Mathematics Grade 11 KZN March Test 2025 Paper & Memo

Testpapers.co.za Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAI RTIFICATE et seeeneann « = = ” . " a " a . a s ¢ ®aesaneeses: This question paper consists of 5 pages. Seseuesseses” saneeeneee®

Mathe: March 2025 Test Testpapers.co.za INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 3 questions. 2. Answer ALL the questions. 3. Number the answers corre: ly. ng ( bering system used in this question paper. determining your answers. Bi Answers only will NOT necess 6. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 7. If necessary, round off answers correct to TWO decimal places, unless stated otherwise. 8. Write neatly and legibly. Copyright Reserved Please Turn Over

Mather Testpapers.co.za March 2025 Test QUESTION 1 Ta Solve for x: 1. x -3x-28=0 GB) 1:42 o(1 + 25) =3x? (answer correct to TWO decimal places) (4) 113 2x? +x-3 20 (4) 2. 12 Given: Jx-3 = vx 1.2.1 Solve for x. _ (5) 2 122 Hence, or otherwise, solve for k if Vk-20 = . 2 vet " (6) 1.4.1 Solve for x in terms of m. (2) 1.4.2 Hence, or otherwise, determine the value(s) of m for which the above equation will have equal roots. (3) [29] Copyright Reserved Please Turn Over

Mathi Testpapers.co.za QUESTION 2 2.1 Evaluate the following expressions, without using a calculator: S74 V121x? —¥27x° 4x stot gh* Dulad on 2.2 Solve for x, without Hiaririorephysics.com 2.3 It is given that 4" =6 and 6” =8. Calculate the value of mn. Copyright Reserved March 2025 Test 3) (4) (3) (3) @G) [16] Please Turn Over

(5) (6) (6) Mathe: Testpapers.co.za March 2025 Test QUESTION 3 a Given that 13cos 8 =5and a+ # =90°. By making use of a sketch and WITHOUT the use of a calculator, calculate the value of Liat . cosa@ 32 Simplify the following to a single trigonometric term: sin (180° +x).cos* (90°—x)+sin(—x).cos* (180°—x). 33 Evaluate WITHOUT cos100°cos315° tan 210° sin 675°cos150°sin170° — jorephysies.com 3.4 Given: (1-cos* »){tans+ ——|, sinx 320 3.4.1 Prove that (1-cos* 2)(tans+ $552) tans sinx 3.4.2 For which value(s) of x in the interval [0°;360°) is the identity in LION 3 .4,1)iiidefined? d 0€ (90°; 270°). If 0<k <3, determine the value of our answer must be a single fraction in its simplest form. Copyright Reserved (4) G) (5) [30] a

Testpapers.co.za BRINE MARKS: 75 Copyright Reserved These marking guidelines consist of 7 pages. Please turn over

Testpapers.co.za Me March 2025 Test CAPS ~ Grade 11 Marking Guidelines QUESTION 1 1.1.1) x7-3x=28=0 2 (x+4)(x-7)=0 OR x= ~(-3)#y(-3) -4()(28) VA correct factors OR 2(1) subst. into correct formula aor x=7 ¥ CA answer VCA answer GB) 2 2 L452 o(t +s) =3x 6+4x =3x° 3x7 -4x-6=0 ¥A standard form _ -b+ Vb —4ac 2a ¥CA subst. into correct a= quadratic formula x=-0,90 or x=2,23 x=-0,90 or x=2,23 CAY answer VCA answer. NOTE: Penalty of 1 mark for incorrect rounding. (4) 1.1.3 2x°+x-320 (2x+3)(x-1)20 ¥A correct factors ¥CA critical values (can be on a sketch or number line) ¥ CAY CA answer (4) 12.1] fg-3=— aa vx. . ” xNx-3=2 x-3= Ea x(x - 3) =2 x 4 P ‘ ~3)=2? -3=-— ¥ A squaring both sides x (x 3) 2 OR x * q is *_3x-4=0 x? —3x-4=0 YA standard form (x-4)(x+1)=0 (x-4)(x+1)=0 ¥CA factors x=4 or x#-1 x=4 or x#-1 ¥CA YCA both answers with rejection (5) Copyright Reserved Please turn over

Testpapers.co.za Me March 2025 Test CAPS — Grade 11 Marking Guidelines 1.22| Jezaoe k-17 k-171=4 YCA equating k-17 to value of x ¥CA answer k=21 (2) 13 | gys2x-3 VA y=2x-3 x?—15=-5x(2x-3)-(2x-3)° VCA substitution x? =15=—10x? +15x—(4x? -12x+9) x? =15=-10x? +15x=4: YCA simplification x? +10x7 +4x° -15x 15x? -27x-6=0 5x? -9x-2=0 (5x+1)(x-2)=0 CA factors 1 on Of oF YCA both x—values y=2(-2)}-3 or y=2(2)-3 17 7 as y= YCA both y—values OR (6) 28 . OR 2 vA xa 2 (=) -15= -5(23)-¥ ¥ CA substitution 2 61,2 y +6y49 15 Sy Isy__ 2 4 2 2 —5y- (0) gas) af SPP) ag y? +6y+9-60=-10)" -30y—4y" ¥CA simplification 15y? +36y—51=0 sy? +12y-17=0 (Sy+17)(y-1)=0 ¥CA factors ¥CA both y—values YCA both x—values (6) Copyright Reserved Please turn over

Testpapers.co.za Me March 2025 Test CAPS — Grade 11 Marking Guidelines LA.l 5x eme+18=0 x" +2mx+36=0 2 1 =m+,|(m) - 43) 18 2 LP ey 2 (8) OR x= ~2m* y(2m) -4(1)(36) | vA correct substitution {3} 2(1) into quadratic formula 2 ~2m+ 4m’ —144 x=—m£m —36 a YCA simplified answer (2) 1.4.2 | A=m? -36 A=4m’? -144 b’ -4ac =0 sm? —36=0 Y¥CA A equated to 0 (m+6)(m-6)=0 YCA factors m=-6 or m=6 ' ¥CA both answers NOTE: Penalise a mark i: i n factors m= +36 (3) [29] Copyright Reserved Please turn over

Me Testpapers.co.za CAPS — Grade I1 March 2025 Test Marking Guidelines QUESTION 2 211) Vi2he 3278 -4°x _ Ily=3x v A numerator: square & cube ~ Ig v A denominator =-8 ¥CA answer d GB) 2.1,2:| igt.6r.o* 161.3% ke k3 _ (2°) Aa} (3° A splitting to prime bases ey _ Beare ah ae ¥CA applying the exponential laws aka 37k correctly = DUES SP ~34+2-2k+k =2134 ill ¥CA simplification ¥CA answer (4) Peer vA isolating the variable vA raising both sides of equation to an exponent of —3 ¥CA answer G3) 2.2.2 vA both equations after transposing YA 5‘ =5° and no solution v A answer (3) 2.3 vA raising both sides to exponent n YA substituting 8 for 6” v A answer (3) [16] Copyright Reserved Please turn over

Testpapers.co.za Me March 2025 Test CAPS ~ Grade 11 Marking Guidelines QUESTION 3 3.1 | 13cosB=5 r 13 12 . y=r—x° [Pythagoras] VA diagram = 3? —5? >x y= 12 v¥ Ay—value 12 tanB 5 ¥CA correct substitution for cosa 12 tan B 13 YCA correct substitution for ~ 12,13 cosa 5 12 = fe or 2 . 5 5 ¥CA answer (5) 3.2 | sin(180°+.x).cos* (90°-x)+sin(—x).cos* (180° —x) F x Ae F 2 VA -sinx VA sin? x =(-sinx)(sinx) +(—sin x)(—cos x) . i VA -sinx VA (—cosx) =~sin x(sin’ x-+cos* x) YCA factors =—sin x(1) =-sinx ¥CA answer (6) 3.3 cos 100° cos 315° tan 210° sin 675° cos150°sin 170° __ (~cos80°) (cos 45°)( tan 30°) vA numerator _ (=8in452)( cos VA denominator Y A cos80° =sinl10° OR sin 10° = cos80° Y CA special angle values ¥CA simplification 2 == ¥CA answer 2 (6) Copyright Reserved Please turn over

Me Testpapers.co.za March 2025 Test CAPS ~ Grade 11 Marking Guidelines cos x 3.4.1 | LHS= (1 =cos* x){ tan x+— } sin x : sinx cosx P sinx 2 sir xf += } YA sin’ x YA cosx sinx cos x .» _( sin? x+cos* x , =sin* x] —————_ VA LCD (cosx.sinx ) cos x.sin x +2 1 =Sin* x] ———— cos x.sin x YA sin? x+cos* x=1 k?4+9 _ sinx cosx : is out of 4 marksin || vA simplification to sins =tanx aper. It should be OES =RHS as in the memo. (5) 3.4.2 | Undefined when sinx =0a “x E {0° ; 90°; AV AY AY (Any 3 values given) Q) 3.5. | x =r?-y? [Pythagoras =(k +9) -(-6k)° =k* +18k? +81-36k? ¥ A diagram in quadrant 3 =k*-18k? +81 -(e-9) x= +(e -9) vx=-(k-9) VAx-— value ie. 1 _ 6k ¥CA correct substitution cosO -(k°-9) —(k?-9) ¥CA simplification to a single fraction Copyright Reserved __(k-3)(k=3) ~ (k=3)(k+3) = (t-3) ¥CA answer (k +3) (5) [30] TOTAL: 75