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GMATHEMATICSxf.kr

2015Rajeev Bansal 's

Class XIICHAPTERWISE

With Solutions

Q UESTION

B ANK

MODEL PAPER(QUESTION-ANSWER WITH MARKING SCHEME)

E SR UU CS C E% S0 S01 NO.1

,l ch ih Mh ifCyds 'kUl

ijh{kk iz'u&iz= osG uohu la'kksf/kr izk:i rFkk ekfo±Gx Ldhe osGvuqlkj lEiw.kZ ikB~;ÿe ij vk/kkfjregRoiw.kZ iz'uksa dk v/;k;@bdkbZ osG ÿe esa ladyufcgkj cksMZ }kjk vk;ksftr xr o"kks± osG ijh{kk iz'u&i= gy lfgrcksMZ ijh{kk 2015 osG fy, pkj izkn'kZ iz'u&i= (gy lfgr)

2015

su

ijh{kk grq ,d acj

gY

osGcqfodih;

'iz uksŸkj lfgr

Strictly according to the latest syllabus prescribed by Central Board of Secondary Education (CBSE), Delhi

and Other State Boards & Navodaya, Kendriya Vidyalayas etc. following CBSE curriculum based on NCERT guidelines.

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Printing and Publishing rights with the Publisher.The material in this publication is copyrighted. No part of this text may be reproduced or

copied in any form or by any means without the written permission of the publisher. Breach of this condition is liable for legal action.

Note : Due care and diligence has been taken while editing and printing the text, the editor, co-ordinator and the publisher of the text hold no responsibility for any mistake that mayhave inadvertently crept in.

Book Code : 8307

© izdk’kd

,l ch ih Mh ifCyds 'kUl] 3/20B, vkxjk&eFkqjk ckbZikl jksM] fudV rqylh flusek] vkxjkµ282 002

nwjHkk"k % (0562) 2854327, 2527707, 3257009, 3208010, 4042977 eksckby % 09358177555, 09412258082-85

QSDl % (0562) 2858183; e-mail : sbpd.publications@gmail.com; website : www.sahityabhawan.com

Rajeev's Model Paper fgUnhµ50 vad

Rajeev's Model Paper fgUnhµ100 vad

Rajeev's Model Paper English : 50 Marks

Rajeev's Model Paper English : 100 Marks

dher % ,d lkS nl #i;k ek= ( 110.00)

Rajeev's Model Paper HkkSfrd foKku

Rajeev's Model Paper jlk;u foKku

Rajeev's Model Paper tho foKku

gekjs vU; ekWMy isij

ISBN : 978-93-5167-995-0

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v/;k; i"B&la[;k1. lEcU/k vkSj Qyu. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1—5

[Relations and Functions]

2. izfrykse f=dks.kferh; Qyu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5—12

[Inverse Trigonometric Functions]

3. vkO;wg . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12—18

[Matrices]

4. lkjf.kd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18—25

[Determinants]

5. lkarR; vkSj vodyuh;rk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25—31

[Continuity and Differentiability]

6. vodyt osG vuqiz;ksx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32—38

[Applications of Derivative]

7. vfuf'pr lekdyu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39—44

[Indefinite Integrals]

8. fuf'pr lekdyu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45—50

[Definite Integrals]

9. ifjcº {ks= dk {ks=Qy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51—58

[Area of Bounded Regions]

10. vody lehdj.k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58—64

[Differential Equations]

11. lfn'k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64—68

[Vectors]

12. f=&foeh; T;kfefr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68—75

[Three-Dimensional Geometry]

13. jSf[kd izksxzkeu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75—81

[Linear Programming]

14. izkf;drk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81—87

[Probability]

lizkn'kZ iz'u&i= : lSV I–IV (OMR 'khV lfgr) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1—21

[Model Paper : Set I–IV] (With OMR Sheet)

lcksMZ ijh{kk iz'u&i= . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (i)

[Board Examination Paper]

fo"k;&lwph

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1 mecyevOe Deewj Heâueve[RELATIONS AND FUNCTIONS]

Yeeie (De) : Jemlegefve… ØeM ve[Objective Type Questions]

