Subject : MATHEMATICS

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1 CCE RF KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE S. S. L. C. EXAMINATION, MARCH/APRIL, 05 : ] MODEL ANSWERS : 8-E Date : ] CODE NO. : 8-E Subject : MATHEMATICS ( / New Syllabus ) ( / Regula Fesh ) ( / English Vesion ) [ : 80 [ Max. : 80 Ans. Key I.. A 8. C. D 6 4. B B 6. D 7. A 5 units 8. C 7 RF-6 [ Tun ove

2 8-E CCE RF II A \ B {,, 4 }. C : V 00 x CV 5. f ( x ) x 4 f ( 4 ) f ( 4 ) AX AY BX CY 4 BX BX 8 4. x 90 III. 5. If possible let us assume + 5 is a ational numbe. + 5 q p whee p, q z, q 0 p 5 q q p q 5 5 is a ational numbe. But, q p q is a ational numbe. 5 is not a ational numbe Ou supposition + 5 is a ational numbe is wong. + 5 is an iational numbe. RF-6

3 CCE RF 8-E 6. Numbe of students who have passed in Physics, n ( P ) 55. Numbe of students who have passed in Mathematics, n ( M ) 67. Numbe of students who have passed in both subjects, n ( P M )? Numbe of students in the classoom, n ( P M ) 00 n ( P ) + n ( M ) n (P M ) + n ( P M ) n ( P M ) n ( P M ) 00 n ( P M ) Numbe of students who have passed in Physics only n ( P ) n ( P M ) 55. n ( ) 700 OR n ( A ) 00 n ( B ) 00 n ( A B ) 00 n ( A ) + n ( B ) n ( A B ) + n ( A B ) n (A B ) n (A B ) n (A B ) 400 n ( A B ) l n (A l B l ) n ( A B ) l 00 n [ \ A B ] n ( ) \ n ( A B ) OR n (A l B l ) Given digits :,,, 7, 8, 9 6 a) 4-digit numbe can be fomed in P 4 ways 6 P P RF-6 [ Tun ove

4 8-E 4 CCE RF b) Even numbes : T H Ten U P 4 P 5 P P Units place can be filled in P ways 5 Tens place can be filled in P ways 4 Hundeds place can be filled in P ways Thousands place can be filled in P ways 8. D 5, n? n D C n Total numbe of ways P 5 P 4 P P n P 5 n! n ( n ) 5 n n n 5 n 5 n n n n 70 0 n 0n + 7n 70 0 n ( n 0 ) + 7 ( n 0 ) 0 n 0 OR n 7 Neglecting n 7 n Numbe of sides S { (, ), (, ), (, 6 ) ( 6, ), ( 6, ), ( 6, 6 ) } n ( S ) 6. RF-6

5 CCE RF 5 8-E a) Same numbe on both faces : Let A be the even. A { (, ), (, ), (, ), ( 4, 4 ), ( 5, 5 ), ( 6, 6 ) } n ( A ) 6 P ( A ) n ( A ) n ( S ) 6 P ( A ). 6 b) Both faces having multiples of five : Let B be the event. B { ( 5, 5 ) } n ( B ) P ( B ) n ( B ) n ( S ) P ( B ) X, 4, 6, 8, 0 X D X x D n 5 D 40 X 0 x 6 n 5 Vaiance Vaiance 8. D n 40 5 x 6 RF-6 [ Tun ove

6 8-E 6 CCE RF. L.C.M. of odes ( Multiplying by RF ) Let P ( x ) x x ax 0 By facto theoem, ( x 5 ) is a facto of P ( x ) iff P ( 5 ) 0 P ( 5 ) ( 5 ) ( 5 ) a ( 5 ) a 0 0 5a a 40 a 8. OR RF-6

7 CCE RF 7 8-E By division algoithm fo polynomials, P ( x ) [ g ( x ). q ( x ) ] + ( x ) P ( x ) ( x ) g ( x ). q ( x ) P ( x ) + { ( x ) } g ( x ). q ( x ) x 4 x x x x x x 4 x x x + x x x x + x + ( x ) x + { ( x ) } x Hence, we should add ( x ) to P ( x ) so that the esulting polynomial is exactly divisible by g ( x ). 4. x 4x + 0 a b 4 c x b b 4 ac a ( 4 ) 6 4 ( ) ( ) ( ) 8 x. Roots ae + OR. RF-6 [ Tun ove

8 8-E 8 CCE RF 5. To pove : Poof : Similaly PE EQ In ABD, PE BD PQ BC AP AQ PE AB AC BD... (i) ( Thale s theoem ) In ADC, AE AQ EQ... (ii) AD AC DC Fom (i) and (ii) PE EQ BD DC 6. But, BD DC ( AD is the median ) PE EQ. In ABC, B 90 AC AB BC 5 BC ( Pythagoas theoem ) BC BC Adjacent side BC cos Hypotenuse AC cos tan Opposite side Adjacent side tan 5. AB BC RF-6

9 CCE RF 9 8-E 7. Let A ( x, y ) (, 0 ) B ( x, y ) ( 5, ) C ( x, y ) ( 4, ) AB ( x x ) ( y y ) ( 5 ) ( 0 ) AB units. Similaly BC (4 5 ) ( ) units AC (4 ) ( 0 ) 4 5 units Peimete AB + BC + AC P units. 8. cm, d 8 cm. PA and PB ae tangents. Dawing cicle of adius cm Bisecting OP Dawing tangents PA, PB. RF-6 [ Tun ove

