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T,Gi
                                                       Reg.No.


                                  L¦Rm /             MATHS
                     (Rªr Utßm Be¡X Y¯ / Tamil and English versions)
                 1
Time Allowed : 2 /2 Hours]                                       [Maximum Marks : 100

úSWm : 21/2 U¦]                                                      [ùUôjR U§lùTiLs: 100

A±ÜûW                : Aû]jÕ ®]ôdLÞm N¬VôL T§Yô¡ Es[Rô GuTRû] N¬TôojÕd
                       ùLôs[Üm, AfÑlT§®p Ïû«Úl©u, Aû LiLô¦lTô[¬Pm
                       EP]¥VôLj ùR¬®dLÜm,
                      Cq®]ôjRôs SôuÏ ©¬ÜLû[ ùLôiPÕ,,
INSTRUCTION          : Check the question paper for fairness of printing. If there is any lack of
                       fairness, inform the Hall Supervisor immediately.
                      This question paper contains Four sections.

                                     ©¬Ü - I / SECTION − I

                               (U§lùTiLs : 15) / (MARKS : 15)

        ϱl× (i)         Cl©¬®p Es[ 15 ®]ôdLÞdÏm ®ûPV°dLÜm,
             (ii)        ùLôÓdLlThÓs[ SôuÏ ®ûPL°p ªLÜm N¬Vô] ®ûPûVj
                         úRokùRÓjÕ GÝRÜm,
        Note :    (i) Answer all the 15 questions.
                  (ii) Choose the correct answer in each question. Each of these questions
                        contains four options with just one correct option.     15 × 1 = 15
  (1) Let A = {1, 3, 4, 7, 11} Utßm B = {− 1, 1, 2, 5, 7, 9}GuL,
      f = {(1, −1, (3, 2), (4, 1), (7, 5), (11, 9)} GuYôß AûUkR Nôo× f : A → B GuTÕ
      (A) JußdÏ Juô] Nôo×                           (B) úUp Nôo×
      (C) CÚ×f Nôo×                                 (D) Nôo× ApX
      Let A = {1, 3, 4, 7, 11}, B = {− 1, 1, 2, 5, 7, 9} and f : A → B be given by
      f = {(1, −1, (3, 2), (4, 1), (7, 5), (11, 9)}. Then f is
      (a) one-one             (b) onto               (c) bijective       (d) not a function
      2     6 18 54
  (2) 5 , 25 , 125 , 625 ..... Gu ùTÚdÏj ùRôPo Y¬ûN«u ùTôÕ ®¡Rm
            2                                 3                4
      (A)                 (B) 5          (C)               (D)
            5                                 5                5
                                   2 6 18      54
      The common ratio of the G.P. ,   ,    ,     ..... is
                                   5 25 125 625
          2                                  3                 4
      (a) 5               (b) 5          (c) 5             (d) 5

                                                                           [§Úl×L / TURN over
a4  3
(3) a1 , a2, a3, .... GuT] JÚ áhÓj ùRôPo Y¬ûN«Ûs[], úUÛm a = 2 G²p, 13YÕ
                                                          7
    Eßl×
       3
   (A) 2                 (B) 0                 (C) 12a1                (D) 14a2

                                           a4  3
   If a1 , a2, a3, .... are A.P. such that a = 2 , then the 13th term of the A.P. is
                                            7

       3
   (a) 2                 (b) 0                 (c) 12a1                (d) 14a2


(4) 6x2y, 9x2yz, 12x2y2z B¡VYt±u Á,ùTô,U,

   (A) 36x2y2z           (B) 48xy2z2           (C) 96x2y2z2            (D) 72xy2z

   The LCM of 6x2y, 9x2yz, 12x2y2z is

   (a) 36x2y2z           (b) 48xy2z2           (c) 96x2y2z2            (d) 72xy2z


(5) b = a + c G²p ax2 + bxc + c = 0 Gu NUuTôh¥tÏ
   (A) ùUnùVi êXeLs EiÓ                        (B) êXeLs CpûX
   (C) NU êXeLs EiÓ                            (D) êXeLs ùUnùViLs ApX

   If b = a + c, then the equation ax2 + bxc + c = 0 has
   (a) real roots        (b) no roots          (c) equal roots         (d) no real roots


        1 1
(6) A ×     = (1 2) G²p, A-u Y¬ûN
        0 2
   (A) 2 × 1             (B) 2 × 2             (C) 1 × 2               (D) 3 × 2

          1 1
   If A ×     = (1 2) then the order of A is
          0 2
   (a) 2 × 1             (b) 2 × 2             (c) 1 × 2               (d) 3 × 2


(7) 7y − 2x = 11 Gu úSodúLôh¥u NônÜ
         7                   7                     2                         2
   (A) − 2               (B) 2                 (C) 7                   (D) − 7

   The slope of the straight line 7y − 2x = 11 is equal to
           7                   7                     2                         2
   (a) −                 (b)                   (c)                     (d) −
           2                   2                     7                         7



                                                     2
(8) (0, 0), (1, 0), (0, 1) Gu ×s°Lû[ Øû]L[ôLd ùLôiP ØdúLôQj§u Ñt[Ü
     (A) 2                 (B) 2                 (C) 2 + 2                (D) 2 − 2
     The perimeter of a triangle formed by the points (0, 0), (1, 0), (0, 1) is
     (a) 2                 (b) 2                 (c) 2 + 2                (d) 2 − 2


 (9) ∆ PQR-p RS GuTÕ ∠R-u Eh× CÚ NUùYh¥, PQ = 6 ùN,Á, QR = 8 ùN,Á, RP = 4
     ùN,Á G²p PS =
     (A) 2 ùN,Á            (B) 4 ùN,Á            (C) 3 ùN,Á               (D) 6 ùN,Á
     In ∆ PQR, RS is the bisector of ∠R. If PQ = 6cm, QR = 8cm, RP = 4 cm then PS is equal to
     (a) 2 cm              (b) 4 cm              (c) 3 cm                 (d) 6 cm


(10) AB Utßm CD Gu CÚ SôiLs JÚ YhPj§u Eh×UôL P Gu ×s°«p
     ùYh¥d ùLôs¡u], AB = 7, AP = 4, CP = 2 G²p, CD =
     (A) 4                 (B) 8                 (C) 6                    (D) 10
     Chords AB and CD cut at P inside the circle. If AB = 7, AP = 4, CP = 2, then CD =
     (a) 4                 (b) 8                 (c) 6                    (d) 10


