SAJC 2026 Prelim P1 Ans

Marking Scheme for Paper 1 Q1 Solutions 1(a) sin cos tan 1 where cos 0, 2 For 0 , 4 For sin cos , 0 , 4 x x x x x x x x x x x                   (b)             0 4 0 4 4 0 4 sin cos d sin cos d sin cos d cos sin cos sin 2 2 2 2 1 0 1 02 2 2 2 2 2 x x x x x x x x x x x x x                                                        

2   2 d 1 e d 1 e 1 e x x xx       Let   2 1, 1, 11 x xxdv e duu x v dx dx e e            ln 3 20 ln 3 ln 3 00 ln 3 ln 3 00 ln 3 0 d 1 1 1 1 d1 1 1 d1 1 ln 3 ln 11 3 ln 3 4 ln ln 24 3 ln 3 ln 4 ln 3 ln 24 3 ln 3 ln 24 x x x x x x x x xe x e x xe e ex xe e e                                                          2 d e d (1 e ) d 1 1d 1 e x x x vu x x u vx      

3(a) Step 1: Replace x with 2x 3( 2) 6 3f ( ) 2 2 x xx e e      Step 2: Replace x with xk 3 f ( ) 2 x kx e  Step 3: Replace y with 1y 3 3 g( ) 2 1 1 x x k kx e e         (b) Since the graph g(x) and 1h( ) ln( 1)2x x  are reflections in y x , then g(x) is the inverse of h(x) 1 2 2 2 3 2 h ( ) : 1 ln( 1)2 2 ln( 1) 1 1 g( ) F 1 Comparing 1 and 1 3 2 3 3 2  or  1.52 ind y y x x xk x y x y x e x x e x e e e x xk kk                  

4(a) Let 2 11 2 3 1 5 5nn n nS u u u u u                   1 2 1 12 1 2 1 2 1 2 2 2 2 5 5 5 5 5 5 15 5 5 5 24 55 24 55 n n n nn n n n n n r r u S S u                       2 1 1 1 5 5 120S     ,  21 24 5 1205u   Since 1 1S u ,  224 5 , 5 rru r    (b)      2 2 1 4 2 42 1 24 24 245 5 55 5 125r r rru                4 2 1 1 1 4 8 4 4 4 4 4 24 5125 24 5 5 5125 5 5 124 125 5 1 5 5 126 n n r r r r n n n u                

(c) 2 1 1 1 2 2 2 24 55 24ln( ) ln (2 ) ln 5 5 24ln 2(ln 5)5 24ln 2(ln 5) (1 )5 2 24ln (ln 5) (ln 5)5 24ln ln 5 (ln 5)5 (ln 24) ( ln 24, ln 5) ln5 r r n n r r r n r a b u u r n r nn n n n n n n n n                                                       Alternative Solution:                 2 1 1 2 1 1 1 1 2 2 24ln ln 5 5 24ln ln 55 24ln 2 ln 55 24ln 2 ln 55 24ln 2 ln 5 15 2 24ln ln 5 15 24ln ln 5 ln 55 ln 24 ln 5 n n r r r r n r r n r n n r r u r r nn n n n n n n n n                                                     ln 24, ln 5a b 

5(a) Since any horizontal line y k , k  cuts the curve of  fy x at most once, f is one-one. Hence f has an inverse. Alternative statement Since any horizontal line y k , (0,3]k cuts the curve of  fy x once, f is one- one. Hence f has an inverse. (b) For 0 1 x  ,  2 2 2x x x        f 3 2x x x   Let     2 3 2 3 2 y x x y x x     Method 1         2 2 2 2 3 2 3 2 3 1 1 3 1 1 y x x x x x x                 2 1 1 3 1 1 3 yx yx       Method 2     2 2 2 3 2 2 0 3 1 1 0 3 1 1 3 y x x yx x yx yx             Method 3     2 2 2 2 3 2 2 0 3 3 2 2 0 3 y x x yx x x y x x yx x                2( 2) ( 2) 4 1 3 2 1 2 4 1 3 2 1 1 3 y x y yx                (1, 3) 0 x y