10 8-E 0 CCE RF 9. Scale : 0 m cm 40 m cm 60 m cm 80 m 4 cm Calculation 00 m 5 cm Field dawing 50 m 7 5 cm 0. F 6 V 8 E F + V E IV.. T, 7 T7 5 T and T 7 of A.P. ae 7, 5 d T p Tq p q T 7 T d T 7 a + d 7 RF-6

11 CCE RF 8-E a + T n a + ( n ) d T T 5 7 a 7 a 8 4 5th tem of HP Note : Fo altenate method full maks may be given.. Let the numbe of books be x Total cost Rs Cost of each book Rs. x If the numbe of books ( x + 5 ) Then cost of each book Rs But, x ( x 5 ) 60 ( x 5 ) 60x x ( x 5 ) 60x 00 60x x 5x x + 5x 00 x + 5x 00 0 x + 0x 5x ( x 5 ) x ( x + 0 ) 5 ( x + 0 ) 0 x 0 OR 5 Neglecting x 0 x 5 No. of books puchased by Aniuddha 5 OR RF-6 [ Tun ove

12 8-E CCE RF. Let the numbe of yeas be x Kavya s and Kathik s ages ae ( + x ) and ( 4 + x ) Poduct of thei ages 04 ( + x ) ( 4 + x ) x + 4x + x + 5x 50 0 x + 0x 5x 50 0 x 04 0 x ( x + 0 ) 5 ( x + 0 ) 0 x 5 OR 0 Neglecting x 0 x 5. i.e. Afte 5 yeas, poduct of thei ages will be 04. In ABC, A 90, AB x, BD CD BC BC BC AC AB B 45 C 45 AB AC x... Pythagoas theoem x x x BC x AD CD. BD BC. BC BC RF-6

13 CCE RF 8-E x x. 4 AD x AD x OR To pove : Poof : P a b In ABC CD AD. BD P AD. BD... (i) CB AB. AD a AB. AD AC AB. AD b AB. AD Adding (ii) and (iii) a b AB. BD AB. AD AB BD AD AD BD AB BD. AD a b P... (ii) a AB. BD... (iii) b AB. AD AB. AB P RF-6 [ fom (i) ] [ Tun ove

14 8-E 4 CCE RF 4. To pove : Poof : LHS LHS sec tan sec tan sec tan sec tan sec. tan + sec tan sec tan sec tan sec tan ( sec sec tan ) tan tan sec tan sec. tan ( sec tan ) sec + + tan + tan sec. tan sec. tan + RHS. cos cos tan sec. tan tan OR ( sec tan ) Multiplying by RF cos cos cos cos ( cos ) cos ( cos ) sin cos sin sin cos sin cosec + cot RHS RF-6

15 CCE RF 5 8-E 5. Data : To pove : BP BQ O is the cente of the cicle B is the extenal point BP and BQ ae the tangents. Poof : In BOP, B P O 90 Similaly in BOQ, B Q O ( adius, tangents at the point of contact ae pependicula ) In ight angled tiangles BOP and BOQ Hypotenuse BO Hypotenuse BO... Common side OP OQ... adii BOP BOQ BP BQ. CSA of cylinde TSA of cylinde h. ( h ) 46 h 54 TSA of cylinde ( + h ) 46 h RF-6 [ Tun ove

16 8-E 6 CCE RF 49 7 cm OR cm, h 0 cm, cm, h? We know, h h 0 h h 60 h 5 cm Volume of fustum h ( ) 5 ( ) 7 0 ( ) Volume of fustum 970 cubic cm. V. 7. Let the thee consecutive tems of the A.P. be ( a d ), a, ( a + d ) Sum 6 a d + a + a + d 6 a 6 a Poduct 0 ( a d ). a ( a + d ) 0 ( a d ) a 0 ( d ) 0 4 d 60 d 64 RF-6

17 CCE RF 7 8-E d 64 d 8 If a, d 8 then the thee numbes ae 6,, 0 If a, d 8, then the thee numbes ae 0,, 6 4 OR Let the thee consecutive tems be Poduct 6 a, a, a a. a. a 6 a 6 a 6 Sum of thei poducts in pais 56 a. (. ) a a a a. a 56 a a a o a) If a 6,, the consecutive numbes ae, 6, 8 b) If a 6,, the consecutive numbes ae 8, 6, + 4 RF-6 [ Tun ove

18 8-E 8 CCE RF 8. Data : In ABC, B 90 To pove : AC AB BC Constn. : Daw BD AC Poof : In tiangles ABC, ADB A B C 90, A D B 90 B A D... common angle ABC ~ ADB AB AD AC AB AB AC. AD.... (i) In tiangles ABC and BDC A B C AC B B D C 90 is common ABC ~ BDC BC AC BC AC. DC... (ii) DC BC Adding (i) and (ii) BC AB ( AC. AD ) + ( AC. DC ) AB AC ( AD + DC ) BC AC OR RF-6 AC AB BC 4

19 CCE RF 9 8-E 9. d 9 cm, R 4 cm, cm, R cm. Dawing AB and making mid-point M Dawing cicles C C and Joining BK, BL, PQ, RS Measuing and witing the length of tangents 8 7 cm 4 Requied tangents PQ RS 8 7 cm. RF-6 [ Tun ove

20 8-E 0 CCE RF 40. x + x 6 0 Let y x + x 6 x 0 4 y Table Dawing paabola Identifying oots 4 Altenate Method : Table + Paabola Staight line Dawing pependiculas and identifying oots + 4 RF-6

Subject : MATHEMATICS

Subject : MATHEMATICS CCE RR 560 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALE 560 00 05 S. S. L. C. EXAMINATION, JUNE, 05 : 5. 06. 05 ] MODEL ANSWERS : 8-E Date : 5. 06. 05 ] CODE NO. : 8-E Subject