➠ yengefJekeâuHeerÙe ØeM ve [Multiple Choice Questions]

efveoxM e—efvecve ØeMvee W keâe Skeâ mener Gòej nw~ efoS ieS efJekeâuHee W ce W mes menere fJekeâuHe hej efveMeeve ueieeÙe W— ieefCele XII (mecyevOe Deewj Heâueve)

Instruction : In the following questions there is onecorrect answer. In each question you have to mark thatcorrect option from the given options :

1. Ùee fo f(x1) = f(x2) ⇒ x1 = x2 ∀ x1. x2 ∈ A lee s f : A → B ke wâmeeHeâueve nesiee? (If f(x1) = f(x2) ⇒ x1 = x2 ∀ x1. x2 ∈ A, then what

type of function is f : A → B?) (B.S.E.B., 2012)

(a) DeeÛÚeokeâ (Onto) (b) Deve ske wâkeâ (Many one)

(c) DeÛej (Constant) (d) Skewâkeâ (One-one)

2. Ùee fo mebe f›eâÙee * hee fjYeee f<ele n w e fkeâ a ∗ b = a2 + b

2 lees (1 * 2) * 5 nw (If

an operation * is defined by a ∗ b = a2 + b

2, then (1 * 2)

* 5 is) : (B.S.E.B, 2012)

(a) 3125 (b) 625 (c) 125 (d) 50.

3. Ùee fo Heâueve f : N → N Fme Øekeâej hee fjYeeef<ele n w e fkeâ f(x) = 2x + 3 leesf

−1(x) = (If f : N → N be defined by f(x) = 2x + 3 then

f −1

(x) =) : (B.S.E.B., 2012)

(a) 2x − 3 (b) x − 3

2

(c) x + 3

2(d) hee fjYeee f<ele veneR (not defined)

4. Ùee fo f(x) = 8x3 Deewj g(x) = x

1/3 lee s fog = (If f(x) = 8x

3 and

g(x) = x1/3

, then fog =) (B.S.E.B., 2012)(a) 3x (b) 9x (b) 4x (d) 8x

5. Ùee fo (If) f(x) + 2f(1 − x) = x2 + 2 ∀ x ∈ R, leye (then) f(x) =

(C.B.S.E., 2010)

(a) x2 − 2 (b) 1

(c) 1

3(x − 2)2

(d) keâe sF& veneR (None)

6. Ùee fo A = {1, 2, 3}, B = {5, 6, 7} leLee f : A → B Skeâ Heâueve S smeen w e fkeâ f(x) = x + 4 lee s f e fvecvee fuee fKele ce W mes e fkeâme Øekeâej keâe Heâueve nw?(If A = {1, 2, 3}, B = {5, 6, 7} and f : A → B is a function

such that f(x) = x + 4, then what type of a function is f )?(a) Deve ske wâkeâ DeeÛÚeokeâ (many-one onto)

(b) DeÛej Heâueve (constant function)

(c) Skewâkeâ DeeÛÚeokeâ (one-one onto)

(d) Devle:#e shee r (into)

7. ceevee e fkeâ A = {1, 2, 3, 4, ...., n} lees e fkeâleve s Skewâkeâ DeeÛÚeefole Heâuevef : A → A heefjYeee f<ele nes mekeâle s n Q ? [Let A = {1, 2, 3, 4, ...., n}.

How many bijective functions f : A → A can be defined] ?

(a) 12

(n!) (b) (n − 1)! (c) n! (d) n

8. x ∗ y = 1 + 12x + xy, ∀ x, y ∈ Q Éeje hee fjYeeef<ele Q hej Skeâ e fÉ-DeeOeejerme be f›eâÙee * keâe r e fJeJe sÛevee keâjW~ leye 2 * 3 keâe ceeve nesiee (Consider the

binary operation * on Q defined by x ∗ y = 1 + 12x + xy,

∀ x, y ∈ Q, then 2 * 3 equals) :(a) 31 (b) 41

(c) 43 (d) 51

9. f : A → B Skeâ DeveeÛÚeokeâ Heâueve nesiee Ùee fo (f : A → B will be an

into function if) :

(a) f(A) ⊂ B (b) f(A) = B

(c) B ⊂ f(A) (d) f(B) ⊂ A

10. Heâueve f(x) = √(x − 1)(3 − x) keâe hejeme nw (The range of the

function f(x) = √(x − 1)(3 − x) is) :

