Practice Test Papers 1 to 14 H

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MATHEMATICS

TARGET IIT JEE 2012 XIII (Y-123)

PRACTICE TEST PAPERS I N D E X PRACTICE TEST PAPER-1 ................................................................. Page-2 PRACTICE TEST PAPER-2 ................................................................. Page-4 PRACTICE TEST PAPER-3 ................................................................. Page-6 PRACTICE TEST PAPER-4 ................................................................. Page-8 PRACTICE TEST PAPER-5 ................................................................. Page-11 PRACTICE TEST PAPER-6 ................................................................. Page-13 PRACTICE TEST PAPER-7 ................................................................. Page-16 PRACTICE TEST PAPER-8 ................................................................. Page-18 PRACTICE TEST PAPER-9 ................................................................. Page-20 PRACTICE TEST PAPER-10 ............................................................... Page-22 PRACTICE TEST PAPER-11 ................................................................ Page-24 PRACTICE TEST PAPER-12 ............................................................... Page-26 PRACTICE TEST PAPER-13 ............................................................... Page-28 PRACTICE TEST PAPER-14 ............................................................... Page-30 ANSWER KEY ........................................................................................ Page-33

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PRACTICE TEST PAPER-1 Time: 60 Min.

M.M.: 56 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

;fn lehdj.k (2 + sin ) (3 + sin ) (4 + sin ) = 6 dks larq"V djus okys   [0, 4] ds lHkh ekuksa dk ;ksx k gks] rks k cjkcj gS (A) 6

Q.2

(B) 5

(C) 4

(D) 2

,d vyekjh esa 6 xf.kr dh iqLrdsa] 3 Hkwxksy dh iqLrdsa rFkk 7 bfrgkl dh iqLrdsa gSaA 4 iqLrdksa dk p;u djus ds mu fofHkUu rjhdksa dh la[;k ftuesa izR;sd p;u esa de ls de ,d xf.kr dh iqLrd vo'; lfEefyr gks] gksxh (ekuk fd leku fo"k; dh iqLrdsa vyx&vyx gS)a (A) 720

Q.3

[3 × 3 = 9]

(B) 1044

(C) 1610

(D) 1820

,d f=?kkrh; cgqin y = f(x) bl izdkj gS fd A(– 1, 3) vkSj B (1, – 1) Øe'k% vkisf{kd mPPkre vkSj vkisf{kd fuEure fcUnq gSaA f(2) dk eku gS (A) 6

(B) 4 (C) 0 [PARAGRAPH TYPE]

(D) 3

Q.4 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Q.4

[3 × 3 = 9]

Paragraph for Question 4 to 6 ekuk P(x) ,d 4 ?kkrh; cgqin gS] tks x = 0 ij 'kwU; gSA fn;k x;k gS fd P(–1) = 55 rFkk P(x) dk x = 1, 2, 3 ij vkisf{kd vf/kdre@vkisf{kd U;wure gSA ml f=Hkqt dk {ks=Qy] ftlds dks.kh; fcUnq P(x) ds pje fcUnq gS] gksxk& (A) 1/2 (B) 1/4 (C) 1/8 (D) 1 1

Q.5

fuf'pr lekdyu

 P( x )  P(x ) dx dk eku cjkcj gS &

1

(A) Q.6

252 15

(B)

452 15

(C)

652 15

(D)

752 15

fuEu esa ls dkSulk ,d dFku lR; gS? (A) P (x) ds nks vkisf{kd vf/kdre rFkk ,d vkisf{kd U;wure fcUnq gSA (B) P(x) ds ifjlj esa 9 _.kkRed iw.kk±d gSA (C) P(x) = 0 ds okLrfod ewyksa dk ;ksxQy 5 gSA (D) P (x) dk Bhd ,d ufr ifjorZu fcUnq gSA [REASONING TYPE]

Q.7 & Q.8 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[2 × 3 = 6]

Q.7

dFku-1:

ekuk A rFkk B dksfV 2 ds nks v'kwU; oxZ vkO;wg bl çdkj gS fd AB = O rc det. (A) = 0 ,oa det.(B) = 0 gSA dFku-2: ;fn M rFkk N dksfV 2 ds nks oxZ vkO;wg bl çdkj gS fd MN = O rks ;g O;Dr ugha djrk gS fd vkO;wgks M rFkk N esa ls de ls de ,d vkO;wg 'kwU; vkO;wg gSA (A) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA

Q.8

dFku-1:

fcUnqvksa A, B, C ftuds funsZ'kkad (1, 1, 1), (1, – 1, 1) ,oa (–1, –3, –5) gS] ls xqtjus okyk lery çR;sd k  R ds fy;s fcUnq (2, k, 4) ls xqtjrk gSA

dFku-2:

rhu vlajs[kh; fcUnqvksa P (x1, y1, z1), Q (x2, y2, z2) ,oa R (x3, y3, z3) ls xqtjus okys lery ( x  x1 )

( y  y1 )

( z  z1 )

dk lehdj.k ( x 2  x1 ) ( y 2  y1 ) (z 2  z1 ) = 0 }kjk fn;k tkrk gSA ( x 3  x1 ) ( y3  y1 ) ( z 3  z1 )

PAGE # 2

(A) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA [MULTIPLE CORRECT CHOICE TYPE] Q.9 to Q.10 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [2 × 4 = 8]

Q.9

 2     x sin  x  12 sin   , x  0, 1  ekuk Qyu g, g(x) =  }kjk ifjHkkf"kr gS] rc fuEu esa ls x  x 1 ;fn x  0, 1 0,

dkSulk@dkSuls dFku lR; gS ? (A)  x  R, g (x) vodyuh; gSA (C) x = 0 rFkk x = 1 nksuksa ij g'(x) vlrr~ gSA Q.10

(B) x = 0 ij g'(x) vlrr~ gS ysfdu x = 1 ij lrr~ gSA (D) vUrjky [0, 1] esa g(x) ds fy, jksys dk izes; ykxw gSA

lfn'kksa 2ˆi  ˆj  kˆ vkSj ˆi  ˆj  kˆ ds lkis{k leryh; rFkk lfn'k 5ˆi  2ˆj  6kˆ ds yEcor~ lfn'k fuEu esa ls fdl lery eas fLFkr gksxk (A) x + y + 3z = 5

(B) 2x + y + 3z = 5

(C) 3x + y + 3z = 5

(D) x + y + 4z = 5

PART-B [MATRIX TYPE] Q.1 has three/four statements (A, B, C OR A, B, C, D) given in Column-I and four/five statements (P, Q, R, S OR P, Q, R, S) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. [3+3+3=9] Q.1

Column-I (A)

ik¡p vadks okyh la[;kvksa dh la[;k ftuesa muds vadks dk ;ksx muds vadks ds oxksZ ds ;ksx ds cjkcj gS] gksxh

(B)

;fn Lim

b

(C)

x

 x 4  1 dx

120L dk eku gksxk 

(Q)

16

;fn lehdj.kksa dk fudk;

(R)

18

ax + 3y – z = a 2ax – y + z = 2 bx – 2y + z = 1 – a vlaxr gks] rks b ds lHkh laHko ekuksa dk ;ksx (tgk¡ a  [1, 8], a  I) gS

(S)

none

b 

dk eku L gks] rks

Column-II (P) 15

1

PART-C [INTEGER TYPE] Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [3 × 5 =15] Q.1

;fn a  (a1, a2) ds ekuksa dk lR; leqPp;] çfrcU/k fd Qyu f(x) = x3 – 3ax2 + 3(a2 – 1) x + 1 ds fy;s LFkkuh; U;wure dk fcUnq ,oa LFkkuh; vf/kdre dk fcUnq Øe'k% 4 ls NksVk rFkk – 2 ls cM+k gS] dks lUrq"V djrk gS] rks (a12 + a22) dk eku Kkr dhft,A

Q.2

;fn fcUnq R (3, 4) ds js[kk niZ.k 2xy + 6 – 4x – 3y = 0 ds lkis{k çfrfcEc P rFkk Q gS] rks f=Hkqt PQR dh vUrf=T;k Kkr dhft,A x

Q.3

ekuk f , R esa ,d vodyuh; Qyu gS vkSj f(x) = –

(x2

–x

+ 1) ex

xy · f ' ( y) dy dks larq"V djrk gSA + e 0

;fn f(1) + f '(1) + f '' (1) = ke, tgk¡ k  N gks] rks k Kkr dhft,A PAGE # 3

PRACTICE TEST PAPER-2 Time: 60 Min.

M.M.: 56

[SINGLE CORRECT CHOICE TYPE] Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

ekuk f vkSj g nks vodyuh; Qyu gS tks R  R+ esa ifjHkkf"kr gSA ;fn x = c ij f(x) dk ,d LFkkuh; vf/kdre eku gS rFkk x = c ij g(x) dk ,d LFkkuh; U;wure eku gks] rks h(x) = (A) x = c ij ,d LFkkuh; vf/kdre eku j[krk gS (C) x = c ij ,dfn"V gS

Q.2

Q.3

[3 × 3 = 9]

f (x ) g( x )

(B) x = c ij ,d LFkkuh; U;wure eku j[krk gS (D) x = c ij ,d ufr ifjorZu fcUnq j[krk gS

f=Hkqt PQR ds vUnj ,d fcUnq O(ewy fcUnq ) bl izdkj gS fd OP  k1 OQ  k 2 OR  0 , tgk¡ k1, k2 fu;rkad bl izdkj gS fd

PQR  dk {ks«kQy  4 gS] rks OQR  dk {ks«kQy

(A) 2

(B) 3

k1 + k2 dk eku gS

(C) 4

(D) 5

x y  z ekuk vkO;wg A =  1 2 3  tgk¡ x, y, z  N gSA ;fn det.(adj.(adj. A)) = 28 · 34 gS] rks bl izdkj ds 1 1 2    vkO;wgksa A dh la[;k gS [fVIi.kh: adj. A oxZ vkO;wg A ds lg[k.Mt dks fu:fir djrk gS ] (A) 91 (B) 45 (C) 55 (D) 110

[PARAGRAPH TYPE] Q.4 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 4 = 12]

Paragraph for Question 4 to 6 fcUnq P (4, 3) ls xqtjus okyh rFkk izo.krk m > 0 dh ,d js[kk L [khaph tkrh gS tks js[kkvksa L1 : 3x + 4y + 5 = 0 rFkk L2 : 3x + 4y + 15 = 0 dks Øe'k% A rFkk B ij feyrh gSA A ls L ds yEcor ,d js[kk [khaph tkrh gS tks js[kk L2 dks A1 ij feyrh gSA blh izdkj B ls L ds yEcor ,d js[kk 3[khaph tkrh gS tks L1 dks B1 ij feyrh gSA bl izdkj ,d lekUrj prqHkqZt AA1BB1 curk gSA js[kk L dk lehdj.k bl izdkj izkIr gS fd lekUrj prqHkqZt AA1BB1 dk {ks=Qy U;wure gSA L

fp= uhps fn;k x;k gS B D B1

A1 



L2

C A

L1

Q.4

js[kk L dk lehdj.k gS & (A) 4x – 3y – 7 = 0

Q.5

(B) x – 7y + 17 = 0

(C) 3x – 4y = 0

(D) 7x – y – 25 = 0

;fn js[kk L o`r x2 + y2 – 6x + 4y – 9 = 0 ds yEcdks.kh; gS] rks  cjkcj gS & (A) – 1

Q.6

P(4, 3)