(11) JÚ úLô×Wj§−ÚkÕ 28.5 Á çWj§p ¨uß ùLôi¥ÚdÏm JÚYo úLô×Wj§u
     Ef£ûV 45° Htd úLôQj§p Lôi¡ôo, AYÚûPV ¡ûP¨ûXl TôoûYd úLôÓ
     RûW«−ÚkÕ 1.5Á EVWj§p Es[Õ G²p, úLô×Wj§u EVWm
     (A) 30 Á              (B) 27.5 Á            (C) 28.5 Á               (D) 27 Á
     A man is 28.5 m away from a tower. His eye level above the ground is 1.5m. The angle of
     elevation of the tower from his eyes is 45°. Then the height of the tower is
     (a) 30 m              (b) 27.5m             (c) 28.5 m               (d) 27 m
           1
(12)               =
     tan θ + cot θ
     (A) sin θ + cos θ     (B) sin θ cos θ       (C) sin θ − cos θ        (D) cosec θ + cot θ
           1
                   =
     tan θ + cot θ
     (a) sin θ + cos θ     (b) sin θ cos θ       (c) sin θ − cos θ        (d) cosec θ + cot θ
                 2
(13) 12π ùN,Á ùUôjR TWl× ùLôiP §iU AûWdúLô[j§u Yû[TWl×
                     2                       2                        2                  2
     (A) 6π ùN,Á           (B) 24π ùN,Á          (C) 26π ùN,Á             (D) 8π ùN,Á
                                                                  2
     If the total surface area of a solid hemisphere is 12π cm then its curved surface area is equal to
                2                      2                      2                      2
     (a) 6π cm             (b) 24π cm            (c) 26π cm               (d) 8π cm
(14) £X ®YWeL°u áhÓfNWôN¬ Utßm §hP ®XdLm ØûúV 48, 12 G²p,
     UôßTôhÓdùLÝ
     (A) 42                (B) 25q               (C) 28                   (D) 48
     Mean and standard deviation of a date are 48 and 12 respectively. The coefficient of variation is
     (a) 42                (b) 25q               (c) 28                   (d) 48
                                                    3
(15) A Utßm B GuT] CWiÓ JuûùVôuß ®XdÏm ¨Lrf£Ls GuL, Ak¨Lrf£«u
                     1
     áßùY° S, P(A) = 3 P(B) Utßm S = A ∪ B G²p P(A) =

         1                     1                    3                      3
     (A) 4                 (B) 2                (C) 4                  (D) 8

                                                                                                1
     If A and B are mutually exclusive events and S is the sample space such that P(A) =          P(B) and S
                                                                                                3
     = A ∪ B, then P(A) =
         1                     1                    3                      3
     (a) 4                 (b) 2                (c) 4                  (d) 8


                                     ©¬Ü - II / SECTION − II

       ϱl× (i)         TjÕ ®]ôdLÞdÏ ®ûPV°dLÜm,
                 (ii)   ®]ô  Gi  30dÏ  Li¥lTôL    ®ûPV°dLÜm,     ØRp                                     14
                        ®]ôdL°−ÚkÕ HúRàm 9 ®]ôdLû[j úRoÜ ùNnVÜm,
                 (iii) JqùYôÚ ®]ô®tÏm CWiÓ U§lùTiLs,                                             10 × 2 = 20
       Note :    (i)    Answer 10 questions.
                 (ii) Question No. 30 is Compulsory. Select any 9 questions from the first 14
                      questions.
                 (iii) Each question carries Two marks                                           10 × 2 = 20


(16) A = {4, 6, 7, 8, 9}, B = {2, 4, 6} Utßm C = {1, 2, 3, 4, 5, 6} G²p A ∪ (B ∩ C) LôiL,
     If A = {4, 6, 7, 8, 9}, B = {2, 4, 6} and C = {1, 2, 3, 4, 5, 6}, then find A ∪ (B ∩ C).


(17) X = {1, 2, 3, 4} GuL. g = {(3, 1), (4, 2), (2, 1)} Gu EÜ X-−ÚkÕ X-dÏ JÚ NôoTôÏUô
     G] BWônL, Eu ®ûPdÏ Ht ®[dLm RÚL,
     Let X = {1, 2, 3, 4}. Examine whether the relation g = {(3, 1), (4, 2), (2, 1)} is a function from X
     to X or not. Explain.


(18) êuß GiL°u ®¡Rm 2 : 5 : 7 GuL, ØRXôm Gi. CWiPôm Gi¦−ÚkÕ 7-Id
     L¯jÕl ùTlTÓm Gi Utßm êuôm Gi B¡V] JÚ áhÓj ùRôPoY¬ûNûV
     HtTÓj§]ôp, AqùYiLû[d LôiL,
     Three numbers are in the ratio 2 : 5 : 7. If 7 is subtracted from the second, the resulting numbers
     form an arithmetic sequence. Determine the numbers.

       2
(19) 2x − 3x − 1 = 0 Gu NUuTôh¥u êXeLs α Utßm β , G²p α − β -u U§lûTd
     LôiL,
                                                2
     If α and β are the roots of the equation 2x − 3x − 1 = 0, find the value of α − β if a > β.
                                                    4
 2 3    1 5 
(20) A =         −       G²p A-u áhPp úSoUôß A¦ûVd LôiL.
           − 9 5   7 − 1
              2 3    1 5 
       If A =      −       , then find the additive iverse of A.
             − 9 5   7 − 1


     2 9 − 3        4 2
(21)         
     4 − 1 0 
                     − 6 7 Gu A¦L°u ùTÚdLûXd LôiL, (ùTÚdL Ø¥ÙUô]ôp)
                      −2 1
                                                   2 9 − 3       
                                                                 4 2
       Find the product of the matrices, if exists          − 6 7
                                                   4 − 1 0 
                                                               −2 1

(22) JÚ YhPj§u ûUVm (− 6, 4). AqYhPj§u JÚ ®hPj§u JÚ Øû]. B§l×s°
     G²p. UtùôÚ Øû]ûVd LôiL,


       The centre of a circle is at (− 6, 4). If one end of a diameter of the circle is at the origin, then find
       the other end.


                          AD 2
(23) ∆ABC-p DE || BC Utßm DB = 3 . AE = 3.7 ùN,Á G²p. EC-I LôiL,

                             AD 2
       In ∆ABC, DE || BC and DB = 3 . If AE = 3.7 cm, find EC.


(24) ÑY¬p NônjÕ ûYdLlThP JÚ H¦Vô]Õ RûWÙPu 60° úLôQjûR
     HtTÓjÕ¡Õ, H¦«u A¥ ÑYt±−ÚkÕ 3.5 Á çWj§p Es[Õ G²p. H¦«u
     ¿[jûRd LôiL,
       A ladder leaning against a vertical wall, makes an angle of 60° with the ground. The foot of the
       ladder 3.5m away from the wall. Find the length of the ladder.


        sin θ    cos θ
(25)           +       = 1 Gu ØtùôÚûUûV ¨ßÜL,
       cosec θ   sec θ
                             sin θ    cos θ
       Prove the identity           +       =1
                            cosec θ   sec θ


(26) JÚ §iU úSo YhP EÚû[«u BWm 14 ùN,Á Utßm EVWm 8 ùN,Á G²p, ARu
     Yû[TWl× LôiL,
       A right circular cylinder has radius of 14cm and height of 8cm. Find its curved surface area.



