81K-A Version_SEP

CCE RR A REVISED — 560 003 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE – 560 003 2020 S.S.L.C. EXAMINATION, SEPTEMBER, 2020 MODEL ANSWERS : 21. 09. 2020 ] : 81-K Date : 21. 09. 2020 ] CODE NO. : 81-K Subject : MATHEMATICS / New Syllabus / Regular Repeater / Kannada Version [ : 80 [ Max. Marks : 80 I. 1. y=p(x) (A) 3 (B) 5 1 (C) 4 (D) 2 [ Turn over (C) 4 RR (A) - 1114 

81-K 2 CCE RR 1 2. sec2 26  tan2 26 (A) 1 (B) 0 2 (C) 2 (D) 1 (D) 1 DE || AC 3. ABC (A) BD AC BC (B) BD DE BE AB  DE  BE AB  AC  BC (C) AB AC BE (D) AD DE BE BD  DE  EC BD  AC  EC (B) BD  DE  BE 1 4. AB AC BC 1 360 cm 3 (A) 120 cm 3 (B) 180 cm 3 (C) 90 cm 3 (D) 360 cm 3 (A) 120 cm 3 RR (A) - 1114 

CCE RR 3 81-K 5. x + 2y – 4 = 0 2x + 4y – 12 = 0 1 1 (A) (B) (C) (D) 1 1 (B) n an  3n  2 [ Turn over 6. (B) 5 9 (D) 25 (A) – 25 (C) – 5 (D) 25 7. P ( A ) = 2 P A 3 (A) 1 (B) 3 3 (C) 1 (D) 3 2 (A) 1 8. 3 7 (B) 616 (A) 154 (D) 308 (C) 616 (C) 616 RR (A) - 1114 

81-K 4 CCE RR II. 81=8 9. a1x  b1y  c1  0 a2x  b2y  c2  0 a1  b1 a2 b2 1 10. cos  = 24 sec  25 sec  = 25 1 24 11. O AC ACB = 50º BAC AC  ABC = 90° ½  ACB  ABC  BAC = 180° ½1 50º + 90º + BAC = 180° BAC = 180° – 140° = 40° RR (A) - 1114 

CCE RR 5 81-K 12. r l = r(r+l) 1 13. =2 P(1) ½ =4 ½1  ( 2, 4 ) = 2 ½ 14. P ( x ) = 2x3  3x2  11x  6 ½1 P ( x ) = 2 x 3 + 3 x 2 – 11x + 6 –4 P ( 1 ) = 2 ( 1 ) 3 + 3 ( 1 ) 2 – 11 ( 1 ) + 6 P ( 1 ) = 2 + 3 – 11 + 6 ½ P(1) = 0 ½1 15. ( x + 4 ) ( x + 3 ) = 0 [ Turn over (x+4)(x+3)=0 –4 x+3=0 x = –3 RR (A) - 1114 

81-K 6 CCE RR 16. sin 2 A = 0 cos A ½ ½1 sin2 A  cos2 A = 1  cos 2 A = 1 – sin 2 A ½ ½ cos A = 1  sin2 A ½ cos A = 1  0 ½2 cos A = 1 = 1 17. 2x + 3y = 11 2x – 4y = – 24 2x + 3y = 11 ... (i) (i) — (ii) ... (ii) 2x – 4y = – 24 (–) (+) (+) 7y = 35 y= 35 7 y=5 y = 5 (i) 2x + 3y = 11 2x + 3 ( 5 ) = 11 2x = 11 – 15 2x = – 4  x=  4 2 x = –2 RR (A) - 1114 

CCE RR 7 81-K 2x + 3y = 11 ... (i) ½ ... (ii) ½ 2x – 4y = – 24 ½ ... (iii) ½2 2x + 3y = 11 ½ y = 11  2x [ Turn over 3 y (ii) 2x – 4y = – 24 2x – 4  11  2x  = – 24 3 6x – 44 + 8x = – 72 14x – 44 = – 72 14x = – 28 x =  28 14 x = –2 x = – 2 (iii) y = 11  2 (  2 ) 3 y = 11  4 3 y = 15  y=5 3 x y1 3 – 11 23 – 4 24 2 –4 RR (A) - 1114 