(B) 1

(C) – 2

(D) 2

;fn js [ kk L nks o` r ks a S rFkk S' dh eq y k{k gS tks bl iz d kj gS fd o` r S ds O;kl ds fljs (0, – 1) rFkk (–2, 0) ij gS rFkk o`r S' fcUnq (1, 2) ls xqtjrk gS] rks o`r S' dk lehdj.k gS & (A) 4x2 + 4y2 + 4x – 8y – 32 = 0 (C) 4x2 + 4y2 – x + 67y – 153 = 0

(B) 4x2 + 4y2 + x – 67y – 11 = 0 (D) 4x2 + 4y2 – 2x + 4y – 10 = 0 PAGE # 4

[MULTIPLE CORRECT CHOICE TYPE] Q.7 to Q.8 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [2 ×4 = 8] Q.7

;fn (2x – 1)20 – (ax + b)20 = (x2 + px + q)10 ,  x  R lR; gS] tgk¡ a, b, p rFkk q okLrfod la[;k,sa gS] rks fuEu esa dkSulk@dkSuls lR; gS\ (A) 2p + 3q = 1

Q.8

(B) a + 2b = 0

(C) a  20 220  1

(D) 4q + p = 0

 3x 2  12 x   gS] rks fuEu esa ls dkSuls dFku lR; gSa? ekuk f (x) =  2  x  16  

(A) Lim f ( x ) fo|eku ugha gSA

(B) f(x) ,dfn"V gSA

x4

(C) k ds Bhd nks ekuksa ds fy, lehdj.k f (x) = k dk dksbZ gy ugha gS (D) f(x) Bhd ,d fcUnq ij vlrr~ gS PART-B [MATRIX TYPE] Q.1 has three/four statements (A, B, C OR A, B, C, D) given in Column-I and four/five statements (P, Q, R, S OR P, Q, R, S) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. [3 + 3 + 3 + 3 = 12] Q.1

Column-I

Column-II

dk eku cjkcj gksxk&

(P)

4

Qyu f (x) = x3 – 3x dk izfrcU/k x4 + 36  13x2 ds lUnHkZ esa vf/kdre eku gS

(Q)

6

(C)

,d o`Ùk fcUnqvksa (2, 2) rFkk (9, 9) ls xqtjrk gS ,oa x-v{k dks Li'kZ djrk gS] rks Li'kZ fcUnq ds x-funsZa'kkad ds vUrj dk fujis{k eku gS

(R) (S)

8 12

(D)

ekuk f(x) = cos–1(3x – 4x3) gS] rks f '   dk eku gS  2 

(T)

18



(A)

2

9 x  x · e dx

0 

x

7

2

· e  x dx

0

(B)

 3

PART-C [INTEGER TYPE] Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [3 × 5 = 15] Q.1

ekuk f(x) vodyuh; Qyu gS tks x x   f ( x )   f ( t ) tan t dt   tan t  x  dt = 0  x   . 2  0

0

dks larq"V djrk gS] rks vUrjky [0, 10] esa lehdj.k f(x) = 0 ds gyksa dh la[;k Kkr dhft,A Q.2

ekuk f (x) ,d f=?kkrh; cgqin gS rFkk f '' (x) = 12x – 4 gSA ;fn f (x) dk x = 1 ij LFkkuh; U;wure eku 0 gS] rks f (x) ds fcUnq M ¼ftldk Hkqt 2 gS½ ij vfHkyEc dk x-vUr%[k.M Kkr dhft,A

Q.3

ekuk f vkSj g, R ij nks okLrfod eku vodyuh; Qyu gSA ;fn f '(x) = g(x) vkSj g'(x) = f(x)  x  R vkSj f(3) = 5, f '(3) = 4 gks] rks  f 2 ( )  g 2 ( )  dk eku Kkr dhft,A PAGE # 5

PRACTICE TEST PAPER-3 Time: 60 Min.

M.M.: 58

[SINGLE CORRECT CHOICE TYPE] Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

10



O;atd

C0

2

10

10

1

2

 C 

C2

10

(A) 10! Q.2

2

  C   (B) 

 ....... 

2

10



C8

2

10

 

C9

2

10

 

C10

(C) – 10C5

5



[3 × 3 = 9]

2

cjkcj gS

(D) 10C5

ekuk f(x) ,d Qyu gS tks f '(x) = f(x) dks larq"V djrk gS] f(0) = 1 gS ,oa g(x), f(x) + g(x) = ex (x + 1)2 1

dks larq"V djrk gS] rks  e  x f ( x ) g( x ) dx dk eku gS 0

(A) e

(B) e – 1 x2



Q.3



fuf'pr lekdyu



(A)

(C)

1  sin x  1  sin 2 x

33 2

(B)

3 3

e 2

(D) e – 2

dx dk eku gS

(C)

2 3 3

(D)

3 6

[PARAGRAPH TYPE] Q.4 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Paragraph for question nos. 4 to 6    ekuk a   log3 x ˆi  2ˆj  kˆ , b  (log3 x ) ˆi + ( log3 x) ˆj  kˆ rFkk c = ˆi  ˆj  kˆ gSA fn;k x;k gS fd   a o b ds e/; izR;sd x (0, ) ds fy;s dks.k vf/kd dks.k gS    x = 3 ds fy;s lfn'kksa a , b ,oa c }kjk fufeZr prq"dQyd ds vk;ru dk ifjlj gS



Q.4



1 1  (B)  ,  3 2       |bc| ;fn a  c = b  c gS] rks   dk eku gS |ac|

1  (C)  , 1 2 

(D) [2, 3]

(A)

(C)

(D) 7

 1 (A) 0,   3

Q.5

Q.6

[3 × 2 = 9]

(B) 3

2  











 

 

5



;fn (a  b)  c = a  2b gS] rks a  b b  c c  a dk eku gS (A) 2

(B) 4

(C) 1

(D) 16

[MULTIPLE CORRECT CHOICE TYPE] Q.7 to Q.8 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [2 × 4 = 8] ex

Q.7

dt  ekuk f : R  R dks f(x) =  1 t2 1

(A) f(x) vkorhZ ugha gS (C) f(1) + f '(1) =

 . 2

e x

 1

dt }kjk ifjHkkf"kr fd;k tkrk gS] rks 1 t2

(B) f f ( x )  = f(x)  x  R. (D) f(x) ifjc) ugha gSA PAGE # 6

Q.8

,d f=Hkqt ABC esa] ;fn a = 4, b = 8 rFkk C = 60° gS] rks fuEu esa ls dkSulk@dkSuls lEcU/k lR; gSa? [Note: iz;ksx fd;s x;s lHkh izrhdksa dk f=Hkqt ABC esa lkekU; vFkZ gSA] (A) f=Hkqt ABC dk {ks=Qy 8 3 gS (C) f=Hkqt ABC dh vUr% f=T;k

(B)

2 3 gS 3 3

 sin 2 A dk eku 2 gS 4 gS 3

(D) dks.k C ds vUr% dks.k lef}Hkktd dh yEckbZ

PART-B [MATRIX TYPE] Q.1 has three/four statements (A, B, C OR A, B, C, D) given in Column-I and four/five statements (P, Q, R, S OR P, Q, R, S) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. [4 + 4 + 4 + 4 = 16 + 1 bonus] Q.1

LrEHk-II

(A)

LrEHk-I ;fn y = 4x – 5 oØ C : y2 = px3 + q ds M (2, 3) ij Li'kZ js[kk gS] rks (p – q) dk eku gS

(P)

1

(B)

lfn'kks V1 = ˆi  ˆj , V2 = ˆi  ( 2 cosec  ) ˆj  kˆ

(Q)

3

(R)

5

(S)

6

(T)

9

RkFkk V3 = ˆj  (2 cosec  ) kˆ }kjk fufeZr lekUrj "kV~Qyd ds vk;ru dk U;wure eku gS tgk¡  (0, ) sin x 1

x 1

cos x x

x = 0 ds gyks dh la[;k gS 1

(C)

lehdj.k

(D)

prqHkqZt ABCD esa ;fn cot A = 4, cot B =

3 rFkk cot C = 5 gS, 2

rks 3 tan D dk eku gS

PART-C [INTEGER TYPE] Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [3 × 5 = 15] Q.1

Q.5

ekuk A1 A2 A3 ............... An ,d lecgqHkqt gS ftldh n Hkqtk,sa gSA fdlh iw.kkZad k < n ds fy,] prqHkqZt A1 A2 Ak Ak + 1 ,d vk;r gS ftldk {ks=Qy 6 gSA ;fn cgqHkqt dk {ks=Qy 60 gks] rks n dk eku Kkr dhft,A ;fn O;atd arc tan  1  x   arc tan ax   arc tan bx    , (a, b  R),  x  R lR; gS, x

rks Q.3

4(a2

+

b2)

8

2

dk eku Kkr dhft,A

,d f=Hkqt ABC esa, ABC dk vkarfjd dks.k lef}Hkktd AC dks K ij feyrk gSA ;fn BC = 2, CK = 1 vkSj BK =

3 2 gks] rks Hkqtk AB dh yEckbZ Kkr dhft,A 2

PAGE # 7

PRACTICE TEST PAPER-4 Time: 60 Min.

M.M.: 58 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. 8

Q.1

O;atd



0

1

dk eku gS

1  tan 3 10 

(A) 5 Q.2

[3 × 3 = 9]

(B)

21 4

(C)

14 3

(D)

9 2

ekuk f ,d vodyuh; Qyu gS tks f '(x) = 2f(x) + 10 dks lUrq"V djrk gS rFkk f(0) = 0 gS] rks lehdj.k f(x) + 5 sec2x = 0 ds (0, 2) esa ewyksa dh la[;k gS (A) 0

(B) 1

(C) 2

(D) 3

2

Q.3

fuf'pr lekdy

tan 1 x  x 2  x  1 dx dk eku cjkcj gS 12

2 (A) 6 3

2 (B) 3 3

2 (C) 12 3

2 (D) 4 3

[PARAGRAPH TYPE] Q.4 to Q.5 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[2 × 3 = 6]

Paragraph for question nos. 4 and 5

ekuk 1 ml lery ds lehdj.k dks fu:fir djrk gS ftlds lfn'k 1, 1, 0 vfHkyEc gS ,oa ftlesa ,d js[kk  L gS ftldk lehdj.k r  ˆi  ˆj  kˆ   ˆi  ˆj  kˆ gSA 2 ml lery ds lehdj.k dks fu:fir djrk gS ftlesa





js[kk L ,oa fLFkfr lfn'k 0, 1, 0 okyk ,d fcUnq gSA Q.4

ifjek.k 6 dk lfn'k] tks leryksa 1 ,oa 2 dh çfrPNsnu js[kk ds vuqfn'k gS ,oa lery 1 ds vfHkyEc lfn'k ds yEcor gS] gksxk (A) ± 2  1, 1, 1

Q.5

(B) ± 2 1,  1, 1

(C) ± 2 1, 1,  1

(D) ± 2 1, 1, 1

1 rFkk 2 ds e/; U;wu dks.k gS (A) tan 1 1  tan 1 (C) tan 1 1  tan 1

(B) tan 1 1  tan 1 2  3

 2  1  2  1

  2  3 

(D) tan 1 1  tan 1

[MULTIPLE CORRECT CHOICE TYPE] Q.6 to Q.7 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [2 × 4 = 8] Q.6

ekuk vUrjky [0, 1] esa ,d lrr~ Qyu g bl izdkj ls gS fd g(0) = 1 vkSj g(1) = 0 gS] rks fuEu esa ls dkSu&dkSuls dFku lnSo lR; gSa\ (A) vUrjky [0, 1] esa ,d la[;k h bl izdkj fo|eku gS fd g(h)  g(x), [0, 1] esa lHkh x ds fy, (B) vUrjky [0, 1] esa ,d la[;k h bl izdkj fo|eku gS fd g(h) = 1/2. (C) vUrjky [0, 1] esa ,d la[;k h bl izdkj fo|eku gS fd g(h) = 3/2. (D) vUrjky (0, 1) esa lHkh h ds fy,, Lim g ( x )  g (h ) . xh