                                                      5
(27) UWj§]Xô] JÚ §iUd ám©u A¥fÑt[Ü 44 Á Utßm ARu EVWm 12 Á G²p
     Aj§iUd ám©u L] A[ûYd LôiL,
    The circumference of the base of a 12m high wooden solid come is 44m. Find the volume.


(28) ØRp 13 CVp GiL°u §hP ®XdLjûRd LQd¡ÓL,
    Calculate the standard deviation of the first 13 natural numbers.


(29) CÚ SôQVeLû[ JúW NUVj§p                            ÑiÓmúTôÕ.        A§LThNUôL         JÚ   RûX
     ¡ûPlTRtLô] ¨LrRL®û]d LôiL,
    Two coins are tossed together. What is the probability of getting at most one head.

                                 2
                              6x − 54
(30) (A) ÑÚdÏL :            2          .
                           x + 7x + 12
                                                 [ApXÕ]
    (B) 2y = 4x + 3 Utßm x + 2y = 10 Gu úSodúLôÓLs JußdùLôuß ùNeÏjÕ G]d
        LôhÓL,
                             2
                        6x − 54
    (a) Simplify :     2          .
                      x + 7x + 12
                                                  [OR]
    (b) Show that the lines 2y = 4x + 3 and x + 2y = 10 are perpendicular.


                                       ©¬Ü - III / SECTION − III

      ϱl× (i)              9 ®]ôdLÞdÏ ®ûPV°dLÜm,
                (ii)        ®]ô  Gi  45dÏ  Li¥lTôL    ®ûPV°dLÜm,     ØRp                          14
                            ®]ôdL°−ÚkÕ HúRàm 8 ®]ôdLû[j úRoÜ ùNnVÜm,
                (iii) JqùYôÚ ®]ô®tÏm IkÕ U§lùTiLs,                                         9 × 5 = 45


      Note :    (i)        Answer 9 questions.
                (ii) Question No. 45 is Compulsory. Select any 8 questions from the first 14
                     questions.
                (iii) Each question carries Five marks                                     9 × 5 = 45

(31) ùYuTPeLû[l TVuTÓj§ A  (B ∩ C) = (A  B) ∪ (A  C) Guàm ¥ UôoL²u LQ
     ®j§VôN ®§«û]f N¬TôodLÜm,
    Use Venn diagrams to verify De Morgan’s law for set difference A  (B ∩ C) = (A  B) ∪ (A  C).


                                                   6
(32) Nôo× f : [− 7, 6 ⇒ R ¸rdLiPYôß YûWVßdLlThÓs[Õ,
                 2
            x + 2x + 1
                                         −7≤x<−5
     f(x) = x + 5                        − 5 ≤ x ≤ 2 . ©uYÚY]Ytûd LôiL,
            x − 1
                                         2<x<6

                                                                            4 f(− 3) + 2 f(4)
     (i) 2 f(− 4) + 3 f(2)               (ii) f(− 7) − f(− 3)       (iii)
                                                                             f(− 6) − 3 f(1)
                                                                                           2
                                                             x + 2x + 1
                                                                                                           −7≤x<−5
     A function f : [− 7, 6 ⇒ R is defined as follows f(x) = x + 5                                         −5≤x≤2 .
                                                             
                                                             x − 1                                         2<x<6

                                                                                        4 f(− 3) + 2 f(4)
     Find (i) 2 f(− 4) + 3 f(2)                  (ii) f(− 7) − f(− 3)       (iii)
                                                                                         f(− 6) − 3 f(1)


      2      2       2       2
(33) 1 − 2 + 3 − 4 + ..... Gu ùRôP¬u ØRp 2n Eßl×L°u áÓRp LôiL,
                                                                        2           2      2    2
     Find the sum of the first 2n terms of the series 1 − 2 + 3 − 4 + .....

                                     3           2
(34) LôW¦lTÓjÕL : x − 5x − 2x + 24
                                             3        2
     Factorize the polynomial x − 5x − 2x + 24

                         2       3          4
(35) m − nx + 28x + 12x + 9x B]Õ JÚ ØÝ YodLm G²p. m, n B¡VYt±u U§l×Lû[d
     LôiL.
                             2       3           4
     If m − nx + 28x + 12x + 9x is a perfect square, then find the values of m and n.

(36) AûNYt ¿¬p JÚ CVk§WlTP¡u úYLm U¦dÏ 15 ¡,Á GuL. AlTPÏ
     ¿úWôhPj§u §ûN«p 30 ¡,Á çWm ùNuß. ©Ï G§oj §ûN«p §Úm© 4 U¦ 30
     ¨ªPeL°p ÁiÓm ×lThP CPj§tÏ §Úm© YkRôp ¿¬u úYLjûRd LôiL,
     The speed of a boat in still water is 15 km / hr. It goes 30 km upstream and return downstream to
     the original point in 4 hrs 30 minutes. Find the speed of the stream.


          5 2              2 − 1              T   T T
(37) A =      Utßm B =           G²p (AB) = B A GuTûR N¬TôodLÜm,
          7 3             − 1 1 
            5 2          2 − 1                  T   T T
     If A =     and B =         verify that (AB) = B A .
            7 3         − 1 1 

(38) (− 4, − 2), (− 3, − 5), (3, − 2) Utßm (2, 3) B¡V ×s°Lû[ Øû]L[ôLd ùLôiP
     SôtLWj§u TWlûTd LôiL,
     Find the area of the quadrilateral formed by the points (− 4, − 2), (− 3, − 5), (3, − 2) and (2, 3).



                                                                    7
(39) ∆ABC-u Øû]Ls A(2, 1), B(6, − 1), C(4, 11) GuL,                          A-«−ÚkÕ YûWVlTÓm
     ÏjÕdúLôh¥u NUuTôhûPd LôiL.
    The vertices of ∆ABC are A(2, 1), B(6, − 1) and C(4, 11). Find the equation of the straight line
    along the altitude from the vertex A.