81-K 8 CCE RR x  y  1 72  44  22  48 86 x  y  1 ½ 28  70  14 ½ ½2 x1 y  1 28   14  70  14 – 14 x = 28 – 14y = – 70 x= 28 y=  70  14  14 x = –2 y=5 18. 5 + 10 + 15 + ..... 20 5 + 10 + 15 + ....... 20 S20 = ? a=5 d=5 Sn  n [ 2a + ( n – 1 ) d ] ½ 2 ½ n = 20 S20  20 [ 2  5 + ( 20 – 1 ) 5 ] ½ 2 ½2 S20 = 10 [ 10 + ( 19 ) 5 ] S20 = 10 [ 10 + 95 ] S20 = 10  105 S20 = 1050 19. P ( x ) = 2x2  6x  k k P ( x ) = 2x 2 – 6x + k P ( x ) = ax 2 + bx + c  a=2 b = –6 c=k RR (A) - 1114 

CCE RR 9 81-K + =  b ½ a ½ ½ + = (6)  + = 3 2 ½2  = c   = k a 2  ( +  ) = 1  ( ) 2 3 = 1  k 2 2 322=k  k = 12 20. 2x2  5x  1  0 2 x 2 – 5x – 1 = 0 a=2 b= –5 c = –1 ½ a x 2 + bx + c = 0  = b 2 – 4ac  = ( – 5 )2 – 4 ( 2 ) ( – 1 ) ½  = 25 + 8  = 33 ½   >0 ½2  21. cosec A ( 1 – cos A ) ( cosec A + cot A ) = 1 tan A  sin A  sec A  1 tan A  sin A sec A  1 cosec A ( 1 – cos A ) ( cosec A + cot A ) = 1 ( LHS ) ( RHS ) RR (A) - 1114  [ Turn over

81-K 10 CCE RR LHS = 1 ( 1  cos A )  1  cos A  ½ sin A sin A sin A ½ ½ = 1  cos A  1  cos A  ½ sin A sin A 2 = 1  cos2 A sin2 A ½ ½ = sin2 A =1 sin2 A ½ ½  LHS = RHS. 2 tan A  sin A sec A  1 tan A  sinA  sec A  1 LHS RHS LHS = tan A  sin A tan A  sin A sin A  sin A cos A = sin A cos A  sin A sin A  1  1   cos A    =  1  sin A  cos A  1    = sec A  1 sec A  1  LHS = RHS. 22. ( 2, 3 ) ( 4, 7 ) ( 2, 3 ) ( 4, 7 ) ( x1 , y1) ( x2 , y2 ) RR (A) - 1114 

CCE RR 11 81-K  =  x1  x2 , y1  y2  ½  2 2    =  2 4 , 37  ½  2   2  =  6 , 10  ½  2 2  = [ 3, 5 ] ½2 23. A B C D E I ( Vowel ) 1, 2, 3, 4, 5, 6, 7, 8 n(S)=6 S = { A, B, C, D, E, I } ½ ½ n(A)=3 A = { A, E, I } ½  P(A) = n(A) ½2 n (S ) [ Turn over P(A) = 3 6 P(A) = 1 2 RR (A) - 1114 

81-K 12 CCE RR n(S)=8 S = { 1, 2, 3, 4, 5, 6, 7, 8 } ½ ½ n(A)=4 A = { 1, 3, 5, 7 } ½  P(A) = n(A) ½2 n (S ) 60º P(A) = 4 8  P(A) = 1 2 24. 4 = 180° – 60° = 120° 120° —½ 2 —½ RR (A) - 1114  —1

CCE RR 13 81-K 25. 3 12 ( 306, 657 ) 3 3  a ab b ab ( b  0 ). ½ 1 a b ½  b 3 =a ½ 3b2  a 2 ½ 3 a2 ½ ½ 3 a 3  a = 3c 1 ( 3c )2  3b2 3b2  9c 2  b2  3c 2 3 b2 3 b a b 3 a b 3 3 RR (A) - 1114  [ Turn over