PAGE # 8

Q.7

lehdj.k arc cos x = arc tan x ds gSa & (A) dsoy ,d gy (C) (–1, 1) esa Bhd nks gy

(B) (0, 1) esa Bhd ,d gy (D) (–1, 0) esa dksbZ gy ugha

[REASONING TYPE] Q.8 & Q.9 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.8

dFku -1:

[2 × 3 = 6]

p dk Bhd ,d ,slk okLrfod eku fo|eku gS fd ftlds fy;s lehdj.k x3 – 3x + p = 0 ds (0, 1) esa nks fHkUu ewy gSA

dFku -2:

;fn Qyu f(x) vUrjky [a, b] esa vodyuh; gS rFkk gS rFkk f(a) = f(b) gS] rks (a, b) esa dqN ,sls c fo|+eku gS fd f '(c) = 0 gksxkA (A) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k ugh gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku& 2 lR; gSA

Q.9

dFku-1:

 x x4 x2      9 3 4  7 x 4 x 13   rFkk f(x) = tr.(A)] tgk¡ tr.(A) vkO;wg A ds ekuk A = x    2  5x 2  x  1  13

Vªsl ¼vuqjs[k½ dks fu:fir djrk gS] rks x > 0 ds fy;s f(x) dk U;wure eku 8 gSA dFku-2:

a, b > 0 ds fy;s

ab  2

ab

(A) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k ugh gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku& 2 lR; gSA PART-B [MATRIX TYPE] Q.1 has three/four statements (A, B, C OR A, B, C, D) given in Column-I and four/five statements (P, Q, R, S OR P, Q, R, S) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. [3 + 3 + 3 = 9] Q.1

LrEHk-I (A) (B)

;fn oØ C1 : y2 = 2ax (a > 0) rFkk C2 : xy = 4 2 yEcdks.kh; izfrPNsn djrs gS] rks a cjkcj gS ;fn f : R  [1, 4] ,d vodyuh; Qyu bl izdkj gS

(P)

0

(Q)

1

fd Lim f ( x )  f ' ( x ) = 3 gS] rks Lim f ( x ) gS

(R)

2

izR;sd m  R ds fy;s oØ y = (m – 1)x + (n + 2) lnSo ,d fLFkj fcUnq P ls xqtjrk gSA ;fn fcUnq P dh dksfV 3 gS] rks n dk eku gS

(S)

3

x 

(C)

LrEHk-II

x 

PAGE # 9

PART-C [INTEGER TYPE] Q.1 to Q.4 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [4 × 5 = 20] Q.1

ekuk K ,d /kukRed iw.kkZad bl izdkj gS fd 36 + K, 300 + K, 596 + K ,d o/kZeku lekUrj Js.kh ds rhu Øekxr inksa ds oxZ gSa rks K dk eku Kkr dhft,A

Q.2

ekuk f (x) = x2 + ax + b gSA ;fn  x  R, y dk ,d okLrfod eku bl izdkj fo|eku gS fd f (y) = f (x) + y, rks 100a dk vf/kdre eku Kkr dhft,A

Q.3

;fn lehdj.k (sin  – 1)(2 sin  – 1)(3 sin  – 1).......(n sin  – 1) = 0 (tgk¡ n  N) dks larq"V djus okys gyksa dh la[;k vUrjky [0, ] eas 9 gS] rks lehdj.k 3 + cosec2x + 2sin Kkr dhft,A

Q.4

2

y

= n, tgk¡ 0  x, y  4 dks larq"V djus okys Øfer ;qXeksa (x, y) dh la[;k

R (lHkh okLrfod la[;kvksa dk leqPp;) esa ifjHkkf"kr ,d okLrfod eku lrr~ Qyu f(x) ds vkjs[k ij fopkj

dhft, tks uhps fn[kk;k x;k gS y 5 (–2,4)

(4,4)

4 3

(–1,2)

2

(2,2)

1 –5

–4

–3

–2

x

–1 O

1

2

3

4

5

–1 –2

lehdj.k f (f(x)) = 4 ds okLrfod gyksa dh la[;k Kkr dhft,A

PAGE # 10

PRACTICE TEST PAPER-5 Time: 60 Min.

M.M.: 56 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Q.1

 sin{cos x} ,   x  ;fn f(x) =  2  1,

x

 2

x

 2

[3 × 3 = 9]

tgk¡ {k}, k ds fHkUukRed Hkkx dks O;Dr djrk gS] rks (A) x =

 ij f(x) lrr~ gSA 2

(B) Lim f(x) fo|eku gS] ijUrq x = x

(C) Lim f ( x ) fo|eku ugha gSA x

Q.2

(D) Lim f ( x )  1 .

 2

x

 2

;fn 1, x, y ,d xq.kksÙkj Js.kh gS rFkk x, y, 3 ,d lekUrj Js.kh gS] rks (x + y) dk vf/kdre eku gS (A) 0

Q.3

 2

 ij f lrr~ ugha gSA 2

(B)

9 2

(C)

15 4

(D) 1

mu N% vadksa okyh la[;kvksa dh la[;k fdruh gksxh ftuesa vadksa ds oxksZa dk ;ksx 9 gks (A) 60

(B) 66

(C) 72

(D) 37

[PARAGRAPH TYPE] Q.4 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 4 = 12]

Paragraph for question nos. 9 to 11

  2 1 3  x  x  4     2  ekuk M(x) =  x  x   x 

    x3   1  2   x  3 

x2  x x

x

1 2

x3

,oa P, Q, Rm, Sk vU; vkO;wg gS tks fuEu çdkj ls ifjHkkf"kr gS P = Lim M (0)  M (0) 2  M (0) 3  ..............  M (0) n n





 k r Sk = M(x)   PQ   . r 1  [fVIi.kh: Tr. (A) oxZ vkO;wg A ds Vsªl dks rFkk adj. A oxZ vkO;wg A ds lg[k.Mt dks fu:fir djrk gSA] 1 1 Q = diag.  , 1,  3 2

Q.9

Tr. adjadj P cjkcj gS (A) 6 (B) 18

Rm = M(x) (PQ)m, m  N

(C) 36

(D) 54

PAGE # 11

 2

Q.10

;fn Tr. (R100) = f(x) gS, rks



  f (sin x )  f (cos x )



0

(A) 2 Q.11

(B) 4

41   dx dk eku gS 6

(C) 6

(D) 10

;fn Tr. (S360) = g(x) gS] rks g(x) dk U;wure eku gS (A) 570

(B) 690

(C)

19 36

(D)

23 36

[MULTIPLE CORRECT CHOICE TYPE] Q.7 to Q.8 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [2 × 4 = 8] Q.7

ekuk f(x) =

( x  1) 2 .e x gS] rks (1  x 2 ) 2

(A) vUrjky (– , – 1 ) esa f(x) fujUrj o/kZeku gS (B) vUrjky (1, ) esa f(x) fujUrj ákleku gSA (C) f(x) ds nks LFkkuh; pje eku ds fcUnq gSaA (D) f(x) dk dqN x  (– 1, 0) ij LFkkuh; U;wure fcUnq gSA Q.8



,d v'kwU; lfn'k a ds fy;s fuEu eas ls dkSulk@dkSuls fu"d"kZ lR; gS       (A) a · b = a · c  b  c           (C) a · b = a · c rFkk a  b = a  c  b  c

      (B) a  b = a  c  b  c

(D) aˆ  bˆ = aˆ  bˆ

 aˆ · bˆ = 0

PART-B [MATRIX TYPE] Q.1 has three/four statements (A, B, C OR A, B, C, D) given in Column-I and four/five statements (P, Q, R, S OR P, Q, R, S) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. [3 + 3 + 3 + 3 = 12] Q.1

Column-I

Column-II

(A)

ekuk cos – sin  = 2 sin  rFkk cos  + sin  = K cos  gSA (P) ;fn K + L =11 gS] rks log3L nks Øekxr iw.kk±dksa ds e/; fLFkr gS ftudk ;ksxQy gS

0

(B)

ewy fcUnq ls 2 f=T;k ds pj o`Ùk ij Li'kZ js[kk,a lnSo yEcor gksrh gSA pj o`Ùk ds dsUnz dk fcUnqiFk ,d o`Ùk gS ftldh f=T;k gSA

(Q)

1

(C)

;fn sin–1 (1 – 2x) =

(R)

2

(D)

fn;k x;k f ,d fo"ke Qyu gS tks loZ= ifjHkkf"kr gS] vkorZukad 2 okyk vkorhZ gS rFkk çR;sd vUrjky esa lekdyuh; gS

(S)

3

(T)

5

 – sin–1 x gS] rks (12x – 12x2) dk eku gS 3

x

ekuk g(x) =  f ( t ) dt gS] rks g(10) dk eku gS 0

PAGE # 12

PART-C [INTEGER TYPE] Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [3 × 5 = 15] Q.1

,d o`r C dh lehdj.k x2 + y2 + kx + (k + 1)y – (k + 1) = 0 gS tks izR;sd okLrfod k ds fy, nks fLFkj fcUnqvksa ls xqtjrk gSA ;fn o`Ùk C dh f=T;k dk U;wure eku

Q.2

ekuk f=Hkqt ABC ds fcUnqvksa A, B, C ls 'kh"kZyEc Øe'k% ha, hb, hc gSA ;fn ha = 8, hb = 8, hc = 10 rks Hkqtk AB dh yEckbZ

Q.3

 gks] rks R (R N) dk eku Kkr dhft,A R

p ds :i esa O;Dr dh tkrh gS] rks (p + q) dk eku Kkr dhft,A (tgk¡ p, q izkÑr la[;k,sa gSa) q

 100  ekuk f(x) ,d Qyu gS tks f(x) = f    x > 0 dks lar"q V djrk gSA ;fn  x 

10

f (x )  x dx = 5 gks] rks 1

100

 1

f (x) dx x

dk eku Kkr dhft,A PRACTICE TEST PAPER-6 Time: 60 Min.

M.M.: 56 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

,d f=Hkqt ABC esa, C = 90º, BC = 3, AC = 4 vkSj AB ij ,d fcUnq D gS rFkk BCD = 30° gS] rks CD dh yEckbZ cjkcj gS (A)

Q.2

Q.3

[3 × 3 = 9]

8 ( 2 3  1) 13

(B)

8 ( 2 3  1) 9

(C)

8 ( 4 3  3) 9

8 ( 4 3  3) 13

(D)

 100  k 3    1 dk eku gS  k  0 k  1!  k  2!  k  3!  103 !   (A) 1 (B) 1/2 (C) 100 (D) 101 ;fn lehdj.kksa ds fudk; x – ky – z = 0, kx – y – z = 0, x + y – z = 0 dk ,d v'kwU; gy gks] rks k ds lEHko

eku gksxsa& (A) –1, – 2

(B) 1, – 2

(C) –1, 2

(D) –1, 1

[PARAGRAPH TYPE] Q.4 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 3 = 9]

Paragraph for Question 4 to 6 ekuk f(x) = + bx + c bl iz d kj gS fd 1, b, c (Øe esa ys us ij) lekUrj Js . kh es a gS rFkk 2, 5b, – 10c (Øe esa ysus ij) xq.kksÙkj Js.kh esa gS] tgk¡ b, c  I (lHkh iw.kkZadksa dk leqPp;) gSA x2

fn;k x;k gS fd g(x) = (a2 + 1) x2 –   a 2  4  x – 3 rFkk h(x) = x2 – (p – 3) x + p, tgk¡ a, p  R 



(lHkh okLrfod la[;kvksa dk leqPp;) gSA Q.4

;fn M rFkk m vUrjky x  [0, 4] esa f(x) dk vf/kdre ,oa U;wure eku gS] rks (M + m) cjkcj gS & (A) 9

(B) – 3

(C) 0

(D) 4 PAGE # 13

Q.5

a ds mu iw.kkZad ekuksa dh la[;k] ftuds fy;s g(x) < 0 de ls de ,d okLrfod x ds fy;s larq"V gksrk gks] gksxh& (A) de ls de 7 (B) vf/kd ls vf/kd 3 (C) 5 (D) 0

Q.6

;fn Qyu y = f(x) + h(x) dk ifjlj [0, ) gS] rks p ds okLrfod ekuksa dk lR; leqPp; gS & (A) {p | p  R, – 3  p < } (C) {p | p  R, –  < p  3}

(B) {p | p  R, 2  p < } (D) 

[REASONING TYPE] Q.7 to Q.9 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.7

[3 × 3 = 9]

dFku-1 :

a, b rFkk c ds fdlh LosfPNd okLrfod ekuksa ds fy;s lehdj.k a cos 3x + b cos 2x + c cos x + d sin x = 0 dk [0, 2] esa de ls de ,d ewy vo'; gSA

dFku-2 :

;fn  f ( x ) dx (p < q) 'kwU; gS] rks lehdj.k f(x) = 0 dk [p, q] esa de ls de ,d ewy vo';

q

p

gSA (A) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA Q.8

dFku&1:

10 leku xasnksa dks 4 fofHkUu ckWDlksa esa bl izdkj forfjr djus ds rjhdksa dh la[;k] fd dksbZ ckWDl [kkyh u jgs] 9C3 gSA dFku&2: 9 fofHkUu LFkkuksa esa ls dksbZ Hkh 3 LFkkuksa ds p;uksa ds rjhdksa dh la[;k 9C3 gSA (A) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA

Q.9

dFku-1:

;fn ,d vuUr xq.kksÙkj Js.kh dk f}rh; in x gS rFkk bldk ;ksxQy 8 gS] rks x dk ifjlj (– 16, 2] gSA dFku-2: ,d vifjfer xq.kksÙkj Js.kh dk ;ksxQy ifjfer gSA fn;k x;k gS 0 < | r | < 1 tgk¡ r xq.kksÙkj Js.kh ds lkoZvuqikr dks fu:fir djrk gSA (A) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS] dFku&2 lR; gS] dFku&2 dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA [MULTIPLE CORRECT CHOICE TYPE]

Q.10 to Q.11 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct.[2 × 4 = 8] Q.10

fuEu esa ls dkSulk@dkSuls dFku lR; gSa\ (A) og fcUnq] tgk¡ Qyu bldh ,dfn"Brk ,oa mUeq[krk ifjofrZr djrk gS] lnSo vodyuh;rk dk fcUnq ugha gSA (B) ;fn Qyu dk x = a ij vodyt 'kwU; gS] rks f(x) ml fcUnq ij o/kZeku ;k âkleku gks ldrk gSA (C) ;fn f(x), R ij ,d f} vodyuh; Qyu gS rFkk bldk x = c ij ,d lkisf{kd vf/kdre gS]rks f "(c) _.kkRed gSA (D) ;fn f(x), R ij vodyuh; Qyu bl izdkj gS fd f '() = 0 rks f(x) dk x =  (  R) ij ,d pje gSA

PAGE # 14

Q.11

ekuk a1, a2, a3 (Øe esa) rhu la[;k;s o/kZeku lekUrj Js.kh esa gS rFkk g1, g2, g3 (Øe esa) rhu la[;k;s xq.kksÙkj 3

Js.kh esa gSA fn;k x;k gS fd a1 + g1 = 85, a2 + g2 = 76, a3 + g3 = 84 rFkk

ai

= 126 gS] rks fuEu esa ls

i 1

dkSulk@dkSuls lR; gS (A) lekUrj Js.kh dk lkoZvUrj 25 gS (C) xq.kksÙkj Js.kh dk lkoZvuqikr

(B) xq.kksÙkj Js.kh dk lkoZvuqikr

1 gS 2

1 gS 4

(D) lekUrj Js.kh dk lkoZvUrj 26 gS

PART-B [MATRIX TYPE] Q.1 has three/four statements (A, B, C OR A, B, C, D) given in Column-I and four/five statements (P, Q, R, S OR P, Q, R, S) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. [3+3+3= 9] Q.1

LrEHk-I (A)

(B)

(C)

LrEHk-II

  1  2  2 ekuk A =  2 1  2 gSA ;fn adj. A = kAT gS] rks 'k' dk eku gS  2  2 1  'k' U;wure /kukRed iw.kkZad gS ftlds fy, Qyu f (x) = (2x + 1)50 (3x – 4)60 vUrjky [k, ) esa o/kZeku gS] rks 'k' dk eku gS  2 | x | 1  ; ekuk f(x) = | x | a  bx  ; | x |  1 

tgk¡ a rFkk b fu;rkad gS ;fn x = 1 ij f(x) vodyuh; gS] rks (2a + b) cjkcj gS

(P)

2

(Q)

3

(R)

4

(S)

5

PART-C [INTEGER TYPE] Q.1 to Q.2 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)[2 × 6 = 12] Q.1

;fn Js.kh 1 · 2 + 2 · 3 + 3 · 4 + .............. ds n inksa dk ;ksxQy Sn }kjk rFkk Js.kh 1 1 1    ............... ds (n – 1)inksa dk ;ksxQy n – 1 }kjk fu:fir fd;k 1· 2 · 3 · 4 2·3· 4·5 3·4·5·6



tkrk gS] rks Sn 18 n  1  1 Q.2



Kkr dhft,A

f ( x )  f ( y) xy ekuk f(x) ,d vodyuh; Qyu gS rFkk f   x, y  R dks larq"V djrk gS rFkk  =

 2 

2

f(0) = 0 gSA ;fn

 f (x)  sin x 

2

2

dx dks U;wure fd;k tkrk gS] rks f(– 42) dk eku Kkr dhft,A

0

PAGE # 15

PRACTICE TEST PAPER-7 Time: 60 Min.

M.M.: 55 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

ekuk m, 5 yM+dks ,oa 5 yM+fd;ks dks ,d js[kk esa ,dkUrj :i ls O;ofLFkr djus ds rjhdks dh la[;k dks fu:fir djrk gS rFkk n, 5 yM+dks ,oa 5 yM+fd;ksa dks ,d o`Ùk bl izdkj O;ofLFkr djus ds rjhdks dks fu:fir djrk gS ftlesa dksbZ nks yM+ds lkFk&lkFk ugh gksA ;fn m = kn gS] rks k dk eku gS (A) 30

Q.2

Q.3

[3 × 4 = 12]

(B) 5 (C) 6   ekuk a  ˆi  ˆj  kˆ rFkk r  xˆi  yˆj  zkˆ ,d pj lfn'k bl çdkj gS fd     iw.kk±d gSA ;fn r · ˆj  3 rFkk r · a  12 gS] rks lEHko r dh dqy la[;k gS (A) 10C3 (B) 11C3 (C) 13C4

(D) 10  ˆ  ˆ  r · i , r · j rFkk r · kˆ /kukRed (D) 13C9

ekuk f ,d ,dSdh Qyu gS ftldk izkUr [a, b] rFkk ifjlj [c, d] gSA ;fn vUrjky (a, b)esa ,d fcUnq  bl izdkj gS fd x =  ij Qyu f dk ck;k¡ vodyt l rFkk nk;k¡ vodyt r gSA l vkSj r nksuksa v'kwU;] fHkUu&fHkUu rFkk _.kkRed gS] rks x = f() ij f–1 dk ck;k¡ vodyt rFkk nk;k¡ vodyt Øe'k% gS 1 1 (A) , r l

1 1 (C) , l r

(B) r, l

(D) l, r

[MULTIPLE CORRECT CHOICE TYPE] Q.4 to Q.6 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [3 × 5 = 15]

Q.4

  x 3  x 2  10 x ;  ekuk f(x) = sin x ;  1  cos x ; 

1  x  0  0x 2  x 2

gS] rks f(x) ds gS  ij LFkkuh; vf/kdre 2 (C) x = 0 ij fujis{k vf/kdre

(A) x =

Q.5

 2

(B) x =

 ij LFkkuh; U;wure 2 (D) x = – 1 ij fujis{k vf/kdre

 2

 

ekuk f(x) =   cos 1 cos x    sin 1 sin x  tgk¡ x  [0, 2] gS] rks fuEu esa ls dkSulk@dkSuls dFku lR; gS? (A) f(x) vUrjky [0, 2] esa lrr~ ,oa vodyuh; gSA

  2 2  (B) f(x) dk ifjlj  ,  gSA 4 4   2

(C) f (x) vUrjky [0, ] esa fujUrj áleku gS

3 (D)  f ( x ) dx = 24 0

PAGE # 16

Q.6

a b c = = lR; gS] rks fuEu esa ls dkSulk@dkSuls 13 7 15

,d f=Hkqt ABC esa lkekU; ladsru ds lkFk ;fn

lEcU/k lR; gS [fVIi.kh :  rFkk s Øe'k% f=Hkqt ds {ks=Qy rFkk f=Hkqt ds v)Zifjeki dks fu:fir djrs gS] BC 4 3 (A) tan  = 11  2 

(B) f=Hkqt vf/kd dks.k f=Hkqt gS

(C) r : r1 = 9 : 35

(D)  : s2 = 3 3 : 1 PART-C [INTEGER TYPE]

Q.1 to Q.6 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [2 × 4 = 8] Q.1

;fn js[kk y = 2 – x o`Ùk S ds fcUnq P(1, 1) ij Li'kZ js[kk gS vkSj o`Ùk S, o`Ùk x2 + y2 + 2x + 2y – 2 = 0 ij yEcdks.kh; gks] rks fcUnq (2, 2) ls o`Ùk S ij [khaph xbZ Li'kZ js[kk dh yEckbZ Kkr dhft,A

Q.2

,d lekUrj vuqØe

a n esa] ekuk a1 > 0 vkSj 3a8 = 5a13 gSA ;fn Sn izFke n inksa dk ;ksx gS] rks n  N dk

eku Kkr dhft, ftlds fy, Sn vf/kdre gksA Q.3 to Q.6 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)[4 × 5 = 20] Q.3

funsZ'kkad lery ij ,d js[kk y = x + 2 [khaph tkrh gSA bl js[kk dks fcUnq (0, 2) ds lkis{k 90° ls nf{k.kkorZ ?kwf.kZr fd;k tkrk gSA ,d js[kk y = – 2x + 10 [khaph tkrh gS vkSj bu rhuksa js[kkvksa ls ,d f=Hkqt cuk;k tkrk gSA ;fn f=Hkqt dk {ks=Qy gks] rks [] dk eku Kkr dhft, tgk¡ [k], k ds cjkcj ;k k ls de egÙke iw.kkZad dks O;Dr djrk gSA

Q.4

   ekuk a  3 i  2 j  4 k ; b  2 i  k vkSj c  4 i  2 j  3 k gSA





   ;fn lehdj.k x a  y b  z c =  x i  y j  z k ds vrqPN gy gks] rks ds lHkh fHkUu&fHkUu laHko ekuksa dk ;ksx Kkr dhft,A



Q.5



  

  



 

  

ekuk r  a  b sin x  b  c cos y  2  c  a  , tgk¡ a , b, c v'kwU; rFkk vleryh; lfn'k gSA ;fn 20     r , a  b  c ds yEcdks.kh; gks] rks 2 (x2 + y2) dk U;wure eku Kkr dhft,A 

Q.6

ekuk f(x) =

x2

+ ax + 3 rFkk g(x) = x + b, tgk¡ F(x) = Lim

n

f (x )  x 2n g(x ) 1  x 2n

gSA

;fn F(x), x = 1 vkSj x = – 1 ij lrr~ gks] rks (a2 + b2) dk eku Kkr dhft,A

PAGE # 17

PRACTICE TEST PAPER-8 Time: 60 Min.