(40) JÚ £ßYu ûYWj§u ÏßdÏ ùYhÓj úRôt Y¥®p. TPj§p Lôh¥VYôß JÚ
     ThPm ùNnRôu, CeÏ AE = 16 ùN,Á, EC = 81 ùN,Á. AYu BD Gu ÏßdÏd
     Ïf£«û]l TVuTÓjR ®Úmסôu. AdÏf£«u ¿[m GqYôß CÚdL úYiÓm?
    A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm.
    He wants to use a straight cross bar BD. How long should it be?
(41) úSodÏjRô] JÚ UWj§u úUpTôLm Lôt±]ôp رkÕ, AmرkR Tϧ ¸úZ
     ®ÝkÕ ®PôUp. UWj§u Ef£ RûWÙPu 30° úLôQjûR HtTÓjÕ¡Õ, UWj§u
     Ef£ ARu A¥«−ÚkÕ 30 Á ùRôûX®p RûWûVj ùRôÓ¡Õ G²p. UWj§u ØÝ
     EVWjûRd LôiL,
    A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle
    30° with it. If the top of the free touches the ground 30m away from its foot, then find the actual
    height of the tree.
(42) L°UiûQl TVuTÓj§ JÚ UôQYu 48 ùN,Á EVWØm 12 ùN,Á BWØm ùLôiP
     úSo YhP §iUd ámûTf ùNnRôo, AdámûT UtùôÚ UôQYo JÚ §iUd
     úLô[UôL Uôt±]ôo, AqYôß UôtlThP קV úLô[j§u BWjûRd LôiL,
    Using clay, a student made a right circular cone of height 48cm and base radius 12cm. Another
    student reshapes it in the form of a sphere. Find the radius of the sphere.
(43) ¸rdLiP AhPYûQ«p ùLôÓdLlThÓs[ ×s° ®YWj§u §hP ®XdLjûRd
     LQd¡ÓL,

                                x       3      8        13    18      23

                                f       7      10       15    10       8

    Calculate the standard deviation of the following data.

                                x       3      8        13    18      23

                                f       7      10       15    10       8


(44) JÚ ×§V U¡rÜkÕ (car) ARàûPV Y¥YûUl©tLôL ®ÚÕ ùTßm ¨LrRLÜ 0.25
     GuL, £kR Øû«p G¬ùTôÚs TVuTôh¥tLô] ®ÚÕ ùTßm ¨LrRLÜ 0.35
     Utßm CÚ ®ÚÕLÞm ùTßYRtLô] ¨LrRLÜ 0.15 G²p. AmU¡rÜkÕ
    (i) ÏûkRÕ HRôYÕ JÚ ®ÚÕ ùTßRp
    (ii) JúW JÚ ®ÚÕ UhÓm ùTßRp B¡V ¨Lrf£LÞdLô] ¨LrRLÜLû[d LôiL,
    The probability that a new car will get an award for its design is 0.25., the probability that it will
    get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is
    0.15. Find the probability that
    (i) it will get atleast one of the two awards (ii) it will get only one of the awards.

                                                    8
(45) (A) JÚ áhÓØj ùRôPo Y¬ûN«u AÓjRÓjR êuß Eßl×L°u áÓRp − 6
         Utßm AYt±u ùTÚdLtTXu 90, Amêuß GiLû[d LôiL,
                                            [ApXÕ]
   (B) 14 ùN,Á ®hPØm. 20 ùN,Á EVWØm EûPV JÚ EÚû[ Y¥Y ØLûY«p
       Tô§V[Ü Ri½o Es[Õ, A§p JúW A[®Xô] 300 DV ÏiÓLû[ êrLf
       ùNnÙm úTôÕ ¿o UhPm 2.8 ùN,Á EVo¡Õ, DVd ÏiÓ JqùYôu±u ®hPm
       LôiL,
   (a) The sum of three consecutive term in an A.P. is − 6 and their product is 90. Find the three
       numbers.
                                              [OR]
   (b) A cylindrical jar of diameter 14cm and depth 20cm is half-full of water. 300 leadshots of same
       size are dropped into the jar and the level of water raises by 2.8 cm. Find the diameter of each
       leadshots.


                                  ©¬Ü - IV / SECTION − IV
     ϱl× (i)         Cl©¬®p Es[ JqùYôÚ ®]ô®Ûm CWiÓ Uôtß ®]ôdLs
                      ùLôÓdLlThÓs[],
               (ii)   JqùYôÚ ®]ô®Ûm Es[ CWiÓ Uôtß ®]ôdL°−ÚkÕ JÚ
                      ®]ôûY úRokùRÓjÕ CÚ ®]ôdLÞdÏm ®ûPV°dLÜm,

               (iii) JqùYôÚ ®]ô®tÏm TjÕ U§lùTiLs,                                          2 × 10 = 20

     Note :    (i)    This section contains Two questions, each with two alternatives.

               (ii) Answer both the questions choosing either of the alternatives.

               (iii) Each question carries Ten marks                                       2 × 10 = 20


(46) (A) 6 ùN,Á BWØs[ JÚ YhPm YûWkÕ ARu ûUVj§−ÚkÕ 10 ùN,Á
         ùRôûX®Ûs[ JÚ ×s°ûVd ϱdL, Al×s°«−ÚkÕ YhPj§tÏ
         ùRôÓúLôÓLs YûWkÕ ARu ¿[eLû[ LQd¡ÓL,
                                            (ApXÕ)
   (B) BC = 5 ùN,Á ∠BAC = 40° Utßm Ef£ A-−ÚkÕ BC-dÏ YûWVlThP
       SÓdúLôh¥u ¿[m 6 ùN,Á Gu A[ÜLs ùLôiP ∆ABC YûWL, A-−ÚkÕ
       YûWVlThP ÏjÕdúLôh¥u ¿[m LôiL,
   (a)   Draw the two tangents from a point which is 10cm away from the centre of a circle of radius
         6cm. Also, measure the lengths of the tangents.
                                              (OR)
   (b)   Construct ∆ABC in which the base BC = 5cm, ∠BAC = 40° and the median from A to BC
         is 6cm. Also measure the length of the altitude from A.


