Mathematics-L3-BASIC-ALGEBRA-AND-TRIGONOMETRY

1 TVET NATIONAL COMPREHENSIVE ASSESSMENT, SCHOOL YEAR 2020/2021 SECTOR: ALL TRADE: ALL RTQF LEVEL: Level 3 MODULE and Title: BASIC ALGEBRA AND TRIGONOMETRY, GENAM301 MARKING SCHEME QUESTION 1. simplify the following a. 𝑖801 /2.5marks b. 𝑖22775 /2.5marks QUESTION 2. A ladder, 3.9 m long, leans against a wall. If its foot is 1.2 m from the wall, how high up the wall does it reach? /5 Marks QUESTION 3. Solve 2 Sinx =√2; between 0≤ x ≤ 2𝜋 /5Marks QUESTION 4. Solve the following simultaneous equations, {3x + 5y = 42 2y – x − 8 = 0 /5 Marks {3x + 5y = 42 −𝑥 + 2𝑦 = 8 QUESTION 5. Given the function f(k) =k2+8k +16. Find the values k. /5Marks.

2 QUESTION 6. a) Convert the following degrees into 𝜋 radians i. 1350 /1.5Marks ii. 2250 /1.5Marks b) Convert the following 𝜋 radians into degrees 5𝜋 6 /2Marks QUESTION 7. Given f(x)= 5𝑥 𝑥+1 a. Find f (2) /2 .5Marks b. Find 𝑓(1/𝑚) / 2.5 Marks 8. Given that z=1+i find a. Re(z) /1.5 Marks b. Im(z) /1.5 Marks c.|z| (Modulus of z) /2 Marks QUESTION 9. Simplify the following sin 𝜃 𝑐𝑜𝑠𝑒𝑐 𝜃 + 𝐶𝑂𝑆 𝜃 𝑠𝑒𝑐𝜃 /5Marks QUESTION10. Gives an example of the network of streets in a modern city, a pedestrian may make a shortcut along the diagonal rather than walk along street M and street A. How much shorter is the shortcut. /5Marks QUESTION 11. Solve 𝑧2 − 𝑖𝑧 = 4 + 2𝑖 /5marks.

3 SECTION B/30MARKS QUESTION12. Given z=i+1 and w=(1 + 3𝑖) Find a) z+w / 2.5 Marks b) z-w /2.5Marks c) 𝑤 𝑧 /2.5 Marks d) z. w /2.5 Marks QUESTION 13. Solve and discuss the parametric equation below (2-3m)x+1=𝑚2(1-x)/10 Marks QUESTION 14. Prove the following trigonometric identities a) 1 1−𝑐𝑜𝑠𝑥 + 1 1+𝑐𝑜𝑠𝑥 = 2𝑐𝑠𝑐2𝑥/5 Marks b) 2𝑡𝑎𝑛𝑥 1−𝑡𝑎𝑛2𝑥 = tan 2𝑥/5 Marks QUESTION 15. Given that z=3+4i. Find the value of the expression z + 25 𝑧 /15Marks QUESTION 16. Find the values of k for which the equation x2 + (k + 1) x + 1 = 0 has: /15Marks a) Two distinct real roots b) No real roots..

4 END MARKING SCHEME QUESTION 1: simplify the following c. 𝑖801 /2.5marks d. 𝑖22775 /2.5marks SOLUTION: a) 𝒊𝟖𝟎𝟏 = 𝒊𝟒∗𝟐𝟎+𝟏 = 𝒊 b) 𝒊𝟐𝟐𝟕𝟕𝟓 = 𝒊𝟒∗𝟓𝟔𝟗𝟑+𝟑 = −𝒊 Reference: Learning Unit 3: Apply Fundamentals of Complex Numbers QUESTION2: A ladder, 3.9 m long, leans against a wall. If its foot is 1.2 m from the wall, how high up the wall does it reach? /5 Marks SOLUTION: A figure below is an illustration of the situation. Note that the ground must be assumed to be horizontal and level and hence at right angles to the wall. By Pythagoras’ theorem, 𝒉𝟐 + 𝟏. 𝟐𝟐 = 𝟑. 𝟗𝟐 𝒉𝟐 = 𝟑. 𝟗𝟐 − 𝟏. 𝟐𝟐 𝒉 = √𝟑. 𝟗𝟐 + 𝟏. 𝟐𝟐 𝒉 = 𝟑. 𝟕𝟏𝒎 Thus, the ladder reaches 3.71 m up the wall. Reference: Learning Unit 2: Apply Fundamentals of trigonometry QUESTION 3: Solve 2 Sinx =√2; between 0≤ x ≤ 2𝜋 /5Marks SOLUTION: 2 Sinx =√𝟐.

5 𝒔𝒊𝒏𝒙 = √𝟐 𝟐 Where √𝟐 𝟐 = 𝟒𝟓𝟎 𝟐𝒙 = Reference Learning Unit1: Solve algebraically or Linear and quadratic Equations or inequalities QUESTION5: Given the function f(k) =k2+8k +16. Find the values k. / 5Marks SOLUTION: By using quadratic formula method 𝑫 = 𝒃𝟐 − 𝟒𝒂𝒄 Where a=1, b=8, c=16 𝑫 = 𝟖𝟐 − 𝟒 ∗ 𝟏 ∗ 𝟏𝟔 = 𝟔𝟒 − 𝟔𝟒 = 𝟎.

6 If D=0 K1 =K2 = −𝒃 𝟐𝒂 = −𝟖 𝟐∗𝟏= -4 Hence, the value of K is -4 Reference Learning Unit1: Solve algebraically or Linear and quadratic Equations or inequalities QUESTION 6a) Convert the following degrees into 𝜋 radians i. 1350 /1.5Marks ii. 2250 /1.5Marks b) Convert the following 𝜋 radians into degrees 5𝜋 6 /2Marks SOLUTION: a. i. 1350 = 𝟏𝟑𝟓 ∗ 𝝅 𝟏𝟖𝟎 = 𝟑𝝅 𝟒 ii. 2250 = 𝟐𝟐𝟓∗𝝅 𝟏𝟖𝟎 = 𝟓𝝅 𝟒 b. 𝟓𝝅 𝟔 = 𝟓𝝅 𝟔 ∗ 𝟏𝟖𝟎 𝝅 = 𝟏𝟓𝟎𝟎 Reference: Learning Unit 2: Apply Fundamentals of trigonometry QUESTION7: Given f(x)= 5𝑥 𝑥+1 a. Find f (2) /2 .5Marks b. Find 𝑓(1/𝑚) / 2.5 Marks SOLUTION: a. f(2) = 𝟓∗𝟐 𝟐+𝟏 = 𝟏𝟎 𝟑 𝒇 ( 𝟏 𝒎) = 𝟓∗𝟏/𝒎 𝟏/𝒎+𝟏 = 𝟓 𝒎 𝟏+𝒎 𝒎 = 𝟓 𝒎 ∗ 𝒎 𝟏+𝒎 = 𝟓 𝟏+𝒎 Reference Learning Unit1: Solve algebraically or Linear and quadratic Equations or inequalities 8. Given that z=1+i find a. Re(z) /1.5 Marks b. Im(z) /1.5 Marks.

7 c.|z| (Modulus of z) /2 Marks SOLUTION: a. Re(z) =1 b. Im(z) =1 c. |z| =√𝟏𝟐 + 𝟏𝟐 = √𝟐 Reference: Learning Unit 2: Apply Fundamentals of trigonometry QUESTION 9: Simplify the following sin 𝜃 𝑐𝑜𝑠𝑒𝑐 𝜃 + 𝐶𝑂𝑆 𝜃 𝑠𝑒𝑐𝜃 /5Marks SOLUTION: We know that 𝒄𝒐𝒔𝒆𝒄𝜽 = 𝟏 𝒔𝒊𝒏𝜽 and 𝒔𝒆𝒄𝜽 = 𝟏 𝒄𝒐𝒔𝜽 𝐬𝐢𝐧 𝜽 𝒄𝒐𝒔𝒆𝒄 𝜽 + 𝑪𝑶𝑺 𝜽 𝒔𝒆𝒄𝜽 = 𝐬𝐢𝐧 𝜽 𝟏 𝒔𝒊𝒏𝜽 + 𝑪𝑶𝑺 𝜽 𝟏 𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝜽 𝟏 𝟏 𝒔𝒊𝒏𝜽 + 𝒄𝒐𝒔𝜽 𝟏 𝟏 𝒄𝒐𝒔𝜽 = 𝐬𝐢𝐧 𝜽 𝟏 ∗ 𝐬𝐢𝐧 𝜽 𝟏 + 𝐜𝐨𝐬 𝜽 𝟏 ∗ 𝒄𝒐𝒔𝜽 𝟏 = 𝐬𝐢𝐧𝟐 𝜽 + 𝐜𝐨𝐬𝟐 𝜽 = 𝟏 Reference: Learning Unit 2: Apply Fundamentals of trigonometry QUESTION10: Gives an example of the network of streets in a modern city, a pedestrian may make a shortcut along the diagonal rather than walk along street M and street A. How much shorter is the shortcut. /5Marks SOLUTION: The network of streets in a modern city is form a rectangle according to the figure above Street M=116m.

8 Street A=86m By using Pythagorean Theorem 𝒄𝟐 = 𝒂𝟐 + 𝒃𝟐 𝒄𝟐 = 𝟏𝟏𝟔𝟐 + 𝟖𝟔𝟐 𝒄𝟐 = 𝟏𝟑. 𝟒𝟓𝟔 + 𝟕. 𝟑𝟗𝟔 𝒄 = √𝟐𝟎. 𝟖𝟓𝟐 𝒄 = 𝟏𝟒𝟒. 𝟒𝒎 Hence the diagonal=144.4m which is the shortcut Reference: Learning Unit 2: Apply Fundamentals of trigonometry QUESTION11. Solve 𝑧2 − 𝑖𝑧 = 4 + 2𝑖 /5marks SOLUTION: 𝒛𝟐 + 𝒊𝒛 − 𝟐𝒊 − 𝟒 = 𝟎 𝑫 = 𝒃𝟐 − 𝟒𝒂𝒄 = (−𝒊)𝟐 − 𝟒 ∗ 𝟏 ∗ (−𝟐𝒊 − 𝟒) = −𝟏 + 𝟖𝒊 + 𝟏𝟔 = 𝟏𝟓 + 𝟖𝒊 √𝑫=±(𝟒 + 𝒊) 𝒁𝟏 = 𝒊 + 𝟒 + 𝒊 𝟐 = 𝟐 + 𝒊 𝒛𝟐 = 𝒊−𝟒−𝒊 𝟐 = −𝟐 Hence, 𝑺 = Reference: Learning Unit 3: Apply Fundamentals of Complex Numbers SECTION B/30MARKS QUESTION12. Given z=i+1 and w=(1 + 3𝑖) Find a) z+w / 2.5 Marks b) z-w /2.5Marks.

9 c) 𝑤 𝑧 /2.5 Marks d) z. w /2.5 Marks SOLUTION: a) z+ w = (i+1) + (1+3i) =1+1+i+3i=2+4i b) z-w = (i+1)- (1+3i) =1-1+i-3i=0-2i =-2i c) 𝒘 𝒛 = 𝟏+𝟑𝒊 𝒊+𝟏 = (𝟏+𝟑𝒊)(𝒊−𝟏) (𝒊+𝟏)(𝒊−𝟏) = 𝒊−𝟏+𝟑𝒊𝟐−𝟑𝒊 𝒊𝟐−𝟏𝟐 = −𝟏−𝟐𝒊−𝟑 −𝟏−𝟏 = −𝟒−𝟐𝒊 −𝟐 = 𝟐 + 𝐢 d) Z.w = (i+1) * (1+3i) =i+3i2 +1 +3i=4i-3+1=4i-2 Reference3: Learning Unit 3: Apply Fundamentals of Complex Numbers QUESTION 13: Solve and discuss the parametric equation below (2-3m)x+1=𝑚2(1-x)/10 Marks SOLUTION: (2 – 3m) x + 1 = m2 (1 – x) = 2x – 3mx + 1 = m2 – m2 x 2x – 3mx + m2 x – m2 + 1 = 0 X (2 – 3m + m2) – m2 + 1 = 0 X (2 – 3m + m2) = m2 – 1 𝒙 = 𝒎𝟐−𝟏 𝟐−𝟑𝒎+𝒎𝟐 = (𝒎−𝟏)(𝒎+𝟏) (𝒎−𝟏)(𝒎−𝟐) = 𝒎+𝟏 𝒎−𝟐 If m = 2, then there is no solution. If m ≠ 2, then the solution is 𝒙 = 𝒎+𝟏 𝒎−𝟐 Reference: Learning Unit1: Solve algebraically or Linear and quadratic Equations or inequalities QUESTION14: Prove the following trigonometric identities a) 1 1−𝑐𝑜𝑠𝑥 + 1 1+𝑐𝑜𝑠𝑥 = 2𝑐𝑠𝑐2𝑥/5 Marks b) 2𝑡𝑎𝑛𝑥 1−𝑡𝑎𝑛2𝑥 = tan 2𝑥/5 Marks.

10 Solution a) 𝟏 𝟏−𝒄𝒐𝒔𝒙 + 𝟏 𝟏+𝒄𝒐𝒔𝒙 = 𝟐𝒄𝒔𝒄𝟐𝒙 LHS= 𝟏 𝟏−𝒄𝒐𝒔𝒙 + 𝟏 𝟏+𝒄𝒐𝒔𝒙 RHS=𝟐𝒄𝒔𝒄𝟐𝒙 By proving that LHS=RHS Let work on the LHS= 𝟏 𝟏−𝒄𝒐𝒔𝒙 + 𝟏 𝟏+𝒄𝒐𝒔𝒙 = 𝟏(𝟏+𝒄𝒐𝒔𝒙)+𝟏(𝟏−𝒄𝒐𝒔𝒙) (𝟏−𝒄𝒐𝒔𝒙)(𝟏+𝒄𝒐𝒔𝒙) = 𝟏+𝟏+𝒄𝒐𝒔𝒙−𝒄𝒐𝒔𝒙 𝟏𝟐−𝐜𝐨𝐬𝟐 𝒙 = 𝟐 𝟏−𝐜𝐨𝐬𝟐 𝒙 We know that 𝟏 − 𝐜𝐨𝐬𝟐 𝒙 = 𝐬𝐢𝐧𝟐 𝒙 We get 𝟐 𝐬𝐢𝐧𝟐 𝒙 As we know 𝟏 𝐬𝐢𝐧𝟐 𝒙 = 𝒄𝒔𝒄𝟐𝒙 Which means that? 𝟐 𝐬𝐢𝐧𝟐 𝒙 = 𝟐𝒄𝒔𝒄𝟐𝒙 Hence LHS=RHS Reference: Learning Unit 2: Apply Fundamentals of trigonometry Question15. Given that z=3+4i. Find the value of the expression z + 25 𝑧 /15Marks SOLUTION: z + 𝟐𝟓 𝒛 =3+4i+ 𝟐𝟓 𝟑+𝟒𝒊= (𝟑+𝟒𝒊)(𝟑+𝟒𝒊)+𝟐𝟓 𝟑+𝟒𝒊 = 𝟗+𝟐𝟒𝒊+𝟏𝟔𝒊𝟐+𝟐𝟓 𝟑+𝟒𝒊 = 𝟗+𝟐𝟒𝒊−𝟏𝟔+𝟐𝟓 𝟑+𝟒𝒊 = 𝟏𝟖+𝟐𝟒𝒊 𝟑+𝟒𝒊 = (𝟏𝟖+𝟐𝟒𝒊)(𝟑−𝟒𝒊) (𝟑+𝟒𝒊)(𝟑−𝟒𝒊).

11 = 𝟓𝟒−𝟕𝟐𝒊+𝟕𝟐𝒊−𝟗𝟔𝒊𝟐 𝟗−𝟏𝟐𝒊+𝟏𝟐𝒊−𝟒𝟔𝒊𝟐 = 𝟓𝟒+𝟗𝟔 𝟗+𝟏𝟔 = 𝟏𝟓𝟎 𝟐𝟓 Therefore z + 𝟐𝟓 𝒛 = 𝟔 Reference: Learning Unit 3: Apply Fundamentals of Complex Numbers 16. Find the values of k for which the equation x2 + (k + 1) x + 1 = 0 has: /15Marks a) Two distinct real roots b) No real roots. SOLUTION: ∆ = (k + 1)2 – 4(1)(1) = (k + 1)2 – 4 = k2 + 2k + 1 – 4 = k2 + 2k – 3 = (k + 3) (k – 1) = 0 then k = –3 or k = 1. Table of sign of ∆ = k2 + 2k – 3 = (k + 3) (k – 1) k Factors – ∞ –3 1 +∞ k + 3 - - - 0 + + + + k – 1 - - - - - 0 + + + (k + 3)(k – 1) + + 0 - 0 + + + a) For two distinct real roots; ∆ > 0 and so k < –3 or k > 1. b) For no real roots ∆ < 0 and so –3< k < 1. Reference: Learning Unit1: Solve algebraically or Linear and quadratic Equations or inequalities.