81-K 14 CCE RR i) ( 306, 657 ) 306 = 3  3  2  17 1½ ( 306, 657 ) 306 = 3  3  73 ½ =3×3=9 ii) 9 12 9 12 = 3 × 3 × 4 = 36 ½ 3 9 12 = 36 ½ i) ( 306, 657 ) 657 = ( 306  2 ) + 45 ½ 306 = ( 45  6 ) + 36 ½ 2 45 = ( 36  1 ) + 9 ½ 306 657 RR (A) - 1114  612 45 6 45 306 270 36 1 36 45 36 9

CCE RR 15 81-K 4 9 36  36 36 = ( 9  4 ) + 0 ½ ii) 9 ½ 0 9 ( 306, 657 ) 12 9 12 = 3 × 3 × 4 = 36 ½ = 36 3 9 12 26. 60 30 6 108 BC = x ½ 60 ½ AC = x + 60 [ Turn over RR (A) - 1114 

81-K 16 CCE RR 30 ½ ½  AB = x + 30 ABC, B = 90° ½ ½ AC 2  AB 2  BC 2 ( x + 60 ) 2 = ( x + 30 ) 2 + x 2 3 x 2 + 120x + 3600 = x 2 + 60x + 900 + x 2 x 2 + 120x + 3600 = 2 x 2 + 60x + 900  2 x 2 – x 2 + 60x – 120x + 900 – 3600 = 0 x 2 – 60x – 2700 = 0 x 2 – 90x + 30x – 2700 = 0 x ( x – 90 ) + 30 ( x – 90 ) = 0 x – 90 = 0 x + 30 = 0 x = 90 x = – 30  BC = x = 90 AB = x + 30 = 90 + 30 = 120 AC = x + 60 = 90 + 60 = 150 ½ BC = x 6   AD = x + 6 RR (A) - 1114 

CCE RR 17 81-K  = 108 ½ ½ A = 1 bh 2 108 = 1 x (x +6 ) 2 108  2 = x 2 + 6x 216 = x 2 + 6x  x 2 + 6x – 216 = 0 ½ ½ x 2 + 18x – 12x – 216 = 0 ½ x ( x + 18 ) – 12 ( x + 18 ) = 0 x + 18 = 0 x – 12 = 0 x = – 18 x = 12  BC = x = 12 AD = x + 6 ½ AD = 12 + 6 = 18 C ( 5, 8 ) 3 27.  ABC A ( 0, 6 ), B ( 8, 0 ) CD  AB CD A ( 8, – 4 ), B ( 9, 5 ) C ( 0, 4 ) RR (A) - 1114  [ Turn over

81-K 18 CCE RR A ( 0, 6 ) B ( 8, 0 ) C ( 5, 8 ) ( x1 y1 )  ABC ( x2 y2 ) ( x3 y3 )  ABC  = 1 ½ 2 x1 ( y2  y3 )  x2 ( y3  y1 )  x3 ( y1  y2 ) ½ = 1 [0(0– 8) +8 (8 –6 )+5(6–0) ] 2 = 1 [ 0 + 16 + 30 ] 2 = 1  46. 2 = 23 A ( 0, 6 ) B ( 8, 0 ) ( x1 , y1 ) ( x2 , y2 ) AB d = ( x2  x1 )2  ( y2  y1 )2 ½ d = ( 8  0 )2  ( 0  6 )2 d = ( 8 )2  ( 6 )2 d = 64  36 ½ d = 100 AB = d = 10   ABC = 1  bh ½ 2 ½ 23 = 1  AB  CD 2 23 = 1  10  CD 2 46 = 10 CD  CD = 46 = 4·6 3 10 RR (A) - 1114 