                                                9
(47) (A) y = x2 − x − 8-u YûWTPm YûWkÕ. ARû]l TVuTÓj§ x2 − 2x − 15 = 0 Gu
          NUuTôhûPj ¾odLÜm,
                                                   (ApXÕ)
     (B) JÚ ª§Yi¥ KhÓTYo A Gu CPj§−ÚkÕ B Gu CPj§tÏ JÚ ºWô]
         úYLj§p JúW Y¯«p ùYqúYß SôhL°p TVQm ùNn¡ôo, AYo TVQm
         ùNnR úYLm. AjçWj§û]d LPdL GÓjÕd ùLôiP úSWm B¡VY]Ytûl
         Tt±V ®YWeLs (úYL – LôX) ©uYÚm AhPYûQ«p ùLôÓdLlThÓs[],
             úYLm (¡,Á,/U¦ x               2         4        6        10    12
             úSWm (U¦«p) y                60        30        20       12    10
     úYL – LôX YûWTPm YûWkÕ A§−ÚkÕ
     (i)    AYo U¦dÏ 5 ¡.Á úYLj§p ùNuôp çWjûRd LPdL BÏm TVQ úSWm
     (ii)   AYo Cdϱl©hP çWjûR 40 U¦ úSWj§p LPdL GkR úYLj§p TV¦dL
            úYiÓm
B¡VY]Ytûd LôiL,
     (a)    Draw the graph of y = x2 − x − 8 and hence find the roots of x2 − 2x − 15 = 0.
                                                    (OR)
     (b)    A cyclist travels from a place A to a place B along the same route at a uniform speed on
            different days. The following table gives the speed of his travel and the corresponding time
            he took to cover the distance.
             Speed in km/hr x         2        4         6        10    12
             Time in hrs             60        30        20       12    10
     Draw the speed-time graph and use it to find.
     (i)    The number of hours he will take if he travels at a speed of 5 km/hr.
     (ii)   The speed with which he should travel if he has to cover the distance in 40 hrs.




                                                     10

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10th Maths model3 question paper

