University:
Stanford University
Course:
MATH 19 | Calculus
Academic year:

2021
Views:

20

Pages:

20
Author:

MathPaladin

Questions and Answers Sheet 9 Fractions Question #1 5 1 8 4 How do you solve 2 − 1 Answer: Step 1 5 1 8 4 The given expression 2 − 1 5 8 and 1 + 1 4 5 1 8 4 consists of mixed fractions. The mixed fractions 2 and 1 can be written as 2 + respectively. Step 2 Now, solve the expression. 5 1 5 1 2 − 1 = (2 + ) − (1 + ) 8 4 8 4 = 2×8+5 1×4+1 − 8 4 = 16 + 5 4 + 1 − 8 4 = 21 8 − 5 4 Step 3 Simplify the expression further. 21 5 21 − 2 × 5 [∵ 𝐿𝐶𝑀(8,4) = 8] − = 8 4 8 = 21 − 10 8 = 11 8 =1 3 8 5 1 3 8 4 8 Hence, 2 − 1 = 1 Question #2 4 1 3 5 2 5 Simplify each numerical expression. − − ⋅ (− ) Answer: Step 1 For subtracting fractions, we have to make denominators equal, then subtract numerators. Multiplication of two fractions - We multiply numerator to numerator and denominator to the denominator. Step 2 Given 4 1 3 − − ⋅ (− ) 5 2 5 4 3 =− + 5 10 =− 8 3 + 10 10 =− 5 10 =− 1 2 Step 3 Answer 4 1 3 1 − − ⋅ (− ) = − 5 2 5 2 Question #3 Express the given integrand as a sum of partial fractions, then evaluate the integrals. ∫ 𝑑𝑥 √25𝑥 2 − 36 Answer: ,𝑥 > 6 5 Question #4 Subtract. 1 4 4 −3 3 9 Answer: Question #5 Simplify cos2 (𝑡) 1−𝑐𝑜𝑠 2 (𝑡) to an expression involving a single trig function with no fractions. Answer: Question #6 Which of the following diophantine equations cannot be solved (1) 6𝑥 + 51𝑦 = 22 (2) 33𝑥 + 14𝑦 = 115 (3) 14𝑥 + 35𝑦 = 93 Answer: Step 1 The diophantine equations are of the form 𝑎𝑥 + 𝑏𝑦 = 𝑐, if c can be divided by the greatest common divisor (gcd) of a and b then this equation has integer solutions. Step 2 6𝑥 + 51𝑦 = 22 gcd of 6 and 51 = 3 22 cannot be divided by 3 without forming fractions, so this equation doesn't have any solutions. Step 3 33𝑥 + 14𝑦 = 115 gcd of 33 and 14 = 1( ' number) 115 can be divided by 1, so this equation has solutions. Step 4 14𝑥 + 35𝑦 = 93 gcd of 14 and 35 = 7 93 cannot be divided by 7 without forming fractions. So this one has no solution. Step 5 So the equations that cannot be solved are the equations in a and c. Question #7 Solve: 1 𝑥 𝑥 2 3 4 ( + 1) = − 2 Answer: Step 1 A linear equation in a variable x is an equation containing the variable only with its first power. Solving an equation in x is finding the value of x that satisfies the equation. For the given equation use the distributive property of multiplication and addition 𝑎(𝑏 + 𝑐) = 𝑎𝑏 + 𝑎𝑐. Find the least common denominator of the fractions and multiply by this least common denominator to eliminate fractions. Step 2 Given equation is 1 𝑥 𝑥 2 3 4 ( + 1) = − 2 Remove the fractions using information from step 1. Then move the x terms to one side and constant terms to other side. 1 𝑥 𝑥 ( + 1) = − 2 2 3 4 𝑥 1 𝑥 + = −2 6 2 4 𝑥 1 𝑥 12 ( + ) = 12 ( − 2) 6 2 4 2𝑥 + 6 = 3𝑥 − 24 2𝑥 + 6 − 2𝑥 = 3𝑥 − 24 − 2𝑥 6 = 𝑥 − 24 6 + 24 = 𝑥 − 24 + 2𝑥 𝑥 = 30 Hence, solution to given equation is 𝑥 = 30. Question #8 How do you integrate: Evaluate the integral.