CCE RR 19 81-K ½ A ( 8, – 4 ), B ( 9, 5 ), C ( 0, 4 ) d = ( x2  x1 )2  ( y2  y1 )2 AB = ( 9  8 )2  ( 5  (  4 ) )2  12  92  1  81  82 ½ BC = ( 9  0 )2  ( 4  5 )2  92  (  1 )2  81  1  82 ½ CA = ( 0  8 )2  ( 4  (  4 ) )2  (  8 )2  82  64  64  128 AB  BC ½ ½ 82 = 82   ABC ½ 3 28.  fi  0—5 8 5 — 10 9 10 — 15 5 15 — 20 3 20 — 25 1  fi = 26 RR (A) - 1114  [ Turn over

81-K 20 CCE RR 0—5  fi  1 5 — 10 10 — 15 8 ½ 15 — 20 9 ½ 20 — 25 5 ½ 3 ½3 1 ∑fx = 26 5 — 10 l=5 f1 = 9 f0 = 8 f2 = 5 h=5 =l+  f1  f 0 f2  h  2f1  f 0     = 5+  98 5   5  2 9  8     = 5+  1 5   5  18  8    = 5+  1  5  18  13    = 5+  1   5  5  = 5+1 =6 RR (A) - 1114 

CCE RR 21 81-K 29. 35 20 2 25 6 30 12 35 16 40 20 45 25 50 35 RR (A) - 1114  [ Turn over

81-K 22 CCE RR X Y- —½ 3 — 1½ RR (A) - 1114  —1

CCE RR 23 81-K 30.  ABD BC : CD = 1 : 2 BD C AD2  7AC2  ABC  ABD BC : CD = 1 : 2  ABC AB = BC = AC AD2  7AC 2 AE  BC 1  ABC BE = EC = a AE = a3 ½ 2 2 ½  ADE AED = 90° [ Turn over AD2  AE 2  ED2 AD 2   a 3 2   2a  a 2  2  2 AD 2  3a 2   5a 2 4 2 RR (A) - 1114 

81-K 24 CCE RR 31. AD 2  3a 2  25a 2 ½ 4 4 ½ AD 2  28a 2 3 4 AD2  7a 2  AC = a AD2  7AC 2 OP PQ ½ PQ = PR OP PR OQ, OR  POR  POQ PRO ½ ½ PQO = ½ OP = OP ½ ½3 OQ = OR   POQ   POR  PQ = PR RR (A) - 1114 

CCE RR 25 81-K OP PQ PR ½ ½ PQ = PR ½ ½ OP, OQ OR ½ OQP ORP ½3 OQP ORP [ Turn over OQ = OR OP = OP   OQP   ORP  PQ = PR RR (A) - 1114 

81-K 26 CCE RR 32. 21 7 O AB CD AOB = 30º ABCD P 11 OAB =   r 2 360 = 30  22  21  21 360 7 = 11  21 1 2 = 231 cm 2 2 RR (A) - 1114 

CCE RR 27 81-K OCD =   r 2  360  = 30  22  7  7 360 7 = 11  7 1 6 ½ = 77 ½3 6 ½ OAB — OCD ½ ½ = [ Turn over = 231  77 2 6 = 693  77 6 = 616  308 6 3 308  3 = 102·6 cm 2 = r 11 = r 11 = 22  r 7  r = 7 = 3·5 cm. ½ 2 2 = r 2 = 22  3·5  3·5 7 = 11  3·5 = 38·5 ABCD  AB = 2  = 2  3·5 AB = 7 cm  ABCD =  = 77 = 49 RR (A) - 1114 

81-K 28 CCE RR  ABCD —2 ½ = ½3 33. 6 7 = 49 – 38·5 = 10·5 8 3 4 4 2 1 A l BC l ½ 3 ½+½ RR (A) - 1114  ½

CCE RR 29 81-K 34. —2 —1 2x + y = 8 —1 x+y =5 2x + y = 8 y = 8 – 2x x01 2 3 4 y86420 x+y =5 y = 5–x x01 2 3 4 y54321 2 4 RR (A) - 1114  [ Turn over