Since 0 1 x  , 1 1 3yx   1 1 f ( ) 1 1 3 f ( ) 1 1 3 yy xx         1 ffD R (0,3]   (c) (d) gf f g Since R (0,3], given that R , gf exist D [0, 4); sSince D .    (e) f gf f gfD (0,1] R (0,3] R ( 2, 2]        gfR ( 2, 2]  6(a) u = a + b and the vector v = a – 2b If u and v have equal magnitudes Then |u|=|v|  |u|2 = |v|2  u  u = v  v ---(1)  (a + b)  (a + b) = (a – 2b)  (a – 2b)  a  a + 2a b + b  b = a  a – 4a b + 4b  b  6 (a b) = 3(b  b) ---(2)  2(|a| |b| cos |) = |b|2 Since b is a non-zero vector,  |b| = 2 |a| cos  (Shown) (b) From part (i), |b| = 2 |a| cos  , Since a is a unit vector, |a| = 1  |b| = 2 cos 30o = 32 3 2      (Ans) (c) Given that AP: PB = 2: 1, (4,0) (2,2) x y (0,-2) 0

1 3OP    (2b + a) by ratio theorem. Since N lies on the line OB, Let ON k   b, where k is a constant. PN   is perpendicular to OB   , PN    b = 0 ( ON OP     )  b = 0 (kb 13 (2b + a))  b = 0 k (b  b) 13 (a b) 23 (b  b) = 0 --- (#) Now, a b = 0 3(1)( 3)cos30 2 From (#), 2 2 1 3 2( 3) ( 3) 03 2 3 5 6 k k        5 6ON    b (d)  1 23 1 = 2( )3 Since 0, 1 = 3 1 = sin 303 1 1 1 = 3 2 6 OP b bPN b a b b bb b a b b b ba b a bb                               

7(a) (b) From the diagram: 1 1tan ( ) tan ( )d h dx x         (Shown) 2 2 2 2 2 tan tantan tan( ) 1 tan tan tan 1 ( )( ) tan (Shown) d h d x x d h d x x h x x dh d x hx x dh d                         (c) 2 2tan hxx dh d    Differentiating with respect to x   2 2 d d tand d dNote that when is maximum, 0 d hx x x x dh d x          

      2 2 2 22 2 2 2 22 2 2 2 2 (2 )dsec d ( ) d 0 since 0 for stationary value of d Since 0, 0 (*) since > 0 ( ) x dh d h hx x x x dh d h d dh x xx dh d h d dh x x d dh x x d d h                           (d) h = 3, d = 2, d 0.5,dxt  Using result in (i), 1 15 2tan ( ) tan ( )x x    2 2 2 2 2 2 2 2 d 1 5 1 2 5 2d 1 ( ) 1 ( ) 5 2 = 25 4 Using chain rule and substituting = 4, d d d 5 2 9 ( 0.5)d d d 4 25 4 4 820 9The rate of change of is rad/sec or 0.0 110 rad/820 x x x x x x x x x t x t                                  sec. 8(a) 2 2 2 2 2 2 ( )( ) 2Re( ) and ( ) 2Re( ) (Sh Since own) z w z w z zw wz ww w w w ww w z w w z w z w z w                      ∣ ∣ ∣ ∣ ∣ ∣ (b)(i) Since p and q are real, complex roots must occur in conjugate pairs. Since w is a root, w* must also be a root. Factors of the polynomial are: 2 2 ( 2),( ) and ( ) ( 2)( 2Re( ) ) 0 (Shown) z z w z w z z w z w         ∣ ∣ The constant term of the polynomial 3 2 16 is 16.z pz qz   2 22 16 8 8 2 2 w w w      ∣ ∣ ∣ ∣ ∣ ∣