(Remember to use absolute balues where appropriate. Use C for the constanr of integration.) ∫ 5𝑥 2 + 5𝑥 − 12 𝑑𝑥 𝑥 3 + 4𝑥 2 Answer: Question #9 Add or subtract as indicated 2 14 − 𝑥𝑧 3 𝑥𝑧 3 Answer: Step 1 The given operation is subtraction of fractions. Step 2 When the denominator of two fractions are same, we can add or subtract the numerators directly and keep the denominator same. That is, the answer is 2 𝑥𝑧 3 − 14 𝑥𝑧 3 = 2−14 𝑥𝑧 4 = −12 𝑥𝑧 4 −12 𝑥𝑧 4 Question #10 write the ratio as a fraction in lowest terms. 2 3 1 to 1 4 2 2 3 1 is___ % of 1 4 2 Answer: Step 1 Since you have submitted two questions, we'll answer the first question. For the second question please submit the question again and specify it. a) Write the mixed fractions as improper fractions. 2 3 1 to 1 4 2 = 11 3 to 4 2 Step 2 Then to simplify it we divide the first fraction by the second fraction. 11 3 to 4 2 11 ) = 4 3 ( ) 2 ( = 11 2 ⋅ 4 3 = 22 12 = 11 6 Answer: = 11 6 Question #11 1 5 7 6 8 48 Consider the following fractions: − , − , − Answer: Given fractions 1 5 7 6 8 48 − ,− ,− Consider − 1 6 Multiply and divide with 8 we get = 6 × 8 8 −8 48 Consider −5 8 Multiply and divide with 6 we get = −1 −5 8 × 6 6 −30 48 similary we have −7 48 Hence, All 3 fractions with the lowest common denominator are Question #12 −8 −30 −7 48 , 48 , 48 Express 1 11 as a repeating decimal, using a slash to indicate the repeating digits. What are the decimal expansions of the following fractions: 2 , 3 , 9 11 11 11 and 11 11 ? Answer: Step 1 Divide 1 by 11 and write the fraction 1 11 as a repeating decimal: 1 = 0.090909090 … 11 1 = 0.090 11 Step 2 Now similarly, write all fractions as decimals: 2 = 0.18181818 … = 0.18 11 3 = 0.27272727 … = 0.27 11 9 = 0.81818181 … = 0.81 11 11 =1 11 Question #13 Use properties of logarithms to expand the logarithmic expression as much as possible. Where possible,evaluate logarithmic expressions without using a calculator. log 2 ( 𝑥𝑦 2 ) 64 Answer: Question #14 Solve the exponential equation 32𝑥 + 3𝑥 − 2 = 0. Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. Answer: Step 1 Given: 32𝑥 + 3𝑥 − 2 = 0 Step 2 Consider, 32𝑥 + 3𝑥 − 2 = 0 we write a given expression as: (3𝑥 )2 + 3𝑥 − 2 = 0 put 3𝑥 = 𝑦 then (3𝑥 )2 + 3𝑥 − 2 = 0 converted into quadratic equation given as: 𝑦2 + 𝑦 − 2 = 0 𝑦 2 − 𝑦 + 2𝑦 − 2 = 0 𝑦(𝑦 − 1) + 2(𝑦 − 1) = 0 (𝑦 + 2)(𝑦 − 1) = 0 𝑦 + 2 = 0 or 𝑦 − 1 = 0 𝑦 = −2 or 𝑦 = 1 Step 3 Consider 𝑦 = −2 but we have 𝑦 = 3𝑥 resubstitute as 𝑦 = 3𝑥 then we get 3𝑥 = −2 Taking both sides natural logarithm we get ln(3𝑥 ) = log(−2) which is not possible because ln(−2) not well defined. If we consider common logarithm we get ln(3𝑥 ) = log(−2) which is not possible because log(−2) not well defined. Hence for 3𝑥 = −2 is not possible solution for given expression. Step 4 Consider 𝑦 = 1 but we have 𝑦 = 