M.M.: 44 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.4 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[4 × 3 = 12]

Q.1

ekuk A (z1), B (z2) ,oa C(z3) f=Hkqt ABC ds 'kh"kZ bl izdkj gS fd z3 + iz2 = (1 + i) z1, tgk¡  ( 1) bdkbZ dk ?kuewy gS] rks f=Hkqt ABC gS (A) leckgq (B) lef}ckgq (C) fo"keckgq (D) ledks.k lef}ckgq

Q.2

nks ik=ksa esa Øe'k% m1 rFkk m2 lQsn xsna as ,oa n1 rFkk n2 dkyh xsna as gSAa çR;sd ik= ls ;kn`PN;k ,d xsna fudkyh tkrh gS ,oa fudkyh xbZ nksuksa xsna ks esa ls ;kn`PN;k ,d xsna yh tkrh gS] rks bl xsna ds lQsn jax ds gksus dh çkf;drk gS

Q.3

Q.4

m2 n 2  1  m1n1 (A) 2  m  n  m  n   1 1 2 2

m2  1  m1 (B) 2  m  n  m  n   1 1 2 2

m 2 n1  1  m1n 2 (C) 2  m  n  m  n   1 1 2 2

n2  1  n1 (D) 2  m  n  m  n   1 1 2 2

1   2011  ds fy;s P(x) = 0 ds gS ;fn P(x) = 2013 – 2012 – 16x + 8 gS] rks x  0, 8   (A) Bhd ,d okLrfod ewy (B) dksbZ okLrfod ewy ugha (C) de ls de ,d ,oa vf/kd ls vf/kd nks okLrfod ewy (D) de ls de nks okLrfod ewy

x2012

x2011

ml o`Ùk dh f=T;k] tks ijoy; y2 = x dks M(1, 1) ij Li'kZ djrk gS rFkk ijoy; y2 = x dh fu;rk blds vfHkyEc gS] gksxh (A)

6 5 4

(B)

7 5 4

(C)

5 5 4

(D)

3 5 4

[PARAGRAPH TYPE] Q.5 to Q.7 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 3 = 09]

Paragraph for Question no. 5 to 7 3 ?kkr ds ,d cgqin f(x) dk vkjs[k fp= esa n'kkZ;k x;k gS rFkk Q (0, 5) ij Li'kZ js[kk dh izo.krk 3 gSA y

P

Q 5 4 3 2 1

–2–1

Q.5

O1 2

R

x

3

lehdj.k f (| x |) = 3 ds gyksa dh la[;k gS (A) 1

(B) 2

(C) 3

(D) 4

PAGE # 18

Q.6

ml fcUnq ij vfHkyEc dk lehdj.k] tgk¡ oØ y-v{k dks dkVrk gS] gksxk (A) 3x + y = 15

Q.7

(B) x + 3y = 15

(C) x + 3y = 5

(D) 3x + y = 5

oØ y = f(x), x-v{k ,oa js[kkvksa x + 1 = 0, x – 1 = 0 }kjk ifjc) {ks=Qy gS (A)

13 2

(B)

15 2

(C)

17 2

(D)

19 2

[MULTIPLE CORRECT CHOICE TYPE] Q.8 and Q.9 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct.[2 × 4 = 08]

Q.8

;fn f(x) = sin–1 (A) f '(–1) =

1 x 2 1 2 1 x2

gS] rks fuEu esa ls dkSulk@dkSuls lR; gS?

1 4

  (B) f (x) dk ifjlj 0,  gS  2

(D) Lim

(C) f '(x) ,d fo"ke Qyu gS Q.9

x0

f (x) 1 = x 2

m ds os laHko okLrfod eku] ftuds fy;s ;qxir lehdj.ksa y = mx + 3 ,oa y = (2m – 1)x + 4 okLrfod la[;kvksa ds de ls de Øfer ;qXe (x, y) ds fy;s larq"V gksrh gks] gS (A) 0 (B) 1 (C) – 2 (D) – 3

PART-C [INTEGER TYPE] Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [3 × 5 = 15] Q.1

,d fpfdRld ekurk gS fd ,d ejht dks rhu chekfj;ksa d1, d2 ;k d3 esa ls ,d gSA fdlh tk¡p ls iwoZ og ekurk gS çR;sd chekjh ds fy;s çkf;drk leku gSA og ,d tk¡p ds ckn ikrk gS fd chekjh d1 gksus dh çkf;drk 0.8, chekjh d2 gksus dh çkf;drk 0.6 rFkk chekjh d3 gksus dh çf;drk 0.4 gSA fn;k x;k gS fd tk¡p dk ifj.kke /kukRed p

gS] rks ejht dks chekjh d1 gksus dh çkf;drk q gS] rc (p + q) dk U;wure eku Kkr dhft,A Q.2





;fn fcUnqvksa P, Q ,oa R ds ewy fcUnq O ds lkis{k fLFkfr lfn'k Øe'k% r1  3ˆi  2ˆj  kˆ , r2  ˆi  3ˆj  4kˆ ,oa  r3  2ˆi  ˆj  2kˆ gS] rks P dh lery OQR ls nwjh Kkr dhft;sA

Q.3

x  y2   1 ij fLFkr fcUnq gS rFkk Q1, Q2, ......., Qn nh?kZo`Ùk ds lgk;d 16 9 o`Ùk ij fLFkr laxr fcUnq gSA ;fn C dks Qi (C nh?kZo`Ùk dk dsUnz gS) ls feykus okyh js[kk fn;s x;s nh?kZo`Ùk ds

ekuk P1, P2, ........, Pn nh?kZo`Ùk

n

lkis{k Pi ij vfHkyEc dks Ki ij feyrh gS rFkk

 C Ki

= 175 gks] rks n dk eku Kkr dhft,A

i 1

PAGE # 19

PRACTICE TEST PAPER-9 Time: 60 Min.

M.M.: 43 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.4 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

 f (x) = x2 ,oa g (x) = cx3 ds vkjs[k nks fcUnqvksa ij izfrPNsn djrs gSA ;fn vUrjky 0,  1 1 2 gS] rks   2  dk eku gS 3 c c  (A) 20 (B) 2

Q.2

[4 × 3 = 12]

x2

vfrijoy;

a2



y2 b2

(C) 6

1 ij {ks= dk {ks=Qy c 

(D) 12

 1 ij nks fcUnqvksa (– 4, 2) rFkk (2, 1) ls [khaph xbZ Li'kZ js[kkvksa dh Li'kZ thok,sa ledks.k

ij gS] rks vfrijoy; dh mRdsUnzrk gksxh (A) Q.3

7 2

(B)

5 3

(C)

3 2

(D)

2

ekuk ,d FkSys esa rhu xsans gS ftuesa ,d dkyh] ,d yky rFkk ,d lQsn gSA ,d xsan FkSys esa ls fudkyh tkrh gS] blds jax dk irk yxkdj bldks FkSys esa izfrLFkkfir fd;k tkrk gSA bl izfØ;k dh 5 ckj iqujko`fÙk gksrh gS rFkk ifj.kkeLo:i de ls de nks yky rFkk de ls de nks lQsn xsna s ik;h tkrh gSA ;fn bl ?kVuk dh Ákf;drk

k 343

gS] rks k dk eku gS

Q.4

(A) 50

(B) 40

a11 a12 ekuk 1 = a 21 a 22 a 31 a 32

a13 b11 b12 a 23 , 1  0; 2 = b 21 b 22 a 33 b 31 b 32

rFkk

c11 c12 3 = c 21 c 22 c 31 c 32

(D) 20 b13 b 23 , tgk¡ bij, aij dk lg[k.M gS  i, j =1,2,3 b 33

c13 c 23 , tgk¡ cij , bij dk lg[k.M gS  i, j = 1, 2, 3. c 33

gS] rks fuEu esa ls dkSulk ,d lnSo lR; gS (A) 1, 2, 3 lekUrj Js.kh esa gS 2 (C) 1 

(C) 25

3

(B) 1, 2, 3 xq.kksÙkj Js.kh esa gS (D) 1 =

2

2 3

[PARAGRAPH TYPE] Q.5 to Q.7 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 3 = 09]

Paragraph for Question no. 5 to 7

ekuk C1 o C2 lfEeJ lery ij nks oØ gS tks fuEu izdkj ifjHkkf"kr gS C1 : z + z = 2 | z – 1 | ; C2 : arg (z + 1 + i) =  tgk¡  vUrjky (0, ) ls bl izdkj lEcfU/kr gS fd oØ C1 ,oa C2 ,d nwljs dks P(z0) ij Li'kZ djrs gSA Q.5

| z0 | dk eku gS (A) 2

(B) 4

(C)

2

(D) 2 2 PAGE # 20

Q.6

,d d.k ,d fcUnq P(z0) ls izkjEHk gksrk gSA ;g ewy fcUnq ls 2 bdkbZ {kSfrt xfr djrk gS rc ewy fcUnq ls 3 bdkbZ m/okZ/kj xfr djrk gS ,oa ,d fcUnq Q(z1) ij igq¡prk gSA ;fn z1 = x1 + i y1 gS] rks (x1 + y1) cjkcj gS (A) 5

Q.7

(B) 7

(C) 8

(D) 9

;fn P(z0) ewy fcUnq ds lkis{k 2 dks.k ls nf{k.kkorZ fn'kk esa ?kqerk gS] rks C1 rFkk P(z0) o Q(z0') dks feykus okyh js[kk }kjk ifjc) {ks=Qy gS (tgk¡ (z0') ?kw.kZu ds i'pkr~ P(z0) dh u;h fLFkfr gS) (A)

2 oxZ bdkbZ 3

(B) 1 oxZ bdkbZ

(C) 2 oxZ bdkbZ

(D)

5 oxZ bdkbZ 6

PART-B [MATRIX TYPE] Q.1 has four statements (A, B, C, D) given in Column-I and five statements (P, Q, R, S, T) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. Q.1

[3+3+3+3 = 12] LrEHk-II (P) 0

LrEHk-I (A)

;fn o`Ùk A(1, 2), B(2, 3) ls xqtjrk gS ,oa U;wure laHko ifjeki okyk gS o`Ùk x2 + y2 + 2x + 2ky = 26 dks yEcdks.kh; izfrPNsn djrk gS] rks k cjkcj gS 1  cos1  cos(1  cos x )  ifjfer gS] rks a dk eku gks ldrk gS xa

(B)

;fn Lim x 0

(C)

1 3 p  2   8  O;wRØe.kh; gS] rks p gks ldrk gS ;fn vkO;wg A = 2 4  3 5 10 

(D)

;fn y (t), ( t  1)

dy  ty  1 , y (0) = – 1 dk ,d gy gS] dt

(Q)

1

(R)

2

(S)

3

(T)

4

rks 2 y(1)  3 cjkcj gS PART-C [INTEGER TYPE] Q.1 and .2 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)[2 × 5 = 10] Q.1

,d lef}ckgq f=Hkqt ABC, ftlesa AB = AC, BC = 4 rFkk ABC = 30° gS] ij fopkj dhft;sA rhu fcUnq A1, B1 rFkk C1 f=Hkqt ABC ds vUr%o`Ùk S1 ftldh f=T;k r1 gS ¼tSlk fd fp= esa n'kkZ;k x;k gS½] ij bl izdkj fy;s tkrs gS fd A1 Bhd A ds uhps gS rFkk A1B1, AB ds lekUrj gS ,oa A1C1, AC ds lekUrj gSA f=Hkqt A1B1C1 ds vUr%o`Ùk S2 dh f=T;k r2 gS] rks (2r2 + 7r1) Kkr dhft;sA 

Q.2

;fn

 0

A A1 I B1 B



C1 C

 8 sin 2 x  2    g ( x )  1 g ( x )  dx = 6, tgk¡ g(x) vUrjky (0, ) esa ,d lrr~ /kukRed Qyu gS] rks  

(0, ) esa g(x) dk vf/kdre eku Kkr dhft;sA

PAGE # 21

PRACTICE TEST PAPER-10 Time: 60 Min.

