 1 −5 ≤ − 4 𝑥 + 3 < 2 Multiplying both sides by 4, we get −20 ≤ −𝑥 + 12 < 2 −20 − 12 ≤ −𝑥 < 2 − 12 −32 ≤ −𝑥 < −10 ZS Now, multiplying both sides by −1 and using result that if 𝑎 < 𝑏 then −𝑏 < −𝑎 , we get 10 ≤ 𝑥 < 32 Therefore, solution is 𝑥 ∈ [10,32) Ans: Solution to inequality is 𝑥 ∈ [10,32) Question #15 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation if possible. 3−𝑥 𝑥+5 ≥1 Answer: Step 1 The given inequality is 3−𝑥 𝑥+5 ≥1 Step 2 Solution: We can also rewrite as 3−𝑥 ≥ 𝑥+5 Substract 5 on both sides, we have 3−𝑥−5 ≥ 𝑥+5−5 −2 − 𝑥 ≥ 𝑥 Add x on both sides, we have −2 − 𝑥 + 𝑥 ≥ 𝑥 + 𝑥 −2 ≥ 2𝑥 Now divide each term by 2 −2 2𝑥 ≥ 2 2 −2 2 ≥𝑥 −1 ≥ 𝑥 i.e. 𝑥 ≤ −1 Therefore, the interval notation for x is [−1, ∞) Question #16 For exercise, solve the equations and inequalities. Write the solutions to the inequalities in interval notation if possible. 1 1 −2 ≤ −4𝑥 − 5 < 2 Answer: Given: Rule used If 𝑎 ≤ 𝑢 < 𝑏 then 𝑎 ≤ 𝑢 and 𝑢 < 𝑏 Calculation: by using above rule 1 1 1 − 2 ≤ − 4 𝑥 − 5 and − 4 𝑥 − 5 < 2 1 1 now, − 2 + 5 ≤ − 4 𝑥 − 5 + 5 1 1 −2+5 ≤ −4𝑥 9 2 9 2 1 ≤ −4𝑥 1 (−4) ≤ − 𝑥(−4) 4 −8 ≥ 𝑥 𝑥 ≤ −18 1 And − 𝑥 − 5 + 5 < 2 + 5 4 1 −4𝑥 < 7 1 − 4 𝑥(−4) < 7(−4) 𝑥 ≻ 28 Thus, the solution set of the inequality is (−28, −18] Question #17 Solve the following rational equations and rational inequalities. Show your complete solution. 1. 𝑥/𝑥 − 3 + 6/𝑥 + 3 = 1 Answer: Our Aim is to solve the following rational and rational inequalities: Considering question −(1), we have: 𝑥 6 𝑥−3 + 𝑥+3 = 1 − (𝑖 ) Considering equation −(𝑖 ) we have: 𝑥 6 𝑥−3 ⇒ + 𝑥+3 = 1 𝑥(𝑥+3)+6(𝑥−3) 𝑥 2 −9 =1 ⇒ 𝑥 2 + 3𝑥 + 6𝑥 − 18 = 𝑥 2 − 9 ⇒ 9𝑥 − 18 = −9 ⇒ 9𝑥 = −9 + 18 ⇒ 9𝑥 = 9 Answer ⇒ 𝑥 = 1 Question #18 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation if possible. √𝑡 − 1 − 5 1 Answer: Step 1 Under the square root, we can not have a negative number so: 𝑡−1 ≥0 𝑡 ≥1 Step 2 Then we solve the inequality √𝑡 − 1 − 5 ≤ 1 √𝑡 − 1 ≤ 5 + 1 √𝑡 − 1 ≤ 6 2 (√𝑡 − 1) ≤ 62 𝑡 − 1 ≤ 36 𝑡 ≤ 36 + 1 𝑡 ≤ 37 Answer: [1,37] Question #19 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation if possible. 1 𝑥 2 −14𝑥+40 ≤0 Answer: Step 1: Analysis Given: 1 𝑥 2 −14𝑥+40 ≤0 To determine the solution sets of the given inequality. Step 2: Simplification Factorising the term 𝑥 2 − 14𝑥 + 40 . Let 𝑝(𝑥 ) = 𝑥 2 − 14𝑥 + 40 𝑝(𝑥 ) = 𝑥 2 − 14𝑥 + 40 = 𝑥 2 − 4𝑥 − 10𝑥 + 40 = 𝑥(𝑥 − 4) − 10(𝑥 − 4) = (𝑥 − 10)(𝑥 − 4) when 𝑥 = 10,4. 𝑝(𝑥 ) = 0 when 𝑥 ∈ (−∞, 4). 𝑝(𝑥 ) > 0 when 𝑥 ∈ (4,10). 𝑝(𝑥 ) < 0 when 𝑥 ∈ (10, ∞). 𝑝(𝑥 ) > 0 Step 3: Solution For 1 𝑥 2 −14𝑥+40 ≤0. 𝑥 2 − 14𝑥 + 40 should be less than 0. 𝑥 2 − 14𝑥 + 40 < 0 when 𝑥 ∈ (4,10) Therefore, the solution set of the given inequality is (4,10) Question #20 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation if possible. 4 √𝑚 + 4 − 5 = −2 Answer: Step 1 Given: Equation below: 4 √𝑚 + 4 − 5 = −2 To find: Solve for m Step 2 Solution: Add 5 on both sides of equation 4 ⇒ √𝑚 + 4 = −2 + 5 = 3 Take 4th power on both sides: 4 ⇒ ((𝑚 + 4)1/4 ) = 34 ⇒ 𝑚 + 4 = 81 Subtract 4 on both sides: ⇒ 𝑚 = 81 − 4 = 77 Verification: Substitute 𝑚 = 77 in the given equation: (77)1/4 − 5 = −2 ⇒ (81)1/4 − 5 = −2 ⇒ (34 )1/4 − 5 = −2 ⇒ 3 − 5 = −2 Step 3 Answer: 𝑚 = 77