TECH MATHEMATICS P1 JUNE 2023_Erratum.pdf

Downloaded from hlayiso.com CHIEF DIRECTORATE: EXAMINATIONS AND ASSESSMENT Steve Vukile Tshwete Complex, Zone 6 Zwelitsha, 5608, Private Bag X0032, Bhisho, 5605 REPUBLIC OF SOUTH AFRICA: Enquiries: Mrs P. Japtha Tel: 040 602 7031 . Fax : 040 608 7026. E-mail: Penny.Japtha@ecdoe.gov.za Website: www.ecdoe.gov.za Ref. no. 13/P Tel.: (040) 608 7026/082 523 7689 Enquire: Mrs P. Japtha Fax: 040 608 7295 NOTIFICATION TO: ALL PRINCIPALS OF TECHNICAL SCHOOLS IN THE FET BAND AND DISTRICT HEADS OF EXAMINATIONS FROM: MRS P. JAPHTA (a) CES: ASSESSMENTS INSTRUMENT DEVELOPMENT AND ITEM BANK MANAGEMENT SUBDIRECTORATE SUBJECT: TECHNICAL MATHEMATICS PAPER 1 JUNE EXAMS ERRATA DATE: 05 JUNE 2023 The Technical Mathematics P1 Grade 12 for June Examinations 2023 was written on Friday, the 02 June 2023. We were made aware of errors and omissions that was discovered in the marking guidelines. The amendment with regards to the marking was prepared in conjunction with the examiner and the moderator of the paper. This amendment addresses the errors and omissions and also ensures that learners are not disadvantaged. The following standardised approach to marking must be adopted across the Province. ERRATA QUES ERROR RECOMMENDATION CODE DESCRIPTORS 2 1.1.2 2 x +13 = 5 x 2 x 2 +13 = 5 x  Standard form 2 x 2 − 15 x + 3 = 0 2 x 2 − 5 x +13 = 0 standaardvorm A −b  b − 4ac 2 −b  b − 4ac 2 x= x=  Formula / Formule A 2a 2a −(−15)  (−15) 2 − 4(2)(3) −(−5)  (−5)2 − 4(2)(13)  Substitution / x= x= Vervanging CA 2(2) 2(2) x = 7, 29 or x = 0, 21 x = imaginary or non − real  Imaginary or non-real Denkbeeldig of nie-reëel CA Page | 1

Downloaded from hlayiso.com QUES ERROR RECOMMENDATION CODE DESCRIPTORS 3.1.3 1 1 log 2 16 + log 3 27 log 2 16 + log 3 27 2 2 1 1 = log 2 24 + log 3 33 = log 2 24 + log 3 33  Log property/eienskap A 2 2 4 4  Log property/eienskap A = log 2 2 + 3log 3 3 = log 2 2 + 3log 3 3 2 2 = 2 1 + 3 1 = 2 1 + 3 1 =2 =5  Simplification / vereenvoudiging CA 4.3.3 x ∈ ℝ, x ≠ 4 𝑦 ∈ ℝ, 𝑦 ≠ 2  𝑦 ∈ ℝ, y ≠ 2 CA 5.2 A = P(1 + i) n A = P(1 + i) n F CA 75 000 = 5 000 (1 + 9,5 ) 75 000 = 5 000 (1 + 9,5% ) n n  SF CA 75 000 75 000 = (1 + 9,5 ) = (1 + 9,5% ) n n 5 000 5 000  Simplification / 15 = (1,95 ) 15 = (1, 095 ) n n vereenvoudiging CA log1,9515 = n log1,09515 = n  log form/vorm CA 4, 0550 = n 29,83 = n  n = 29,83 CA 6.2.1 2 2 y= − 15 x + 7 m y= − 15 x + 7 m x3 x3  2x−3 A = 2 x −3 − 15 x + 7 m = 2 x −3 − 15 x + 7 m dy dy  −6x− CA = −6 x −2 − 15 = −6 x −4 − 15  −15 CA dx dx We request that this must be brought to the attention of all educators marking these papers and sincerely apologise for the inconvenience. Yours in education. 05 June 2023 MRS P.E. JAPHTHA DATE (a) CES: ASSESSMENTS INSTRUMENT DEVELOPMENT AND ITEM BANK MANAGEMENT SUBDIRECTORATE Page | 2