M.M.: 45 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.3 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 3 = 09]

28  1 r2 n C  ;fn   gS] rks n dk eku gS  r 6 r  0  r 1  n

Q.1

(A) 4

(B) 5

(C) 6

(D) 7

1

Q.2

fuf'pr lekdy

 tan

1

1

1

1

sin cos x   cot cossin x  dx dk eku gS

1

(A) 0 Q.3

(B) – 1

(C) 1

(D) 

;fn thok] tks fd ijoy; y2 = 4x ds fcUnq (, ),   0 ij vfHkyEc gS] }kjk ukfHk ij vUrfjr dks.k  gS] rks x-v{k dh ?kukRed fn'kk ds lkFk dks.k cukus okyh ,oa (1, 2) ls xqtjus okyh js[kk dk lehdj.k gS (A) y = 2

(B) x + 2y = 5

(C) x + y = 3

(D) x = 1

[PARAGRAPH TYPE] Q.4 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 3 = 09]

Paragraph for question nos. 4 to 6 vkB izfrfuf/k gS ftuesa 4 vesfjdu, 1 fczfV'k, 1 pkbuht, 1 Mp rFkk 1 bftfif'k;u gSA bu izfrfuf/k;ksa ds ;kn`PN;k

;qXe cuk;s tkrs gSA Q.4

og Ákf;drk ftlesa dksbZ Hkh nks izfrfuf/k leku ns'k ds ;qXe esa ugha gks] gksxh (A)

Q.5

(B)

8 35

(C)

16 35

(D)

24 35

(D)

2 35

og Ákf;drk ftlesa izfrfuf/k leku ns'k ds nks ;qXe cukrs gS] gksxh (A)

Q.6

6 35

3 35

(B)

6 35

(C)

8 35

og Ákf;drk ftlesa Bhd nks izfrfuf/k leku ns'k ds lkFk&lkFk ;qXe cukrs gS] gksxh (A)

5 35

(B)

8 35

(C)

24 35

(D) dksbZ ugha

PAGE # 22

PART-B [MATRIX TYPE] Q.1 has four statements ( A, B, C, D) given in Column-I and five statements (P, Q, R, S, T) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. Q.1

[3+3+3+3 = 12] Column-II

Column-I 4

(A) (B)

zi lehdj.k    1 ds gyksa dh la[;k i   1 gS



(P)

1

,d thok PQ ijoy; y2 = 4ax ds P ij vfHkyEc gS rFkk 'kh"kZ ij ledks.k vUrfjr djrh gSA ;fn SQ = SP, tgk¡ S ukfHk gS] rks dk eku gS

(Q)

2

(R)

3

(S)

4

(T)

5

1

(C)

t

;fn

sin x

(D)



 z i 

2

 1    f ( t ) dt  1  sin x , x   0,  gS] rks f   cjkcj gS  2  2

lehdj.k 4y2 + 2 cos2x = 4y – sin2x dks larq"V djus okys Øfer ;qXeksa (x, y) dh la[;k] tgk¡ x, y  [0, 2] gS, gksxh

PART-C [INTEGER TYPE] Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)[3 × 5 = 15] Q.1

Re(z) – 2 = | z – 7 + 2i | }kjk fn;s x;s vkxZUM lery esa lfEeJ la[;k z ds fcUnqiFk ij fopkj dhft;sA ekuk P(z1) ,oa Q(z2) nks lfEeJ la[;k,sa gS tks fn;s x;s fcUnqiFk dks larq"V djrh gS  z  ( 2  i) 



= ,oa arg  1 (  R) dks Hkh larq"V djrh gS] rks PQ dk U;wure eku Kkr dhft;sA   z 2  ( 2   i)  2

[fVIi.kh : Re(z), lfEeJ la[;k z ds okLrfod Hkkx dks fu:fir djrk gS rFkk i2 = – 1 gS]

Q.2

ekuk oØ y = xn (n 1) ,oa js[kk x = 0, y = 0 ,oa x = n

;fn



2n A n

n 1

Q.3

n



1 }kjk ifjc) {ks=Qy An gSA 2

1 gS] rks n dk eku Kkr dhft;sA 3

  

;fn u , v, w v'kwU; rFkk vleryh; lfn'k gS] rks Øfer ;qXeksa (p, q) dh la[;k Kkr dhft;s rkfd

3u

        pv pw  – pv w qu  – 2w qv qu  = 0  p, q  R gksA

PAGE # 23

PRACTICE TEST PAPER-11 Time: 60 Min.

M.M.: 49 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.4 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

[4 × 3 = 12]

;fn Qyu f(x) = x3 – ax2 + 2x vUrjky [0, 2] esa LMVT ds izfrcU/k dks larq"V djrk gS rFkk x =

1 ij oØ 2

y = f(x) dh Li'kZ js[kk oØ ds] x = 0 ,oa x = 2 ij dksfV;ksa ds lkFk izfrPNsnu fcUnqvksa dks feykus okyh thok ds lekUrj gS] rks a dk eku gS (A) Q.2

9 4

(B)

13 4

15 4

(D)



(B) 2

(C) 3

(D) 4

ekuk z rFkk w lfEeJ la[;k;sa bl izdkj gS fd z + w = 0 rFkk z2 + w2 = 1 gS] rks z  w (A) 1

Q.4

(C)

lery 2x – y – z = 4 ds yEcor~ rFkk ewy fcUnq ls xqtjus okyh ljy js[kk ds lkFk lery r · (3ˆi  5ˆj  2kˆ )  6 dk izfrPNsn fcUnq (x0, y0, z0) gks] rks (2x0 – 3y0 + z0) dk eku gS (A) 0

Q.3

11 4

(B)

2

(C) 2 2

cjkcj gS

(D) 2

;fn z ,d lfEeJ la[;k bl izdkj gS fd |z – 2| + |z – 4| = 5 lSno ,d nh?kZo`Ùk dks iznf'kZr djrk gS] tgk¡   R+ gS] rks  ds iw.kkZad ekukas dh la[;k gksxh (A) 2

(B) 3

(C) 4

(D) 5

[PARAGRAPH TYPE] Q.5 to Q.7 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Q.5

Paragraph for question nos. 5 to 7 okLrfod eku Qyu f : R  R tks f(x) = x2 e–x }kjk ifjHkkf"kr gS] ij fopkj dhft;sA fuEu esa ls dkSulk ,d dFku lR; gS ? (A) f(x) dk x = 0 ij LFkkuh; vf/kdre ,oa x = 2 ij LFkkuh; U;wure gSA (B) f(x) dk x = 0 ij LFkkuh; U;wure ,oa x = 2 ij LFkkuh; vf/kdre gSA (C) Lim f ( x ) = 1.

(D) f(x) ,d le Qyu gSA

x  2e x

Q.6

[3 × 3 = 09]

ekuk g(x) =

 0

f ' (t ) dt gks] rks 1  t2

(A) g(x) vUrjky (–, 0) esa o/kZeku ,oa (0, ) esa áleku gS (B) g(x) dk x = 0 ij LFkkuh; U;wure gS (C) g(x) vUrjky (–, 0) esa áleku ,oa (0, ) esa o/kZeku gS (D) g(x) dk x = 0 ij uk rks vf/kdre gS uk gh U;wure Q.7

lehdj.k 4x2 e–x – 1 = 0 ds gyksa dh la[;k gS (A) 0

(B) 1

(C) 2

(D) 3

PAGE # 24

[MULTIPLE CORRECT CHOICE TYPE] Q.8 and Q.9 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct.[2 × 4 = 08] Q.8

fuEufyf[kr esa ls dkSulk@dkSuls dFku nks ?kVukvksa A rFkk B ds fy, lR; gS? (A) ekuk fd P(A/B) = P(B/A), P(A  B) = 1 rFkk P(A  B) > 0 gS] rks P(A) > (B) ;fn P(BC) =

1 . 2

1 1 5 rFkk P(A/B) = gS] rks P(A) dk vf/kdre eku gSA 4 2 8

1 1 2 rFkk P(B) = gS] rks P(AC  BC)C + P(AC  BC)C = . 2 3 3 (D) ;fn A, B dk mileqPp; gS] rks P(B/A) dk eku 1 gS] tgk¡ P(A)  0 gSA

(C) ;fn P(A) =

Q.9





ekuk vkO;wg A ds izFke ,oa f}rh; iafDr lfn'k r1  1 1 3 ,oa r2  2 1 1 gS ,oa ekuk rhljk iafDr lfn'k    r1 ,oa r2 ds lery esa gS ,oa r2 ds yEcor~ gS rFkk ifjek.k

5 okyk gS] rks fuEu esa ls dkSulk@dkSuls lR; gS\

[fVIi.kh : Tr. (P) vkO;wg P ds VSªl dks fu:fir djrk gSA]     (A) Tr. (A) = 3 (B) r2 , r3 , oa r2  r3 }kjk cus lekUrj "kBQyd dk vk;ru 30 gS       (C) iafDr lfn'k js[kh; ijrU= gS (D) r1  r2 r2  r3 r3  r1  = 0 PART-C [INTEGER TYPE] Q.1 to Q.4 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [4 × 5 = 20] Q.1

k > 0 dk eku Kkr dhft;s ftlds fy;s ijoy;ks y = x – kx2 ,oa y =

x2 ds e/; ifjc) {ks= dk {ks=Qy k

vf/kdre gSA Q.2

fcUnq P (3h – 2, 3k) dk fcUnqiFk] tgk¡ (h, k) o`Ùk x2 + y2 – 2x – 4y – 4 = 0 ij fLFkr gS] ,d vU; o`Ùk gS] rks bldh f=T;k Kkr dhft;sA

Q.3

ekuk an (n  1), x dk eku gS ftlds fy;s

2x

e

tn

dt (x > 0) vf/kdre gSA

x

;fn L = Lim ln (a n ) gS] rks e– L dk eku Kkr dhft;sA n 

Q.4

;fn A dksfV 3 dk oxZ vkO;wg gS bl izdkj gS fd det.(A) = 2 gS] rks det.((adj. A–1)–1) Kkr dhft;sA [fVIi.kh : adj. P oxZ vkO;wg P ds lg[k.Mt dks fu:fir djrk gS]

PAGE # 25

PRACTICE TEST PAPER-12 Time: 60 Min.