(a) [1, 3] (b) [0, 1]

(c) [− 2, 2]

(d) Fvece W mes keâe sF& veneR (None of these)

11. Ùee fo R Skeâ mecyevOe nw A hej, peneB A = {1, 2, 3} Deewj R = {(2, 2),

(3, 3), (2, 3), (3, 2), (3, 1), (2, 1)} lee s R nw (If R be a relation

on A such that A = {1, 2, 3} and R = {(2, 2), (3, 3), (2, 3),

(3, 2), (3, 1), (2, 1)} then R is) :

(a) mJeleguÙe (Reflexive) (b) meceefcele (Symmetric)

(c) le guÙelee (Equivalence) (d) meb›eâecekeâ (Transitive)

12. ceevee R Øeeke =âle mebKÙeeDee W ke sâ mece gÛÛeÙe N hej efvecve Øekeâej heefjYeee f<ele Heâueven w (Let R be the relation in the set of natural number

defined as) R = {(x, y) ∈ N × N : 3x + y = 11}, leye (then)

R− 1

nesiee (is given by) :

(a) {(0, 1), (1, 8), (2, 5), (3, 2)}

(b) {(1, 8), (2, 5), (3, 2)}

(c) {(11, 0), (8, 1), (5, 2), (2, 3)}

(d) {(8, 1), (5, 2), (2, 3)}

13. Skeâ Dee fjke äle mecegÛÛeÙe A hej hee fjYeee f<ele Skeâ mecyevOe, efpemeceW n DeJeÙeJe nQ,jKelee nw (A relation defined in a non-empty set A, having

n elements, has) :

(a) n mecyevOe (n relations) (b) 2 mecyevOe (2 relations)

(c) n2 mecyevOe (n

2 relations) (d) 2

n2

mecyevOe (2n

2

relations)

14. Ùee fo f : R → R peneB f(x) = x2 lee s f kewâmee Heâueve n w? (If f : R → R

such that f(x) = x2, then what type of a function is f ?)

(a) Skewâkeâer DeeÛÚeefole (bijective)

(b) ve Skewâkeâ Dee wj ve DeeÛÚeoer (neither injective nor surjective)

(c) Skewâkeâ, DeeÛÚeoer veneR (injective, not surjective)

(d) DeeÛÚeoer, Skewâkeâ veneR (surjective, not injective)

15. Jeemlee fJekeâ mebKÙeeDeeW ke sâ mece gÛÛeÙe R ce W efÉÛej mebe f›eâÙee * Fme Øekeâej heefjYeee f<ele

n w e fkeâ a ∗ b = 1

2(a + b)~ Fme mebe f›eâÙee ke sâ e fueS keâewve-mee efveÙece DemelÙe

n w? (The binary operation * defined on the set R of real

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numbers is a ∗ b = 1

2(a + b). Which of the following

laws do not hold for this operation ?)

(a) meenÛeÙe& e fveÙece (associative law)

(b) mebJejkeâ efveÙece (closure law)

(c) ›eâce-efJee fvece sÙe efveÙece (commutative law)

(d) Fvece W mes keâe sF& veneR (none of these)

16. Ùee fo he tCee Ëkeâe W kesâ mece gÛÛeÙe Z hej xRy, (x − y) = Skeâ e fJe<ece he tCee Ëkeâ Éejehee fjYeee f<ele Skeâ mecyevOe nes lees R nw (If Z is the set of integers and

a relation R is defined by xRy, (x − y) = an odd integer,

then R is) :

(a) meb›eâecekeâ veneR (not transitive)

(b) le guÙelee mecyevOe (equivalence relation)

(c) mecee fcele (symmetric)

(d) mJele guÙe (reflexive)

17. Øeeke=âe flekeâ mebKÙeeDee W ke sâ mecegÛÛeÙe ce W mecyevOe `mes Úesše' n w (The relation

‘less than’ in the set of natural numbers is) :

(a) kesâJeue mJele guÙe (only reflexive)

(b) kesâJeue mecee fcele (only symmetric)

(c) le guÙelee mecyevOe (equivalence relation)