  • 1. T,Gi Reg.No. L¦Rm / MATHS (Rªr Utßm Be¡X Y¯ / Tamil and English versions) 1 Time Allowed : 2 /2 Hours] [Maximum Marks : 100 úSWm : 21/2 U¦] [ùUôjR U§lùTiLs: 100 A±ÜûW : Aû]jÕ ®]ôdLÞm N¬VôL T§Yô¡ Es[Rô GuTRû] N¬TôojÕd ùLôs[Üm, AfÑlT§®p Ïû«Úl©u, Aû LiLô¦lTô[¬Pm EP]¥VôLj ùR¬®dLÜm, Cq®]ôjRôs SôuÏ ©¬ÜLû[ ùLôiPÕ,, INSTRUCTION : Check the question paper for fairness of printing. If there is any lack of fairness, inform the Hall Supervisor immediately. This question paper contains Four sections. ©¬Ü - I / SECTION − I (U§lùTiLs : 15) / (MARKS : 15) ϱl× (i) Cl©¬®p Es[ 15 ®]ôdLÞdÏm ®ûPV°dLÜm, (ii) ùLôÓdLlThÓs[ SôuÏ ®ûPL°p ªLÜm N¬Vô] ®ûPûVj úRokùRÓjÕ GÝRÜm, Note : (i) Answer all the 15 questions. (ii) Choose the correct answer in each question. Each of these questions contains four options with just one correct option. 15 × 1 = 15 (1) Let A = {1, 3, 4, 7, 11} Utßm B = {− 1, 1, 2, 5, 7, 9}GuL, f = {(1, −1, (3, 2), (4, 1), (7, 5), (11, 9)} GuYôß AûUkR Nôo× f : A → B GuTÕ (A) JußdÏ Juô] Nôo× (B) úUp Nôo× (C) CÚ×f Nôo× (D) Nôo× ApX Let A = {1, 3, 4, 7, 11}, B = {− 1, 1, 2, 5, 7, 9} and f : A → B be given by f = {(1, −1, (3, 2), (4, 1), (7, 5), (11, 9)}. Then f is (a) one-one (b) onto (c) bijective (d) not a function 2 6 18 54 (2) 5 , 25 , 125 , 625 ..... Gu ùTÚdÏj ùRôPo Y¬ûN«u ùTôÕ ®¡Rm 2 3 4 (A) (B) 5 (C) (D) 5 5 5 2 6 18 54 The common ratio of the G.P. , , , ..... is 5 25 125 625 2 3 4 (a) 5 (b) 5 (c) 5 (d) 5 [§Úl×L / TURN over
  • 2. a4 3 (3) a1 , a2, a3, .... GuT] JÚ áhÓj ùRôPo Y¬ûN«Ûs[], úUÛm a = 2 G²p, 13YÕ 7 Eßl× 3 (A) 2 (B) 0 (C) 12a1 (D) 14a2 a4 3 If a1 , a2, a3, .... are A.P. such that a = 2 , then the 13th term of the A.P. is 7 3 (a) 2 (b) 0 (c) 12a1 (d) 14a2 (4) 6x2y, 9x2yz, 12x2y2z B¡VYt±u Á,ùTô,U, (A) 36x2y2z (B) 48xy2z2 (C) 96x2y2z2 (D) 72xy2z The LCM of 6x2y, 9x2yz, 12x2y2z is (a) 36x2y2z (b) 48xy2z2 (c) 96x2y2z2 (d) 72xy2z (5) b = a + c G²p ax2 + bxc + c = 0 Gu NUuTôh¥tÏ (A) ùUnùVi êXeLs EiÓ (B) êXeLs CpûX (C) NU êXeLs EiÓ (D) êXeLs ùUnùViLs ApX If b = a + c, then the equation ax2 + bxc + c = 0 has (a) real roots (b) no roots (c) equal roots (d) no real roots 1 1 (6) A ×   = (1 2) G²p, A-u Y¬ûN 0 2 (A) 2 × 1 (B) 2 × 2 (C) 1 × 2 (D) 3 × 2 1 1 If A ×   = (1 2) then the order of A is 0 2 (a) 2 × 1 (b) 2 × 2 (c) 1 × 2 (d) 3 × 2 (7) 7y − 2x = 11 Gu úSodúLôh¥u NônÜ 7 7 2 2 (A) − 2 (B) 2 (C) 7 (D) − 7 The slope of the straight line 7y − 2x = 11 is equal to 7 7 2 2 (a) − (b) (c) (d) − 2 2 7 7 2
  • 3. (8) (0, 0), (1, 0), (0, 1) Gu ×s°Lû[ Øû]L[ôLd ùLôiP ØdúLôQj§u Ñt[Ü (A) 2 (B) 2 (C) 2 + 2 (D) 2 − 2 The perimeter of a triangle formed by the points (0, 0), (1, 0), (0, 1) is (a) 2 (b) 2 (c) 2 + 2 (d) 2 − 2 (9) ∆ PQR-p RS GuTÕ ∠R-u Eh× CÚ NUùYh¥, PQ = 6 ùN,Á, QR = 8 ùN,Á, RP = 4 ùN,Á G²p PS = (A) 2 ùN,Á (B) 4 ùN,Á (C) 3 ùN,Á (D) 6 ùN,Á In ∆ PQR, RS is the bisector of ∠R. If PQ = 6cm, QR = 8cm, RP = 4 cm then PS is equal to (a) 2 cm (b) 4 cm (c) 3 cm (d) 6 cm (10) AB Utßm CD Gu CÚ SôiLs JÚ YhPj§u Eh×UôL P Gu ×s°«p ùYh¥d ùLôs¡u], AB = 7, AP = 4, CP = 2 G²p, CD = (A) 4 (B) 8 (C) 6 (D) 10 Chords AB and CD cut at P inside the circle. If AB = 7, AP = 4, CP = 2, then CD = (a) 4 (b) 8 (c) 6 (d) 10 (11) JÚ úLô×Wj§−ÚkÕ 28.5 Á çWj§p ¨uß ùLôi¥ÚdÏm JÚYo úLô×Wj§u Ef£ûV 45° Htd úLôQj§p Lôi¡ôo, AYÚûPV ¡ûP¨ûXl TôoûYd úLôÓ RûW«−ÚkÕ 1.5Á EVWj§p Es[Õ G²p, úLô×Wj§u EVWm (A) 30 Á (B) 27.5 Á (C) 28.5 Á (D) 27 Á A man is 28.5 m away from a tower. His eye level above the ground is 1.5m. The angle of elevation of the tower from his eyes is 45°. Then the height of the tower is (a) 30 m (b) 27.5m (c) 28.5 m (d) 27 m 1 (12) = tan θ + cot θ (A) sin θ + cos θ (B) sin θ cos θ (C) sin θ − cos θ (D) cosec θ + cot θ 1 = tan θ + cot θ (a) sin θ + cos θ (b) sin θ cos θ (c) sin θ − cos θ (d) cosec θ + cot θ 2 (13) 12π ùN,Á ùUôjR TWl× ùLôiP §iU AûWdúLô[j§u Yû[TWl× 2 2 2 2 (A) 6π ùN,Á (B) 24π ùN,Á (C) 26π ùN,Á (D) 8π ùN,Á 2 If the total surface area of a solid hemisphere is 12π cm then its curved surface area is equal to 2 2 2 2 (a) 6π cm (b) 24π cm (c) 26π cm (d) 8π cm (14) £X ®YWeL°u áhÓfNWôN¬ Utßm §hP ®XdLm ØûúV 48, 12 G²p, UôßTôhÓdùLÝ (A) 42 (B) 25q (C) 28 (D) 48 Mean and standard deviation of a date are 48 and 12 respectively. The coefficient of variation is (a) 42 (b) 25q (c) 28 (d) 48 3