M.M.: 44 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.4 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

[4 × 3 = 12]

x 2 y2   1 (a > b) ds fdlh fcUnq ij Li'kZ js[kk dk mRdsUnzh; dks.k 60° gS rFkk tks lgk;d o`Ùk dks a 2 b2

nh?kZo`Ùk

L rFkk M ij feyrh gSA ;fn LM dsUnz ij ledks.k cukrh gS] rks nh?kZo`Ùk dh mRdsUnzrk gksxh (A) Q.2

(B)

51 101

(B)

(C)

3 7

50 101

(C)

(D)

51 100

;fn A, B rFkk C fu''ks"kh ?kVuk,sa gS rFkk tks P(A  B  C ) = P(A  C) = (A)

Q.4

2 7

1 2

,d ckWDl esa 100 xsna s gSA ckWDl esa lHkh 'osr (white) ,oa v'osr (non-white) xsanksa dh la[;k le laHkkoh gSA ckWDl esa ,d 'osr (white) xsan Mkyh tkrh gS ,oa ckWDl dks fgyk;k tkrk gSA vc ckWDl ls ,d xsan fudkyh tkrh gS] rks fudkyh xbZ xsan ds 'osr (white) gksus dh izkf;drk gS (A)

Q.3

1 7

(D)

1 51

1 1 , P(B  C) – P(A  B  C) = rFkk 5 15

1 dks larq"V djrh gS] rks P(C  (A  B)' ) cjkcj gS 10

17 30

(B)

18 30

(C)

19 30

(D)

20 30

ekuk  bdkbZ dk lfEeJ ?kuewy gS rFkk 0 < arg() < 2gSA ,d fu"i{kikrh ikalk rhu ckj mNkyk tkrk gSA ;fn a, b, c ikals ij izkIr la[;k,sa gS] rks izkf;drk] fd (a + b + c2) (a + b2 + c) = 1, cjkcj gS (A)

1 18

(B)

1 9

(C)

5 36

(D)

1 6

[PARAGRAPH TYPE] Q.5 to Q.7 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 3 = 09]

Paragraph for question nos. 9 to 11     8ˆi ekuk L1 : r  2ˆi  ˆj  kˆ   ˆi  2ˆj  kˆ rFkk L2 : r    3ˆj  kˆ    2 ˆi  ˆj  kˆ lef"V (space)  3 



 







esa nks js[kk;sa gS] tgk¡ ,  R gSA ewy fcUnq ls xqtjus okyh ,d js[kk L ] js[kkvksa L1 ,oa L2 dks Øe'k% P rFkk Q ij feyrh gSA Q.9

ekuk M (1, 2, 3) lef"V (space) esa fLFkr ,d fcUnq gS rFkk js[kk L1 ij fLFkr dksbZ fcUnq N gS] rks  ds fdl  eku ds fy, lfn'k MN lery r · ˆi  4ˆj  3kˆ = 1 ds lekUrj gS





3 7 5 2 (B) (C) (D) 4 12 8 9  lery r · ˆi  ˆj  kˆ + 1 = 0 ds yEcor~ rFkk fcUnqvksa P rFkk Q ls xqtjus okys lery dk lehdj.k gS     (A) r · 2ˆi  3ˆj = 0 (B) r · 3ˆi  2ˆj = 0 (C) r · ˆi  ˆj = 0 (D) r · ˆi  ˆj = 0

(A) Q.10





















PAGE # 26

Q.11

;fn O ewy fcUnq gS rFkk A (0, – 1, 3) , B (2, 0, 5) gks] rks prq"Qyd OPAB dk vk;ru gksxk (A)

32 3

(B)

51 2

(C)

17 2

(D)

13 3

[MULTIPLE CORRECT CHOICE TYPE] Q.8 and Q.9 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [2 × 4 = 8] Q.8

,d flDdk rhu ckj mNkyk tkrk gSA ?kVukvksa ij fopkj dhft;s : A : izFke mNky esa fpÙk (Heads) B : f}rh; mNky esa iV (Tails) C : r`rh; mNky esa fpÙk (Heads) D : Bhd ,d fpÙk (Head) vk;s E : lHkh 3 ifj.kke leku gS fuEu esa ls dkSu ls dFku lR; gS? (A) P(C), P(D) ,oa P(E) (blh Øe es)a lekUrj Js.kh esa gSA (B) A, B, C fu''ks"k ?kVuk,sa gSA (C) A, B, C LorU= gSA (D) A, B, C le laHkkoh gSA

Q.9

ekuk  > 0,  > 0 lehdj.k x2 + px + q = 0 ds ewy gSA ,

1 ,  lehdj.k x2 + p1x + q1 = 0 ds ewy gS rFkk 

1 lehdj.k x2 + p2x + q2 = 0 ds ewy gS] rks fuEu esa ls dkSulk@dkSuls lEcU/k lR; gS 

(A) q1 q2 = 1 (C)

(B) p1 + p2 =

p (q + 1) q

(D) (qp1 – qp2)2 = (p2 – 4q) (q + 1)2

q q2  p q2  q = 0

PART-C [INTEGER TYPE] Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)[3 × 5 = 15] Q.1

ekuk ABC ,d U;wu dks.k f=Hkqt bl izdkj gS fd a = 14, sin B =

12 rFkk c, a, b (blh Øe esa) 13

,d lekUrj Js.kh cukrs gS] rks f=Hkqt ABC ds vUrfuZfgr o`Ùk dh f=T;k Kkr dhft,A [fVIi.kh : iz;qDr fd;s x;s lHkh izrhdksa dk f=Hkqt ABC esa lkekU; vFkZ gSA]

Q.2

n n

n n

 n  n

ekuk Sn =  0   1  +  1   2  + ... +  n 1  n  , tgk¡ n N gSA          Sn 1

15

;fn S = gS] rks n ds lHkh laHko ekuksa dk ;ksxQy Kkr dhft,A 4 n Q.3

;fn lery r .(ˆi  ˆj  kˆ ) = 1, r .(ˆi  2aˆj  kˆ ) = 2 ,oa r .(aˆi  a 2ˆj  kˆ ) = 3 ,d js[kk esa izfrPNsn djrs gS] 





rks a ds okLrfod ekuksa dh la[;k Kkr dhft,A

PAGE # 27

PRACTICE TEST PAPER-13 Time: 60 Min.

M.M.: 49 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.4 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

ekuk x, y R lehdj.k tan–1x + tan–1y + tan–1(xy) = (A) 1+

Q.2

3 2

(B) –1 +

3 2

[4 × 3 = 12]

11 dy dks lar"q V djrs gS] rks x = 1 ij dk eku gS 12 dx

(C) –1 –

3 2

3 2

(D) 1 –

 1  3  sin  | x | ; x  0 ekuk f(x) =  . x  0 ; x0

gS] rks x = 0 ij f dk gS (A) LFkkuh; vf/kdre (B) LFkkuh; U;wure (C) uk rks LFkkuh; vf/kdre uk gh LFkkuh; U;wure (D) vlkarR;rk dk fcUnq Q.3

vody lehdj.k ds fy;s x dy = y (dx + y dy), y > 0, y (1) = 1 rFkk y () = – 3 gS] rks  cjkcj gS (A) – 5

Q.4

(B) 5

(C) 15

(D) –15

tan 2 x  4 tan x  9 ds vf/kdre ,oa U;wure eku gS] rks (M + m) cjkcj gS 1  tan 2 x (B) 14 (C) 10 (D) 8

;fn M ,oa m Qyu f (x) = (A) 20

[PARAGRAPH TYPE] Q.5 to Q.7 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[3 × 3 = 09]

Paragraph for question nos. 5 to 7 ekuk ,d vfrijoy; nh?kZo`Ùk 16x2 + 25y2 = 400 dh ukfHk ls xqtjrk gSA bl vfrijoy; dh vuqizLFk rFkk

Q.5

la;qXeh v{k fn;s x;s nh?kZo`Ùk ds nh?kZ v{k rFkk y?kq v{k ds lkFk lEikrh gSA vfrijoy; dh mRdsUnzrk] nh?kZo`Ùk dh mRdsUnzrk dk O;qRØe gSA fuEufyf[kr esa ls dkSulk dFku lR; gS ? (A) vfrijoy; ds 'kh"kZ (±3, 0) gSA (B) vfrijoy; dh ukfHk;ksa ds e/; nwjh 6 gSA (C) vfrijoy; dh fu;rkvksa dh lehdj.ksa x = 

Q.6

(D) dksbZ ugha

vfrijoy; ij fdlh fcUnq ls nh?kZo`Ùk ds lgk;d o`Ùk ij Li'kZ js[kk,¡ [khaph tkrh gS] rks Li'kZ thok ds e/; fcUnq dk fcUnqiFk gksxk 2

Q.7

5 gSA 9

2

 x 2 y2  x 2  y2     (A)   25  9 16 

 x2 y 2   x 2  y 2     (B)  16   25   9

 x2 y 2  x 2  y 2  (C)  9  16   25  

 x2 y 2   x 2  y 2    (D)  16   25   9

2

2

vfrijoy; rFkk blds la;qXeh vfrijoy; dh ukfHk;ksa dks feykus ls cus prqHkqZt dk {ks=Qy gS (A) 337

(B) 674

(C) 50

(D) 25 PAGE # 28

[MULTIPLE CORRECT CHOICE TYPE] Q.8 and Q.9 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [2 × 4 = 8] Q.8

leqPp; {1, 2, 3, 4 ,..., n} ls ;kn`PN;k ,d la[;k dk p;u fd;k tkrk gSA ekuk fudkyh xbZ la[;k ds 2 ls HkkT; gksus dh ?kVuk E1 gS rFkk fudkyh xbZ la[;k ds 3 ls HkkT; gksus dh ?kVuk E2 gS] rks (A) E1 ,oa E2 lnSo Lora= gS (B) E1 ,oa E2 Lora= gS ;fn n = 6k (k N) (C) E1 ,oa E2 Lora= gS ;fn n = 6k + 2 (k N) (D) E1 ,oa E2 ijra= gS ;fn n = 10

Q.9

;fn 2xy dy = (x2 + y2 + 1)dx, y(1) = 0 ,oa y (x0) = 3 gS] rks x0 gks ldrk gS (A) 2

(B) – 2

(C) 3

(D) – 3

PART-C [INTEGER TYPE] Q.1 to Q.4 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [4 × 5 = 20] Q.1

,d FkSys esa 18 flDds gSA buesa ls 13 flDds fu"i{kikrh gS rFkk 'ks"k ikap flDds Hkkjh gS ,oa bl izdkj i{kikrh gS fd muds 'kh"kZ (head) vkus dh izkf;drk vkus dh izkf;drk

1 1 , iqPN (tail) vkus dh izkf;drk gS rFkk buds dksj (edge) ds vuqfn'k 2 3

1 gSA FkSys ls ,d flDdk ;kn`PN;k fudkyk tkrk gS ,oa rhu ckj mNkyk tkrk gSA blds rhuksa 6

p

voljksa ij 'kh"kZ n'kkZus dh izkf;drk q ds cjkcj gS] tgk¡ p o q lg vHkkT; gS] rks (p + q) dk eku Kkr dhft,A Q.2

;fn 2y2 – 2x + 1 = 0 ,oa 2x2 – 2y + 1 = 0 ds e/; y?kqÙke nwjh d gS] rks lehdj.k | sin  | = 2 2 d ds vUrjky [– , 2] esa gyksa dh la[;k Kkr dhft,A

Q.3

,d Qyu y = f(x) tks x f '(x) – 2f(x) = x4 f 2(x),  x > 0 ,oa f(1) = – 6 dks larq"V djrk gS] rks f '  35  dk

  

1

  

eku Kkr dhft,A Q.4

y = x2 ds vkjs[k ij fopkj dhft,A ekuk A izFke prqFkkZa'k esa vkjs[k ij fLFkr dksbZ fcUnq gSA ekuk x-v{k rFkk y = x2 ds vkjs[k ds fcUnq A ij Li'kZ js[kk dk izfrPNsn fcUnq B gSA ;fn y = x2 ds vkjs[k rFkk js[kk[k.M OA }kjk  p  q

fufeZr vkÑfr dk {ks=Qy] f=Hkqt OAB (tgk¡ O ewy fcUnq gS ) ds {ks=Qy dk   xquk gS] rks (p + q) dk U;wure eku Kkr dhft,A tgk¡ p, q  N.