3𝑥 resubstitute as 𝑦 = 3𝑥 then we get 3𝑥 = 1 Taking both sides natural logarithm we get ln(3𝑥 ) = ln(1) 𝑥 ln(3) = 0 𝑥=0 If we consider common logarithm we get log(3𝑥 ) = log(1) 𝑥 log(3) = 0 𝑥=0 Hence the solution set in terms of natural logarithms and common logarithms for 32𝑥 + 3𝑥 − 2 = 0 is 𝑥 𝑥 =0 Question #15 Use the properties of logarithms, given that ln(2) ≈ 0.6931 and ln(3) ≈ 1.0986 to approximate the logarithm. Use a calculator to confirm your approximations. (a) ln(0.75) (b) ln(288) 3 (c) ln( √24) 1 (d) ln ( ) 6 Answer: Question #16 Rewrite each number in the Simple form Drag the numbers to create your fractions. 3 1. −7 = 2. −2 = 3. 1 = 4 1 2 3 5 Answer: Step 1 Use the formula below to change the fractions into rational numbers. Assume a fraction in the form of 𝑎 𝑏 𝑐 So, rational number for the fraction is (𝑎×𝑐)+𝑏 𝑐 Given fraction = −7 3 4 3 (7×4)+3 4 4 Thus, − (7 ) = 28 + 3 = −( ) 4 31 = −( ) 4 3 31 4 4 Hence, −7 = − Step 2 Similarly, determine the rational numbers for the remaining fractions. Given fraction = −2 (2×2)+1 2 2 4+1 = −( ) 2 5 = −( ) 2 Hence, 1 5 2 2 Again, 3 (1×5)+3 5 5 1 = = 5+3 5 = 8 5 2 1 Thus, − (2 ) = −2 = − 1 3 8 5 5 Hence, 1 = Question #17 Give the 4 cases of partial fractions with examples. Answer: Question #18 The equation 𝑥 𝑥−3 =2+ 4 3 can be solved by multiplying both sides by the__________ of the fractions, which is________ Answer: Step 1 Given, the equation, 𝑥 4 =2+ 𝑥−3 3 Step 2 The given equation is solved by multiplying both sides by Least Common Denominator (L.C.D) of the fractions, which is 12. In addition, solution can be given as follows: 𝑥 4 =2+ 𝑥−3 3 Now, Least Common Denominator of the fractions is, LCD of (4,3) which is 12. 𝑥 × 12 4 = (2 + 𝑥−3 ) × 12 3 3𝑥 = 24 + 4(𝑥 − 3) 3𝑥 = 24 + 4𝑥 − 12 −𝑥 = 12 𝑥 = −12 is the solution. Question #19 (a) How do we express a rational function r as a partial frac- tion decomposition? (b) Give the form of the partial fraction decomposition. i. ii. iii. 2𝑥 (𝑥−5)(𝑥−1)2 2𝑥 (𝑥−5)(𝑥 2 +1) 3𝑥+1 𝑥(𝑥 2 +1)2 Answer: Step 1 (a) A rational function is a function of the form 𝑟(𝑥) = 𝑃(𝑥) 𝑄(𝑥) Where 𝑃(𝑥) and 𝑄(𝑥) are polynomials in x with 𝑄(𝑥) ≠ 0 Rational fractions can be integrated by splitting into partial fractions. Step 2 Consider the partial fraction, 2𝑥 (𝑥 − 5)(𝑥 − 1)2 To write the given fraction in partial fraction decomposition, 2𝑥 (𝑥 − 5)(𝑥 − 1)2 = 𝐴 𝐵 𝐶 + + 𝑥 − 5 𝑥 − 1 (𝑥 − 1)2 Where A, B and C are constant. Consider the partial fraction, 2𝑥 (𝑥 − 5)(𝑥 2 + 1) To write the given fraction in partial fraction decomposition, 2𝑥 (𝑥 − 5)(𝑥 2 + 1) = 𝐴 𝐵𝑥 + 𝐶 + 𝑥 − 5 (𝑥 2 + 1) Where A, B and C are constant. Step 3 Consider the partial fraction, 3𝑥 + 1 𝑥(𝑥 2 + 1)2 To write the given fraction in partial fraction decomposition, 3𝑥 + 1 𝑥(𝑥 2 + 1)2 = 𝐴 𝐵𝑥 + 𝐶 𝐷𝑥 + 𝐸 + 2 + 2 (𝑥 (𝑥 𝑥 + 1) + 1)2 Where A, B and C are constant. Question #20 Integrate using partial fraction 5𝑥+6 ∫ (𝑥−2)(3𝑥+1)2 𝑑𝑥 Answer: