Namma Kalvi 10th Answer_key_maths_march_2017.pdf

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g¤jh« tF¥ò bghJ¤ nj®Î kh®¢ - 2017 éil¡ F¿¥ò ÃçÎ - 𝐈 1) M) 21 2) M) 3 3) m) 8 4) <) 31 𝑐+𝑎 5) m) 2𝑏 −8 −2 6) Ï) ( ) 1 −7 7) Ï) −3 8) <) 8 9) <) 6 br. Û 10) M) 16 br. Û 11) Ï) 60° 12) <) −9 13) Ï) 5 br. Û

𝑙𝑥2 +𝑚𝑥1 𝑙𝑦2 +𝑚𝑦1 , 𝑙+𝑚 ) 𝑙+𝑚 𝑃(𝑥, 𝑦) = 𝑃(−2, 3) 24) ϧF 𝑎 > 0, k‰W« (𝑥, 𝑦) = (0, 𝑎). ⇒ 𝑥 = 0, 𝑦 = 𝑎 𝑥 = 0 v‹gJ 𝑦 − m¢Á‹ rk‹ghL. nkY«, 𝑦 = 𝑎 v‹gjhš (0, 𝑎) v‹w òŸëahdJ 𝑦 − m¢Áš mikÍ«. vdnt, 𝑎− ‹ všyh äif kÂ¥òfS¡F« (0, 𝑎) v‹w òŸëahdJ 𝑦 − m¢Áš mikÍ«. vdnt, bfhL¡f¥g£LŸs T‰W bkŒašy. 25) ∆𝑃𝐴𝐵 k‰W« ∆𝑃𝑄𝑅 − š ∠𝑃𝐴𝐵 = ∠𝑃𝑄𝑅 ⇒ ∆𝑃𝐴𝐵~∆𝑃𝑄𝑅 vdnt, 𝐴𝐵 𝑃𝐵 = 𝑄𝑅 𝑃𝑅 𝐴𝐵 × 𝑃𝑅 3 × 6 ⇒ 𝑄𝑅 = = = 9 𝑐𝑚 𝑃𝐵 2 26) gl¤ÂèUªJ, br§nfhz ∆𝐷𝐸𝐶 − š, 𝐶𝐷 tan 𝜃 = 𝐸𝐶 ⇒ 𝐶𝐷 = 30 Û nfhòu¤Â‹ cau«: 𝐵𝐷 = 𝐵𝐶 + 𝐶𝐷 = 1.5 + 30 = 31.5 Û 27) cUisæ‹ bkh¤j¥ òw¥ gu¥ò = 1540 r. br. Û 4 ℎ = 4𝑟 ⇒ 𝑟 = ℎ 2𝜋𝑟(ℎ + 𝑟) = 1540 ℎ = 28 𝑐𝑚

23) 𝑃 (𝑥, 𝑦) = 𝑃 (

14) Ï) 6√2 11 15) M) 13 ÃçÎ - II 16) 𝐴 ∪ (𝐵 ∪ 𝐶 ) = {1, 2, 4, 5, 6, 7, 8} (𝐴 ∪ 𝐵) ∪ 𝐶 = {1, 2, 4, 5, 6, 7, 8} 17) 𝑎 = 9, 𝑏 = 15 𝑡2 𝑡3 18) = 𝑡1 𝑡2 7 (𝑚 + 2) 𝑚 = −2 2 𝑚 −7 𝑚2 − 𝑚 − 2 = 0 𝑚 = −1 (𝑜𝑟) 𝑚 = 2 19) (1) + (2) ⇒ 24𝑥 + 24𝑦 = 144 ⇒ 𝑥 + 𝑦 = 6 → (3) (1) − (2) ⇒ 2𝑥 − 2𝑦 = −4 → (4) Solving (3) and (4), 𝑥 = 2, 𝑦 = 4 2 6𝑥 +9𝑥 3𝑥(2𝑥+3) 20) 2 = 3𝑥 −12𝑥 3𝑥 (𝑥−4) 2𝑥 + 3 = 𝑥−4 1 0 21) 𝐴 = ( ) 3 2 2×2 3 2 8 −1 22) 𝐶 = 2 ( )+( ) 5 1 4 3 6+8 4−1 14 3 𝐶=( ) =( ) 10 + 4 2 + 3 14 5