PAGE # 29

PRACTICE TEST PAPER-14 Time: 60 Min.

M.M.: 55 [SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.4 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

,d js[kk L o`Ùk x2 + y2 = 1 rFkk ijoy; y2 = 4x dh mHk;fu"B Li'kZ js[kk gSA ;fn ;g js[kk /kukRed x-v{k ds lkFk dks.k  cukrh gS] rks tan2 cjkcj gS (A) 2 sin 18°

Q.2

[4 × 3 = 12]

(B) 2 sin 15°

(C) cos 36°

(D) 2 cos 36°

ekuk f : R R ,d f}vodyuh; Qyu gS] tks f(2) = –1, f '(2) = 4 ,oa 3

 (3  x ) f ''(x) dx = 7 dks larq"V djrk gS] rks

f(3) dk eku fuEu esa ls fdl vUrjky esa fLFkr gS

2

(A) (0, e) Q.3

Q.4

(B) (e, e2)

(C) (e2, e3)

(D) (e3, e4)

dksfV 3 × 3 ds lHkh laHko lefer vkO;wgksa] ftudh izR;sd izfof"V;k¡ 0 ;k 1 gS rFkk ftudk vuqjs[k (trace) 1 gS] dh la[;k gS (A) 24

(B) 48

(C) 192

;fn z oØ arg(z + i) =

 ij fLFkr gks] rks | z + 4 – 3i | + | z – 4 + 3i | dk U;wure eku gksxk 4

[uksV: i2 = – 1] (A) 5

(B) 10

(C) 15

(D) 512

(D) 20

[REASONING TYPE] Q.5 and Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. [2 × 3 = 06]      Q.5 dFku-1: ekuk a  ˆi  2ˆj  3kˆ rFkk b  2ˆi  ˆj  kˆ gS] rks a  x  a  b rFkk a · x  0 dks larq"V  djus okys lfn'k x dh yEckbZ 10 gSA           dFku-2: ;fn p, q, r v'kwU; fofHkUu lfn'k bl izdkj gks fd p  q  p  r gS] rks p ,  q  r  ds

lekUrj gSA (A) dFku&1 lR; gS] dFku&2 lR; gSA dFku&2 dFku&1dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS dFku&2 lR; gS dFku&2 dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; dFku&2 vlR;A (D) dFku&1 vlR; rFkk dFku&2 lR;A Q.6

ekuk f(x) = x  1 sin (x) gSA dFku-1: f(x) lHkh okLrfod x ds fy, vodyuh; gSA dFku-2: f(x) dk x = 1 ij u rks LFkkuh; vf/kdre gS vkSj uk gh LFkkuh; U;wure gSA (A) dFku&1 lR; gS] dFku&2 lR; gSA dFku&2 dFku&1dk lgh Li"Vhdj.k gSA (B) dFku&1 lR; gS dFku&2 lR; gS dFku&2 dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; dFku&2 vlR;A (D) dFku&1 vlR; rFkk dFku&2 lR;A

PAGE # 30

[MULTIPLE CORRECT CHOICE TYPE] Q.7 to Q.9 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [3 × 4 = 12] Q.7

ekuk A(–1, 0) ,oa B(2, 0) x-v{k ij nks fcUnq gSA ,d fcUnq 'M', xy-lery esa (x-v{k ds vU;) bl izdkj xfreku gS fd MBA = 2MAB gS] rks fcUnq 'M' ,d 'kkado ds vuqfn'k xfr djrk gS ftldh (A) mRdsUnzrk 2 gS (B) 'kh"kZ (±3, 0) gS (C) ukfHkyEc dh yEckbZ 6 gS

Q.8

(D) fu;rkvksa dk lehdj.k x = ±





1  x  · x · 1  cos 2x  , 0,

ekuk f(x) = 

1 gS 2

0  x 1 . x0

;fn x [0, 1] ds fy;s f (x) ij jksys izes; ykxw gksrh gS] rks  gS (A) – 2 Q.9

(B) – 1

(C)

1 2

(D) 1

ekuk 'L' ,d fcUnq (t, 2) gS rFkk 'M', y-v{k ij ,d fcUnq bl izdkj gS fd 'LM' dh izo.krk –t gS] rks 'LM' ds e/; fcUnq dk fcUnqiFk] tSls t okLrfod ekuksa ij ifjofrZr gksrk gS] ,d ijoy; gS] ftldk (A) 'kh"kZ (0, 2) gS (B) ukfHkyEc dh yEckbZ 2 gS  17   gS (C) ukfHk  0, 8 

(D) fu;rk dk lehdj.k 8y – 15 = 0 gS

PART-B [MATRIX TYPE] Q.1 has three statements (A, B, C) given in Column-I and four statements (P, Q, R, S) given in Column-II. Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II. Q.1 (A)

LrEHk-1 ekuk z ,d v'kwU; lfEeJ la[;k gS vkSj  ( 1) bdkbZ dk vokLrfod ?kuewy gSA ;fn A(z), B(z) rFkk C (2z) }kjk cus f=Hkqt dk {ks=Qy 48 3 gks] rks

[3+3+3+3 = 12] LrEHk-II (P) 2

z cjkcj gksxk

(B)

o`Ùk x2 + y2 = 4 dh ,d Li'kZ js[kk vfrijoy; x2 – 2y2 = 2 dks P rFkk Q fcUnqvksa ij izfrPNsn djrh gSA ;fn PQ ds e/; fcUnq dk fcUnqiFk (x2 – 2y2)2 = (x2 + 4y2) gks] rks  cjkcj gS

(C)

ukfHk;ksa S rFkk S' ls nh?kZo`Ùk

x 2 y2  1 4 9

(Q)

4

(R)

6

(S)

8

dh fdlh Li'kZ js[kk ij yEc dh yEckbZ;k¡ Øe'k% a rFkk c gS] ac

rks

 {2x}dx

dk eku cjkcj gS

ac

[uksV : {k}, k dk fHkUukRed Hkkx Qyu gS]

PAGE # 31

PART-C [INTEGER TYPE] Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [3 × 5 = 15] Q.1

ekuk B okLrfod izfof"V;k okyh dksfV 3 × 3 dh fo"ke lefer vkO;wg gSA fn;k gS fd I – B ,oa I + B O;qRØe.kh; vkO;wg gSA ;fn A = (I + B) (I – B)–1, tgk¡ det.(A) > 0 gS, rks det.(2A) – det. (adj A) dk eku Kkr dhft,A [fVIi.kh : det.(P) oxZ vkO;wg P ds lkjf.kd dks ,oa det.(adj (P) oxZ vkO;wg P ds lg[k.Mt ds lkjf.kd dks fu:fir djrk gSA]

Q.2

ekuk fcUnq B(– 3, 5, 2) ls xqtjus okyh js[kk ls fcUnq A(– 2, 3, 1) dh yEcor~ nwjh p gS rFkk js[kk x, y rFkk z-v{k dh /kukRed fn'kk ds lkFk cjkcj dks.k cukrh gS] rks 30p2 dk eku Kkr dhft,A

Q.3

ekuk F(x) = max. (sin x, cos x) gS] rks

 4 2

10

 F(x ) dx

dk eku Kkr dhft,A

10

PAGE # 32

ANSWER

Q.1 Q.5 Q.9 Q.1 Q.1

Q.1 Q.1 Q.1

Q.1 Q.6

KEY

PRACTICE TEST PAPER-1 B Q.2 C Q.3 D Q.4 D B Q.6 B Q.7 B Q.8 A ACD Q.10 ABC PART-B (A) Q; (B) P; (C) R PART-C 0010 Q.2 0001 Q.3 0009

PRACTICE TEST PAPER-2 A Q.2 B Q.3 C Q.4 B Q.5 A Q.6 C Q.7 BCD Q.8 BC PART-B (A) P ; (B) T; (C) S; (D) Q 0001

Q.2

C B

PRACTICE TEST PAPER-3 A Q.3 B Q.4 B Q.5 D BC Q.8 AB

Q.2 Q.7

0086

PART-C Q.3 0009

PART-B Q.1

(A) T (B) P (C) Q (D) R

Q.1

0040

Q.2

0003

PART-C Q.3 0003

Q.1

PRACTICE TEST PAPER-4 A Q.2 A Q.3 A Q.4 A Q.5 B Q.6 ABD Q.8 D Q.9 A PART-B (A) R ; (B) S ; (C) Q PART-C 0925 Q.2 0050 Q.3 0020 Q.4 0005 or 0007

Q.1 Q.7

C AC

Q.1 Q.7 Q.1

Q.2 Q.8

C CD

PRACTICE TEST PAPER-5 Q.3 D Q.4 C Q.5 B Q.6

ABD

A

PART-B Q.1 Q.1

(A) T ; (B) R ; (C) Q ; (D) P 0008

Q.2

0061

PART-C Q.3 0010

PAGE # 33

Q.1 Q.8 Q.1

PRACTICE TEST PAPER-6 D Q.2 B Q.3 D Q.4 B Q.5 C Q.6 A Q.9 D Q.10 AB Q.11 AC PART-B (A) Q ; (B) P ; (C) S

D

Q.7

C

Q.5

D

C

Q.5

C

PRACTICE TEST PAPER-10 PART-A Q.2 D Q.3 D Q.4 B

Q.5

A

PART-C Q.1

Q.1 Q.1

0002

D 0002

Q.1 Q.6

B B

Q.1

0013

Q.2

Q.2 Q.2

A

0003

PRACTICE TEST PAPER-7 Q.3 A Q.4 AD Q.5 BCD Q.6

0020

Q.3

PART-C 0021 Q.4 0007

Q.5

0025

Q.6

PRACTICE TEST PAPER-8 PART-A Q.2 B Q.3 D Q.4 C Q.7 D Q.8 AC Q.9 ACD PART-C Q.2 0003 Q.3 0025

BC 0017

PRACTICE TEST PAPER-9 PART-A Q.1 Q.6

C B

Q.2 Q.7

C A

Q.3

A

Q.4

PART-B Q.1 Q.1

(A) S; (B) P, Q, R, S, T; (C) P,Q,R,S; (D) R PART-C 0004 Q.2 0004

Q.1 Q.6

B C

Q.1

(A) R ; (B) R ; (C) Q ; (D) Q

PART-B PART-C Q.1

10

Q.2

0002

Q.3

1

PAGE # 34

PRACTICE TEST PAPER-11 PART-A Q.2 D Q.3 B Q.4 B Q.7 D Q.8 ABD Q.9 BCD PART-C

Q.1 Q.6

C A

Q.1

0001

Q.2

Q.1 Q.6

B D

Q.1

0004

PRACTICE TEST PAPER-12 PART-A Q.2 A Q.3 C Q.4 C Q.7 C Q.8 ACD Q.9 ABCD PART-C Q.2 0006 Q.3 0000

Q.1 Q.6

C B

Q.1

0009

Q.1 Q.6 Q.1 Q.1

0009

Q.3

0002

Q.4

Q.5

B

Q.5

B

Q.5

A

Q.5

D

0004

PRACTICE TEST PAPER-13 PART-A Q.2 B Q.3 D Q.4 C Q.7 C Q.8 BCD Q.9 AB PART-C Q.2 0003 Q.3 0008 Q.4 0005

PRACTICE TEST PAPER-14 PART-A A Q.2 C Q.3 A Q.4 B B Q.7 ACD Q.8 CD Q.9 ACD PART-B (A) S, (B) Q, (C) Q PART-C 0007 Q.2 0140 Q.3 0005

PAGE # 35

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