  • 4. (15) A Utßm B GuT] CWiÓ JuûùVôuß ®XdÏm ¨Lrf£Ls GuL, Ak¨Lrf£«u 1 áßùY° S, P(A) = 3 P(B) Utßm S = A ∪ B G²p P(A) = 1 1 3 3 (A) 4 (B) 2 (C) 4 (D) 8 1 If A and B are mutually exclusive events and S is the sample space such that P(A) = P(B) and S 3 = A ∪ B, then P(A) = 1 1 3 3 (a) 4 (b) 2 (c) 4 (d) 8 ©¬Ü - II / SECTION − II ϱl× (i) TjÕ ®]ôdLÞdÏ ®ûPV°dLÜm, (ii) ®]ô Gi 30dÏ Li¥lTôL ®ûPV°dLÜm, ØRp 14 ®]ôdL°−ÚkÕ HúRàm 9 ®]ôdLû[j úRoÜ ùNnVÜm, (iii) JqùYôÚ ®]ô®tÏm CWiÓ U§lùTiLs, 10 × 2 = 20 Note : (i) Answer 10 questions. (ii) Question No. 30 is Compulsory. Select any 9 questions from the first 14 questions. (iii) Each question carries Two marks 10 × 2 = 20 (16) A = {4, 6, 7, 8, 9}, B = {2, 4, 6} Utßm C = {1, 2, 3, 4, 5, 6} G²p A ∪ (B ∩ C) LôiL, If A = {4, 6, 7, 8, 9}, B = {2, 4, 6} and C = {1, 2, 3, 4, 5, 6}, then find A ∪ (B ∩ C). (17) X = {1, 2, 3, 4} GuL. g = {(3, 1), (4, 2), (2, 1)} Gu EÜ X-−ÚkÕ X-dÏ JÚ NôoTôÏUô G] BWônL, Eu ®ûPdÏ Ht ®[dLm RÚL, Let X = {1, 2, 3, 4}. Examine whether the relation g = {(3, 1), (4, 2), (2, 1)} is a function from X to X or not. Explain. (18) êuß GiL°u ®¡Rm 2 : 5 : 7 GuL, ØRXôm Gi. CWiPôm Gi¦−ÚkÕ 7-Id L¯jÕl ùTlTÓm Gi Utßm êuôm Gi B¡V] JÚ áhÓj ùRôPoY¬ûNûV HtTÓj§]ôp, AqùYiLû[d LôiL, Three numbers are in the ratio 2 : 5 : 7. If 7 is subtracted from the second, the resulting numbers form an arithmetic sequence. Determine the numbers. 2 (19) 2x − 3x − 1 = 0 Gu NUuTôh¥u êXeLs α Utßm β , G²p α − β -u U§lûTd LôiL, 2 If α and β are the roots of the equation 2x − 3x − 1 = 0, find the value of α − β if a > β. 4
  • 5.  2 3 1 5  (20) A =   −   G²p A-u áhPp úSoUôß A¦ûVd LôiL. − 9 5 7 − 1  2 3 1 5  If A =   −   , then find the additive iverse of A. − 9 5 7 − 1 2 9 − 3  4 2 (21)   4 − 1 0  − 6 7 Gu A¦L°u ùTÚdLûXd LôiL, (ùTÚdL Ø¥ÙUô]ôp)  −2 1 2 9 − 3   4 2 Find the product of the matrices, if exists   − 6 7 4 − 1 0   −2 1 (22) JÚ YhPj§u ûUVm (− 6, 4). AqYhPj§u JÚ ®hPj§u JÚ Øû]. B§l×s° G²p. UtùôÚ Øû]ûVd LôiL, The centre of a circle is at (− 6, 4). If one end of a diameter of the circle is at the origin, then find the other end. AD 2 (23) ∆ABC-p DE || BC Utßm DB = 3 . AE = 3.7 ùN,Á G²p. EC-I LôiL, AD 2 In ∆ABC, DE || BC and DB = 3 . If AE = 3.7 cm, find EC. (24) ÑY¬p NônjÕ ûYdLlThP JÚ H¦Vô]Õ RûWÙPu 60° úLôQjûR HtTÓjÕ¡Õ, H¦«u A¥ ÑYt±−ÚkÕ 3.5 Á çWj§p Es[Õ G²p. H¦«u ¿[jûRd LôiL, A ladder leaning against a vertical wall, makes an angle of 60° with the ground. The foot of the ladder 3.5m away from the wall. Find the length of the ladder. sin θ cos θ (25) + = 1 Gu ØtùôÚûUûV ¨ßÜL, cosec θ sec θ sin θ cos θ Prove the identity + =1 cosec θ sec θ (26) JÚ §iU úSo YhP EÚû[«u BWm 14 ùN,Á Utßm EVWm 8 ùN,Á G²p, ARu Yû[TWl× LôiL, A right circular cylinder has radius of 14cm and height of 8cm. Find its curved surface area. 5
  • 6. (27) UWj§]Xô] JÚ §iUd ám©u A¥fÑt[Ü 44 Á Utßm ARu EVWm 12 Á G²p Aj§iUd ám©u L] A[ûYd LôiL, The circumference of the base of a 12m high wooden solid come is 44m. Find the volume. (28) ØRp 13 CVp GiL°u §hP ®XdLjûRd LQd¡ÓL, Calculate the standard deviation of the first 13 natural numbers. (29) CÚ SôQVeLû[ JúW NUVj§p ÑiÓmúTôÕ. A§LThNUôL JÚ RûX ¡ûPlTRtLô] ¨LrRL®û]d LôiL, Two coins are tossed together. What is the probability of getting at most one head. 2 6x − 54 (30) (A) ÑÚdÏL : 2 . x + 7x + 12 [ApXÕ] (B) 2y = 4x + 3 Utßm x + 2y = 10 Gu úSodúLôÓLs JußdùLôuß ùNeÏjÕ G]d LôhÓL, 2 6x − 54 (a) Simplify : 2 . x + 7x + 12 [OR] (b) Show that the lines 2y = 4x + 3 and x + 2y = 10 are perpendicular. ©¬Ü - III / SECTION − III ϱl× (i) 9 ®]ôdLÞdÏ ®ûPV°dLÜm, (ii) ®]ô Gi 45dÏ Li¥lTôL ®ûPV°dLÜm, ØRp 14 ®]ôdL°−ÚkÕ HúRàm 8 ®]ôdLû[j úRoÜ ùNnVÜm, (iii) JqùYôÚ ®]ô®tÏm IkÕ U§lùTiLs, 9 × 5 = 45 Note : (i) Answer 9 questions. (ii) Question No. 45 is Compulsory. Select any 8 questions from the first 14 questions. (iii) Each question carries Five marks 9 × 5 = 45 (31) ùYuTPeLû[l TVuTÓj§ A (B ∩ C) = (A B) ∪ (A C) Guàm ¥ UôoL²u LQ ®j§VôN ®§«û]f N¬TôodLÜm, Use Venn diagrams to verify De Morgan’s law for set difference A (B ∩ C) = (A B) ∪ (A C). 6
  • 7. (32) Nôo× f : [− 7, 6 ⇒ R ¸rdLiPYôß YûWVßdLlThÓs[Õ, 2 x + 2x + 1  −7≤x<−5 f(x) = x + 5 − 5 ≤ x ≤ 2 . ©uYÚY]Ytûd LôiL, x − 1  2<x<6 4 f(− 3) + 2 f(4) (i) 2 f(− 4) + 3 f(2) (ii) f(− 7) − f(− 3) (iii) f(− 6) − 3 f(1) 2 x + 2x + 1  −7≤x<−5 A function f : [− 7, 6 ⇒ R is defined as follows f(x) = x + 5 −5≤x≤2 .  x − 1 2<x<6 4 f(− 3) + 2 f(4) Find (i) 2 f(− 4) + 3 f(2) (ii) f(− 7) − f(− 3) (iii) f(− 6) − 3 f(1) 2 2 2 2 (33) 1 − 2 + 3 − 4 + ..... Gu ùRôP¬u ØRp 2n Eßl×L°u áÓRp LôiL, 2 2 2 2 Find the sum of the first 2n terms of the series 1 − 2 + 3 − 4 + ..... 3 2 (34) LôW¦lTÓjÕL : x − 5x − 2x + 24 3 2 Factorize the polynomial x − 5x − 2x + 24 2 3 4 (35) m − nx + 28x + 12x + 9x B]Õ JÚ ØÝ YodLm G²p. m, n B¡VYt±u U§l×Lû[d LôiL. 2 3 4 If m − nx + 28x + 12x + 9x is a perfect square, then find the values of m and n. (36) AûNYt ¿¬p JÚ CVk§WlTP¡u úYLm U¦dÏ 15 ¡,Á GuL. AlTPÏ ¿úWôhPj§u §ûN«p 30 ¡,Á çWm ùNuß. ©Ï G§oj §ûN«p §Úm© 4 U¦ 30 ¨ªPeL°p ÁiÓm ×lThP CPj§tÏ §Úm© YkRôp ¿¬u úYLjûRd LôiL, The speed of a boat in still water is 15 km / hr. It goes 30 km upstream and return downstream to the original point in 4 hrs 30 minutes. Find the speed of the stream. 5 2  2 − 1 T T T (37) A =   Utßm B =   G²p (AB) = B A GuTûR N¬TôodLÜm, 7 3 − 1 1  5 2  2 − 1 T T T If A =   and B =   verify that (AB) = B A . 7 3 − 1 1  (38) (− 4, − 2), (− 3, − 5), (3, − 2) Utßm (2, 3) B¡V ×s°Lû[ Øû]L[ôLd ùLôiP SôtLWj§u TWlûTd LôiL, Find the area of the quadrilateral formed by the points (− 4, − 2), (− 3, − 5), (3, − 2) and (2, 3). 7
