MODEL SSLC EXAMINATION KEY FOR MATHEMATICS

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1 MODEL SSLC EXAMINATION 018 KEY FOR MATHEMATICS SECTION I Q. Key Aswer Q. Key Aswer No No 1. () A \ B = A B 9. (4) 60 m. (4) {,4,5}. (1) a A.P 10. (4) 4.5 cm 4. (4) a k () ta θ 5. () cx + bx + a = 0 1. () ta θ 6. (1) 1, 0, 0, 1 1. (4) 7 : (4) (4) 7 8. () x y = () SECTION-II 16. B C = {, 4, 6} A (B C) = {, 4, 6, 7, 8, 9} 17. f (5) = 9, a = 9 f (8) = 15, b = = (1 4) + (9 16) +. = ( ) + ( 7) +.. a =, d = 4, = 10 S = [a + ( 1) d] 15 x 1 = 15 marks 10 x = 0 marks x 8 x 6x x 4 x x 4 = x α + β = 1, β = k, k = 1. a b = 10 a + b = 5 a =, b = 4 10 = [ 6 + (9 4)] = 10 ( x ) ( x x 4) ( x 4)( x ) ( x )( x ) x x 4 010

2 . Let a = k, b = k x y 1 k k It passes (, 4), k = x y (or) x +y 18 = 0.. I ABC, DE BC AD AE DB EC EC = 5.55 cm 4. LHS = si 6 x + cos 6 x = (si θ) + (cos θ) = (si θ + cos θ) si θ cos θ (si θ + cos θ) = 1 si θ cos θ = RHS BC 5. si 0 = AC AC = 1.8 m The legth of the ramp is 1.8 m r = = 7 cm h = 14 cm v = 1 π r h 156 v = (or) cm 7. π r = 77 cm π r = 186 TSA = π r = 4158 cm 8. Cv = 57, σ = 6.84 Cv = 100 x 684 x (s) = 4, (A) = ( A) P(A) = ( S) 4 00

3 1 0. (a) A I = A IA = A 9 6 AI = IA = A. 1 k 4 k (b) Area of ABC = 0, 0 k 1 k 7 k = SECTION-III 1. (Each diagram carries 1 mark) - 5 marks. f ( ) = 5, f ( ) = 15, f ( 1) =, f (1) = (i) f (5) + f (6) = 16 (ii) f (1) f ( ) = (iii) f ( ) f (4) = 5 (iv) f () f ( 1) 10 f (6) f (1) 15. a(1 r ) a(1 r S1, S 1 r 1 r a r (1 r ) S 1 = (S -S ) = (1 r) ), S a(1 r 1 r ar S S 1 = (1 r ) 1 r (S S 1 ) a r = (1 r ) S1( S S) (1 r) ) - marks - marks 00

4 4. 4x x + 4x 16x 4 4x + 5x 1x x 4 8x x 4x + 5x 4x + 9x 8x 6x + 16x 1x x 1x marks 4 16x 4x 5x 1x 4 4x x 5. The speed of the boat i the dowstream ad i the upstream are (15 + x) km/hr ad (15 x) km/hr respectively. 0 T1 ad T 15 x x 15 x 5 x = 00 x = ± x The speed of the stream is 5 km/ hr. 6. AB = marks (AB) T = B T A T = marks Equatio of PQ is x y + 9 = 0 At P, x = 0, y = P (0, ) 9 At Q, y = 0, x = 9 Q, 0 The equatio of the lie alog PR is x + y 9 = 0 040

5 8. a + b = 5, b = 5 a x y 1 a 5 a (5 a) x + ay = a (5 a) It passes (6, ), a = (or) a = 10 Whe a =, x +y = 6 Whe a = 10, x y 10 = Diagram Statemet Give To prove Costructio - marks Note: No diagram o mark. 40. LHS = (1 + cot θ cosec θ) (1 + ta θ + sec θ) cos 1 si 1 = 1 1 si si cos cos si cos 1 (cos si 1) = si cos (si cos ) 1 = si cos 1 si cos 1 = si cos = = RHS. 41. Let R, r be the radii ad h be the height of the frustum respectively. π R= 44 cm, πr = 8.4 π cm, h = 14 cm. R = 7 cm, r = 4. cm. 1 Volume of the frustum = h( R r Rr) = cm. - marks 4. Let r 1 ad h be the radius ad height of a right circular coe. Let r be the radius of the spherical shaped clay. Give that r 1 = 1 cm, h = 48 cm 4 1 r r1 h r - marks 1 The radius of the spherical clay = 1 cm 050

6 1 4. x x 7 x d = x x d d =154 - marks σ = d σ = σ = marks 44. (s) = 0 10 P(W) = 0 5 P(B) = 0 P (G) = 0 P(W U B U G ) = P(W) +P(B) +P (G) 10 5 = = (a) 108, 117,16., 999 This is a A.P, a = 108, d = 9, l = 999 l a 1 d = 100 S ( a l) S 100 = (b) Let α ad β = α be the roots of the equatio x + kx 81 = 0 a =, b = k, c = 81 α + β α + α k = α β α (α ) = 7 α = k =

7 SECTION IV 46. (a) Rough diagram First Circle Lie Segmet OP Perpedicular bisector Secod circle Two tagets lies Measurig the legth 46. (b) Rough diagram Draw a lie segmet PQ Draw arcs with radii 7 cm ad 5.5 cm. Joi PR ad QR Draw perpedicular bisectors of PQ ad QR Draw a circumcircle Joi PS ad RS - marks - marks -1mark -1mark - marks - marks -1mark 47. (a) y = x + x 1 x x x y Plot the poits ( 4, 0), (, 6), (, 10), ( 1, 1), (0, 1), (1, 10), (, 6), (, 0) Joi the poits by a smooth curve Scale ad drawig x ad y axis It has o real roots. - marks -4 marks - marks (b) y x y = kx y = x y x k Formatio of equatio Plottig the poits ad drawig the curve Drawig x ad y axes, scale Solutio Set x = 4, y = 8 y = 1, x = 6-5 marks - marks - marks YouTube:mathstimes_thirumuruga K.Thirumuruga PGT i maths GHSS Vazhuthavur VIllupuram Dt *************** 070

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