28) Å¢R= 𝐿 − 𝑆 59 = 𝐿 − 12 ⇒ 𝐿 = 71 29) 𝑆 = {𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇} 𝑛 (𝑆 ) = 4 𝑛 (𝐴 ) 1 𝑖)𝑃(𝐴) = = 𝑛 (𝑆 ) 4 𝑛 (𝐵 ) 1 𝑖𝑖)𝑃(𝐵) = = 𝑛 (𝑆 ) 2

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g¤jh« tF¥ò bghJ¤ nj®Î kh®¢ - 2017 éil¡ F¿¥ò 1 f. br. Û 7 4 3 50688 𝜋𝑟 = 3 7 𝑟 = 12 br. Û (𝑂𝑅) 2 2 M) 𝑥 − 𝑦 = (𝑎 sec 𝜃 + 𝑏 tan 𝜃 )2 −(𝑎 tan 𝜃 + 𝑏 sec 𝜃 )2 = 𝑎2 (sec2 𝜃 − tan2 𝜃 ) + 𝑏2 (tan2 𝜃 − sec2 𝜃 ) = 𝑎2 − 𝑏 2 . ÃçÎ - III 31) 𝐵 ∪ 𝐶 = {1, 2, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, 𝑦} 𝐴\(𝐵 ∪ 𝐶) = {𝑎, 𝑏, 𝑥, 𝑧} 𝐴\𝐵 = {𝑎, 𝑏, 𝑓, 𝑔, 𝑥, 𝑦, 𝑧} 𝐴\𝐶 = {𝑎, 𝑏, 𝑐, 𝑥, 𝑧} (𝐴\𝐵) ∩ (𝐴\𝐶 ) = {𝑎, 𝑏, 𝑥, 𝑧}.

30) m) cUisæ‹ fdsÎ = 7241

𝑥−3 3 𝑓 (6) = 1; 𝑓 (9) = 2; 𝑓 (15) = 4; 𝑓 (18) = 5; 𝑓 (21) = 6 𝑖) m«ò¡F¿ gl«: 𝑖𝑖) tçir nrhofë‹ fz«: 𝑓 = {(6, 1), (9,2), (15, 4), (18, 5), (21, 6)} 𝑖𝑖𝑖) m£ltiz:

32) 𝑓 (𝑥 ) =

𝑥 6 𝑓(𝑥) 1 𝑖𝑖𝑖) tiugl«:

9 2

15 4

18 5

21 6

33) 𝑆𝑛 = 1 − 4 + 9 − 16 + 25 − 36 + ⋯ 2𝑛 terms = (1 − 4) + (9 − 16) + (25 − 36) + ⋯ 𝑛 terms = −3 − 7 − 11 − ⋯ 𝑛 terms 𝑎 = −3, = −4 𝑛 𝑆𝑛 = [2𝑎 + (𝑛 − 1)𝑑] 2 𝑛 = [2(−3) + (𝑛 − 1)(−4)] 2 = −𝑛(2𝑛 + 1)

34) 𝑆𝑛 = 7 + 77 + 777 + ⋯ 𝑛 terms 7 = [9 + 99 + 999 + ⋯ 𝑛 terms] 9 7 = [(10 + 100 + 1000 + ⋯ 𝑛 terms) − 𝑛] 9 7 = [(10 + 102 + 103 + ⋯ 𝑛 terms) − 𝑛] 9 𝑎(𝑟 𝑛 − 1) 𝑆𝑛 = ,𝑟 > 1 𝑟−1 7 10(10𝑛 − 1) 𝑆𝑛 = [ − 𝑛] 9 9

(OR) 70 7 (10𝑛 − 1) − 𝑛 𝑆𝑛 = 81 9 35) Úç‹ ntf« = 𝑥 ».Û/kâ gl»‹ ntf« = 15 ».Û/kâ 30 30 𝑇1 = , 𝑇2 = 15+𝑥 15−𝑥 1 𝑇1 + 𝑇2 = 4 kâ 2 30 30 9 + = 15+𝑥 15−𝑥 2 1800 = 9(225 − 𝑥 2 ) 𝑥 2 = 25 ⇒ 𝑥 = ±5 gl»‹ ntf«= 5 ».Û/kâ 36) 4𝑥 2 − 3𝑥 + 7 4𝑥 2