  • 8. (39) ∆ABC-u Øû]Ls A(2, 1), B(6, − 1), C(4, 11) GuL, A-«−ÚkÕ YûWVlTÓm ÏjÕdúLôh¥u NUuTôhûPd LôiL. The vertices of ∆ABC are A(2, 1), B(6, − 1) and C(4, 11). Find the equation of the straight line along the altitude from the vertex A. (40) JÚ £ßYu ûYWj§u ÏßdÏ ùYhÓj úRôt Y¥®p. TPj§p Lôh¥VYôß JÚ ThPm ùNnRôu, CeÏ AE = 16 ùN,Á, EC = 81 ùN,Á. AYu BD Gu ÏßdÏd Ïf£«û]l TVuTÓjR ®Úmסôu. AdÏf£«u ¿[m GqYôß CÚdL úYiÓm? A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be? (41) úSodÏjRô] JÚ UWj§u úUpTôLm Lôt±]ôp رkÕ, AmرkR Tϧ ¸úZ ®ÝkÕ ®PôUp. UWj§u Ef£ RûWÙPu 30° úLôQjûR HtTÓjÕ¡Õ, UWj§u Ef£ ARu A¥«−ÚkÕ 30 Á ùRôûX®p RûWûVj ùRôÓ¡Õ G²p. UWj§u ØÝ EVWjûRd LôiL, A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30° with it. If the top of the free touches the ground 30m away from its foot, then find the actual height of the tree. (42) L°UiûQl TVuTÓj§ JÚ UôQYu 48 ùN,Á EVWØm 12 ùN,Á BWØm ùLôiP úSo YhP §iUd ámûTf ùNnRôo, AdámûT UtùôÚ UôQYo JÚ §iUd úLô[UôL Uôt±]ôo, AqYôß UôtlThP קV úLô[j§u BWjûRd LôiL, Using clay, a student made a right circular cone of height 48cm and base radius 12cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere. (43) ¸rdLiP AhPYûQ«p ùLôÓdLlThÓs[ ×s° ®YWj§u §hP ®XdLjûRd LQd¡ÓL, x 3 8 13 18 23 f 7 10 15 10 8 Calculate the standard deviation of the following data. x 3 8 13 18 23 f 7 10 15 10 8 (44) JÚ ×§V U¡rÜkÕ (car) ARàûPV Y¥YûUl©tLôL ®ÚÕ ùTßm ¨LrRLÜ 0.25 GuL, £kR Øû«p G¬ùTôÚs TVuTôh¥tLô] ®ÚÕ ùTßm ¨LrRLÜ 0.35 Utßm CÚ ®ÚÕLÞm ùTßYRtLô] ¨LrRLÜ 0.15 G²p. AmU¡rÜkÕ (i) ÏûkRÕ HRôYÕ JÚ ®ÚÕ ùTßRp (ii) JúW JÚ ®ÚÕ UhÓm ùTßRp B¡V ¨Lrf£LÞdLô] ¨LrRLÜLû[d LôiL, The probability that a new car will get an award for its design is 0.25., the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is 0.15. Find the probability that (i) it will get atleast one of the two awards (ii) it will get only one of the awards. 8
  • 9. (45) (A) JÚ áhÓØj ùRôPo Y¬ûN«u AÓjRÓjR êuß Eßl×L°u áÓRp − 6 Utßm AYt±u ùTÚdLtTXu 90, Amêuß GiLû[d LôiL, [ApXÕ] (B) 14 ùN,Á ®hPØm. 20 ùN,Á EVWØm EûPV JÚ EÚû[ Y¥Y ØLûY«p Tô§V[Ü Ri½o Es[Õ, A§p JúW A[®Xô] 300 DV ÏiÓLû[ êrLf ùNnÙm úTôÕ ¿o UhPm 2.8 ùN,Á EVo¡Õ, DVd ÏiÓ JqùYôu±u ®hPm LôiL, (a) The sum of three consecutive term in an A.P. is − 6 and their product is 90. Find the three numbers. [OR] (b) A cylindrical jar of diameter 14cm and depth 20cm is half-full of water. 300 leadshots of same size are dropped into the jar and the level of water raises by 2.8 cm. Find the diameter of each leadshots. ©¬Ü - IV / SECTION − IV ϱl× (i) Cl©¬®p Es[ JqùYôÚ ®]ô®Ûm CWiÓ Uôtß ®]ôdLs ùLôÓdLlThÓs[], (ii) JqùYôÚ ®]ô®Ûm Es[ CWiÓ Uôtß ®]ôdL°−ÚkÕ JÚ ®]ôûY úRokùRÓjÕ CÚ ®]ôdLÞdÏm ®ûPV°dLÜm, (iii) JqùYôÚ ®]ô®tÏm TjÕ U§lùTiLs, 2 × 10 = 20 Note : (i) This section contains Two questions, each with two alternatives. (ii) Answer both the questions choosing either of the alternatives. (iii) Each question carries Ten marks 2 × 10 = 20 (46) (A) 6 ùN,Á BWØs[ JÚ YhPm YûWkÕ ARu ûUVj§−ÚkÕ 10 ùN,Á ùRôûX®Ûs[ JÚ ×s°ûVd ϱdL, Al×s°«−ÚkÕ YhPj§tÏ ùRôÓúLôÓLs YûWkÕ ARu ¿[eLû[ LQd¡ÓL, (ApXÕ) (B) BC = 5 ùN,Á ∠BAC = 40° Utßm Ef£ A-−ÚkÕ BC-dÏ YûWVlThP SÓdúLôh¥u ¿[m 6 ùN,Á Gu A[ÜLs ùLôiP ∆ABC YûWL, A-−ÚkÕ YûWVlThP ÏjÕdúLôh¥u ¿[m LôiL, (a) Draw the two tangents from a point which is 10cm away from the centre of a circle of radius 6cm. Also, measure the lengths of the tangents. (OR) (b) Construct ∆ABC in which the base BC = 5cm, ∠BAC = 40° and the median from A to BC is 6cm. Also measure the length of the altitude from A. 9
  • 10. (47) (A) y = x2 − x − 8-u YûWTPm YûWkÕ. ARû]l TVuTÓj§ x2 − 2x − 15 = 0 Gu NUuTôhûPj ¾odLÜm, (ApXÕ) (B) JÚ ª§Yi¥ KhÓTYo A Gu CPj§−ÚkÕ B Gu CPj§tÏ JÚ ºWô] úYLj§p JúW Y¯«p ùYqúYß SôhL°p TVQm ùNn¡ôo, AYo TVQm ùNnR úYLm. AjçWj§û]d LPdL GÓjÕd ùLôiP úSWm B¡VY]Ytûl Tt±V ®YWeLs (úYL – LôX) ©uYÚm AhPYûQ«p ùLôÓdLlThÓs[], úYLm (¡,Á,/U¦ x 2 4 6 10 12 úSWm (U¦«p) y 60 30 20 12 10 úYL – LôX YûWTPm YûWkÕ A§−ÚkÕ (i) AYo U¦dÏ 5 ¡.Á úYLj§p ùNuôp çWjûRd LPdL BÏm TVQ úSWm (ii) AYo Cdϱl©hP çWjûR 40 U¦ úSWj§p LPdL GkR úYLj§p TV¦dL úYiÓm B¡VY]Ytûd LôiL, (a) Draw the graph of y = x2 − x − 8 and hence find the roots of x2 − 2x − 15 = 0. (OR) (b) A cyclist travels from a place A to a place B along the same route at a uniform speed on different days. The following table gives the speed of his travel and the corresponding time he took to cover the distance. Speed in km/hr x 2 4 6 10 12 Time in hrs 60 30 20 12 10 Draw the speed-time graph and use it to find. (i) The number of hours he will take if he travels at a speed of 5 km/hr. (ii) The speed with which he should travel if he has to cover the distance in 40 hrs. 10