81-K 30 CCE RR 35. A B A 60º 10 B 30º RR (A) - 1114 

CCE RR 31 81-K 648 / 3 = 1·73  648 km/h 648  1000 ½ 10  3600 ½  180 m/sec. [ Turn over = 180  10 = 1800 m OC = x CD = 1800 m/s OD = 1800 + x RR (A) - 1114 

81-K 32 CCE RR 1 OAC C = 90° tan  = AC 1 ODB D = 90° OC ½ tan 60° = h ½4 x 3  h x h= x 3 ... (i) tan  = BD OD tan 30° = h 1800  x 1  h 3 1800  x h 3 = 1800 + x ... (ii) (i) (ii) x 3  3 = 1800 + x x + 3 = 1800 + x 3x = 1800 + x 3x – x = 1800 2x = 1800 x= 1800 = 900 2  h= x 3 h = 900  3  900  1·73  h = 1557 m. RR (A) - 1114 

CCE RR 33 81-K 36. ½  ABC  DEF BAC = EDF ½ ½ ABC = DEF ½ AB  BC  AC DE EF DF [ Turn over AG = DE AH = DE AB G GH AC H RR (A) - 1114 

81-K 34 CCE RR  AGH  DEF AG = DE GAH = EDF AH = DF ½  AGH   DEF ½ AGH = DEF ½ ABC = DEF ½4  AGH = ABC -1  GH || BC   ABC AB  BC  AC AG GH HA AB BC AC  AGH   DEF. DE  EF  FD ½ RR (A) - 1114 

CCE RR 35 81-K  ABC  DEF A= D, B = E C= F DP = AB DQ = AC PQ ½ PQ || EF  ABC   DPQ 1 B= P= E 1 14 DP DQ 14  PE  QF AB AC DE  DF AB  BC DE EF AB  BC  AC DE EF DF 37. 5 30 1 27 RR (A) - 1114  [ Turn over

81-K 36 CCE RR  =5 ½  = 2·5  = 14 ½ ½+½  h = 14 – 5 h=9 ½ ½ = 2rh + 2 ( 2r 2 ) ½ = 2r [ h + 2r ] ½ = 2  22  2·5 [ 9 + 2  2·5 ] 7 = 2  22  2·5  14 7 = 2 22  2·5 2 7 = 88  2·5 = 220 4 RR (A) - 1114 

CCE RR 37 81-K r1  h1 ... (i) ½ r2 30 ½ ½ = 1 ½ 27  ½ 1 r12  h1  1  1    r22  h2 ½ 3 27 3 ½ r12  h1  1  r22  h2 ½4 27 [ Turn over r12  h1  1  r22  30 27 r12  h1  10 ... (ii) r22 9 (i) (ii)  h1 2  h1  10  30  9 h13  10 900 9 h13 = 1000 h1  31000 AB = h1 = 10  BP = AP – AB = 30 – 10 BP = 20 cm RR (A) - 1114 

81-K 38 CCE RR 38. 3 29 7 28 8 a = b+3 ... (i) ½ a7 = 28 ... (ii) ½ ... (iii) ½ a + 6d = 28 ½ b8  29 ... (iv) ½ b + 7d = 29 ½ (i) (ii) ½ a + 6d = 28 b + 3 + 6d = 28 b + 6d = 25 (iii) (iv) b + 7d = 29  d=4 b + 6d = 25 (–) (–) (–) d=4 d = 4 (ii) a + 6d = 28 a + 6(4) = 28 a + 24 = 28 a = 28 – 24 a=4 RR (A) - 1114 

CCE RR 39 81-K d = 4 (iii) ½ ½ b + 7d = 29 ½5 b + 7(4) = 29 b + 28 = 29 b=1 I a, a + d, a + 2d, .............. 4, 4 + 4, 4 + 2 (4), ............. 4, 8, 12, ............  II b, b + d, b + 2d, .............. 1, 1 + 4, 1 + 2 (4), ............. 1, 5, 9, ............ RR (A) - 1114  [ Turn over