16𝑥 4 − 24𝑥 3 + (𝑎 − 1)𝑥 2 + (𝑏 + 1) + 49

16𝑥 4 8𝑥 2 − 3𝑥

− 24𝑥 3 + (𝑎 − 1)𝑥 2 −24𝑥 3 + 9𝑥 2

8𝑥 2 − 6𝑥 + 7

(𝑎 − 10)𝑥 2 + (𝑏 + 1)𝑥 + 49 56𝑥 2 −

42 𝑥 + 49 0

bfhL¡f¥g£l gšYW¥ò¡nfhit xU KG t®¡fkhjyhš, 𝑎 − 10 = 56, 𝑏 + 1 = −42 ⇒ 𝑎 = 66 k‰W« 𝑏 = −43. 8 −3 37) 𝐴𝐵 = ( ) 11 −4 (𝐴𝐵)𝑇 = ( 8 11 ) −3 −4 5 7 𝑇 𝐴 =( ) 2 3 2 −1 𝐵𝑇 = ( ) −1 1 8 −3 𝐵 𝑇 𝐴𝑇 = ( ). 11 −4 1 𝑥1 𝑥2 𝑥3 𝑥4 𝑥1 38) eh‰fu¤Â‹ gu¥ò = {𝑦 𝑦 𝑦 𝑦 𝑦 } r.m 2 3 4 1 2 1 1 −4 −3 3 2 −4 } = { 3 −2 2 −2 −5 −2 1 = {(20 + 6 + 9 − 4) − (6 − 15 − 4 − 12)} 2 1 = {56} = 28 r. myFfŸ 2 39) Ãjhfu° nj‰w« gl«

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g¤jh« tF¥ò bghJ¤ nj®Î kh®¢ - 2017 éil¡ F¿¥ò ã%áf, mik¥ò ã%gz«.

= 24.59 (𝑜𝑟) 24.6

40) br§nfhz ∆𝐶𝐴𝐵 − š, tan 45° =

𝐵𝐶 𝐴𝐵

⇒ 𝑥 = ℎ → (1) br§nfhz ∆𝐷𝐴𝐵 − š, tan 60° = ⇒𝑥=

ℎ + 10

𝐵𝐷 𝐴𝐵

44) 𝑆 = {(1, 1), … , (6, 6)} 𝑛(𝑆) = 36 𝑛(𝐴) 18 𝑃 (𝐴 ) = = 𝑛(𝑆) 36 𝑛(𝐵) 5 𝑃 (𝐵 ) = = 𝑛(𝑆) 36 𝑛(𝐴 ∩ 𝐵) 3 𝑃 (𝐴 ∩ 𝐵 ) = = 𝑛(𝑆) 36 𝑃 ( 𝐴 ∪ 𝐵 ) = 𝑃 (𝐴 ) + 𝑃 ( 𝐵 ) − 𝑃 ( 𝐴 ∩ 𝐵 ) 18 5 3 = + − 36 36 36 20 5 = = . 36 9 45) m) 𝑓 (𝑥 ) = 3𝑥 4 + 6𝑥 3 − 12𝑥 2 − 24𝑥 = 3𝑥 (𝑥 3 + 2𝑥 2 − 4𝑥 − 8) 𝑔(𝑥 ) = 4𝑥 4 + 14𝑥 3 + 8𝑥 2 − 8𝑥 = 2𝑥 (2𝑥 3 + 7𝑥 2 + 4𝑥 − 4) 2

→ (2) √3 (1) k‰W« (2) − èUªJ, ℎ + 10 ℎ= √3 √3 ℎ − ℎ = 10 ⇒ ℎ = 13.66 Û. 41) 2𝜋𝑅 = 44 𝑐𝑚 ⇒ 𝑅 = 7𝑐𝑚, 2𝜋𝑟 = 8.4𝜋 𝑐𝑚 ⇒ 𝑟 = 4.2 𝑐𝑚, ℎ = 14 𝑐𝑚 1 𝑉 = 𝜋ℎ(𝑅 2 + 𝑅𝑟 + 𝑟 2 ) 𝑐𝑢𝑏𝑖𝑐 𝑢𝑛𝑖𝑡𝑠. 3 1 22 = × × 14(72 + 7 × 4.2 + 4.22 ) 3 7 = 1408.57 𝑐𝑚3 𝑥 3 + 2𝑥 2 − 4𝑥 − 8 2𝑥 3 + 7𝑥 2 + 4𝑥 − 4 42) br›tf«: 𝑙 = 44 𝑐𝑚, 𝑏 = 21 𝑐𝑚, ℎ1 = 12 𝑐𝑚 2𝑥 3 + 4𝑥 2 − 8𝑥 − 16 T«ò: ℎ2 = 24 𝑐𝑚 fd¢ br›tf tot¡ f£oæ‹ fdsÎ = ©k¡ 3𝑥 2 + 12𝑥 + 12 T«Ã‹ fdsÎ 3(𝑥 2 + 4𝑥 + 4) 1 𝑙 × 𝑏 × ℎ1 = 𝜋𝑟 2 ℎ2 Û ≠ 0 3 1 22 44 × 21 × 12 = × × 𝑟 2 × 24 3 7 𝑥−2 𝑟 2 = 441 𝑟 = ±21 𝑥 2 + 4𝑥 + 4 𝑥 3 + 2𝑥 2 − 4𝑥 − 8 𝑟 = 21 𝑐𝑚 𝑥 3 + 4𝑥 2 + 4𝑥 𝑑 = 2𝑟 ⇒ 𝑟 = 42 𝑐𝑚 − 2𝑥 2 − 8𝑥 − 8 43) 𝑥̅ = 18, 𝜎=√

