Gr 11 Maths 05 Term 1 Revision & Consolidation

GRADE 11 Maths: TERM 1 REVISION Terminology: Quadratic inequalities Nature of roots Number patterns Analytical geometry 1. Solve for x: 3x  20  7 x 2 (4) 2. The two roots of the equation 4𝑥 2 + 𝑝𝑥 − 9 = 0 differ by 5. Calculate the value(s) of p. (5) 3. A quadratic pattern has a second term equal to 1, a third term equal to -6 and a fifth term equal to -14. Calculate the second difference of this pattern and hence calculate the first term. (10) 4. In the diagram, 𝑃 −4; 4 , 𝑄 1; −2 and 𝑅 𝑘; −4 are three points in a Cartesian plane. Y P(-4;4) X Q(1;-2) R ( k ; -4 ) 4.1 Determine the length of PQ, in surd form. (2) 4.2 Determine the co-ordinates of M, the midpoint of PQ. (2) 4.3 Determine the equation of the perpendicular bisector of PQ. (4) 4.4 Calculate the value of 𝑘 if PR = QR. (4) 1 4.5 If 𝑘 = −7 calculate the size of RPQ, correct to ONE decimal place. (5) 2 [17]

Lesson 5: Term 1 Revision and Consolidation 3x 2  7 x  20  0 1. (3x  5)( x  4)  0 5  x   or x  4 3  p  p 2  4(4)(9) x 2(4)  p  p 2  144  8  p  p 2  144   p  p 2  144   5 8  8    2.  p  p 2  144  p  p 2  144  40 2 p 2  144  40 p 2  144  400  p  16 3. x; 1;  6; y;  14 Fd: 1  x; -7; 6  y;  14  y Sd: 8  x; 13+y;  20  2 y 20  2 y  13  y 33  3 y  y  11 8  x  13  11  x  10  Second difference =10 and T1  10 4.1 𝑃𝑄 = −4 − 1 2 + 4 + 2 2  Y ∴ 𝑃𝑄 = 61 (2) P(-4;4) 3 4.2 𝑀 − ; 1 (2) 2 4+2 6 4.3 𝑚𝑃𝑄 = −4−1 = − 5 X 5 ∴ 𝑦 = 6 𝑥 + 𝑐 Q(1;-2) R ( k ; -4 )

5 3 ∴ 1 = 6 × − 2 + 𝑐 5 9 ∴ 𝑦 = 6 𝑥 + 4 (4) 4.4 𝑘 + 4 2 + −4 − 4 2 = 𝑘 − 1 2 + −4 + 2 2  ∴ 𝑘 2 + 8𝑘 + 16 + 64 = 𝑘 2 − 2𝑘 + 1 + 4 ∴ 10𝑘 = −75 15 ∴𝑘 =− 2 (4) 4+4 16 4.5 𝑚𝑃𝑅 = 1 =  −4+7 7 2 16 ∴ tan 𝛼 =  7 ∴ tan 𝛼 = 66,4° 6 tan 𝛽 = − 5 ∴ 𝛽 = 129,8° ∴RPQ = 63,4° (5) [17]