=√

− 2𝑥 2 − 8𝑥 − 8

∑𝑑 2 𝑛 98 = √19.6 5

≃ 4.427 𝜎 𝐶. 𝑉 = × 100 𝑥̅ 4.427 = × 100 18

ÛÂ = 0 Û.bgh.t (𝑓(𝑥), 𝑔(𝑥 )) = 𝑥(𝑥 2 + 4𝑥 + 4). (mšyJ) M) ne®¡nfhL 𝑥 − m¢ir bt£L« òŸë: 𝐴(𝑥1 , 𝑦1 ) = 𝐴(𝑎, 0) ne®¡nfhL 𝑦 − m¢ir bt£L« òŸë: 𝐵(𝑥2 , 𝑦2 ) = 𝐵(0, 𝑏)

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g¤jh« tF¥ò bghJ¤ nj®Î kh®¢ - 2017 éil¡ F¿¥ò {(−3, 9), (−2, 0), (−1, −5), (0, −6), (1, −3), (2, 4), (3,15)}

𝐴𝐵 − æ‹ eL¥òŸë: 𝑀 (𝑥, 𝑦) = 𝑀(3, 2)

òŸëfis¡ F¿¤jš gutisa« tiujš ԮΠfz« = {−2, 1.5}

𝑥1 + 𝑥2 𝑦1 + 𝑦2 ) , 2 2 𝑎 𝑏 (3, 2) = ( , ) 2 2 ⇒ 𝑎 = 6, 𝑏 = 4 𝐴𝐵 − æ‹ rk‹ghL: 𝑥 𝑦 + =1 𝑎 𝑏 𝑥 𝑦 + =1 6 4 ⇒ 2𝑥 + 3𝑦 − 12 = 0 𝑀 (𝑥, 𝑦) = 𝑀 (

ÃçÎ - 𝐈𝐕 46) m) cjé¥gl« c©ik¥gl«: Kjš t£l«: 𝑟 = 6 𝑐𝑚 òŸë: 𝑂𝑃 = 10 𝑐𝑚 ika¡F¤J¡ nfhL Ïu©lhtJ t£l« bjhL nfhLfŸ bjhL nfh£o‹ Ús«= 8 𝑐𝑚

(2) (2) (1) (1) (1) (2) (1)

(mšyJ)

M) cjé¥gl«

(2)

(mšyJ) M) 𝑥, 𝑦 − m¢R tiujš (1) msΤ £l« (1) m£ltiz: (2) x 1 2 4 5 10 20 y 20 10 5 4 2 1 òŸëfŸ: (1) {(1, 20), (2, 10), (4, 5), (5, 4), (10, 2), (20, 1)} òŸëfis¡ F¿¤jš (2) gutisa« tiujš (2) Ô®Î: (1) 𝑥 = 5 våš 𝑦 = 4 𝑦 = 10 våš 𝑥 = 2.

PREPARED BY M. GOKULKUMAR, M. Sc., B. Ed., P.G.D.C.A B.T. Assistant in Mathematics, Sunstars High School, Vadakaraiattur, Namakkal (Dt) – 637 213 Phone: 8675635687 E-mail: [email protected]

c©ik¥gl«: nfh£LJ©L 𝐴𝐵 = 6 𝑐𝑚

(1)

K¡nfhz« 𝐴𝐵𝐶

(2)

R‰W t£l«

(2)

∠𝐴𝐶𝐷 = 30°

(1)

eh‰fu« 𝐴𝐵𝐶𝐷.

(2)

47) m) 𝑥, 𝑦 − m¢R tiujš msΤ £l« m£ltiz: x - 3 -2 -1 0 y 9 0 -5 -6 òŸëfŸ:

1 -3

(2) (2) (1)

(1) (1) (2) 2 3 4 15 (1)

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