(d) ke sâJeue meb›eâecekeâ (only transitive)

18. Ùee fo Heâueve f : R → R Fme Øekeâej hee fjYeee f<ele n w e fkeâ (If function

f : R → R be defined by) f(x) = (3 − x3)

1

3 lee s (then) fof(x) nw(is) :

(a) x1

3 (b) x3

(c) (3 − x3) (d) x

19. Ùee fo (2x, x + y) = (6, 2) nes, lee s x Deewj y %eele keâjW ? [Find x and

y if (2x, x + y) = (6, 2).](a) x = 3, y = − 1 (b) x = 1, y = 5

(c) x = − 1, y = 3 (d) x = 5, y = 1

20. Skeâ efÉ-DeeOeeje r mebe f›eâÙee ∗ , R × R hej efvecve Øekeâej heefjYeee f<ele nw (A binary

composition * is defined on R × R by) (a, b) ∗ (c, d) =

(ac, bc + d), peneB (where) a, b, c, d ∈ R~ %eele keâe re fpeS (Find):

(2, 3) ∗ (1, − 2)~(a) (1, 2) (b) (2, 1)

(c) (1, 1) (d) (2, 2)

[Gòej—1. (d), 2. (d), 3. (b), 4. (d), 5. (d), 6. (c), 7. (c), 8. (a),

9. (a), 10. (b), 11. (d), 12. (d), 13. (d), 14. (b), 15. (a),

16. (c), 17. (d), 18. (d), 19. (a), 20 (b).]

Yeeie-ye : iewj-Jemlegefve… ØeM ve [Non-Objective Type Questions]

➠ ueIeg GòejerÙe ØeMve (Short Answer Type Questions)

ØeM ve 1 Ùeefo f : R → R, f(x) = x2 − 3x + 2 Éeje heefjYeeef<ele nes lees

f[f(x)] Øeehle keâjW~

If f : R → R is defined by f(x) = x2 − 3x + 2, find f[f(x)].

(B.S.E.B., 2010)

nue (Solution)

ÛetBe fkeâ f(x) = x2 − 3x + 2

∴ f[f(x)] = f(x2 − 3x + 2)

= (x2 − 3x + 2)2

− 3(x2 − 3x + 2) + 2

= x4 + 9x2 + 4 − 6x3 − 12x + 4x2 − 3x2 + 9x − 6 + 2

= x4 − 6x

3 + 10x

2 − 3x.

ØeM ve 2. Heâueve f(x) = |x − 1|

x − 1 keâe heefjmej keäÙee nw?

What is the range of the function f(x) = |x − 1|

x − 1?

(C.B.S.E., Delhi, 2010)

nue (Solution)

f(x) = |x − 1|

x − 1

|x − 1|

x − 1 =

x − 1

x − 1 Ùee fo x − 1 > 0

= 1 Ùee fo x > 1

|x − 1|x − 1

= − (x − 1)x − 1

Ùeefo x − 1 < 0

= − 1 Ùee fo x < 1

DeefheÛe, f(x) ke sâJeue leYee r hee fjYeee f<ele nw peyex − 1 ≠ 0

⇒ x ≠ 1

∴ f(x) keâe hee fjmej = (− ∞, 1) ∪ (1, ∞) Ùee – {1}

ØeM ve 3. Ùeefo N Hej R = {(x, y) : x + 2y = 8} Skeâ mecyevOe nw leye Rkeâe Heefjmej %eele keâere fpeS~

If R = {(x, y) : x + 2y = 8} is a relation on N, then writethe range of R. [C.B.S.E., (All India), 2014]

nue (Solution)

ØeMveeve gmeej x + 2y = 8

⇒ y = 8 − x

2

Ùee fo x = 2, leye y = 8 − 2

2 = 3

Ùee fo x = 4, leye y = 8 − 4

2 = 2

Ùee fo x = 6, leye y = 8 − 6

2 = 1

∴ R = (2, 3), (4, 2), (6, 1)Dele: R keâe Hee fjmej = {3, 2, 1}

ØeM ve 4. %eele keâjW (Find) x Deewj (and) y, Ùeefo (If) :

x

3 + 1, y −

2

3

=

5

3,

1

3

(J.A.C., 2011)

nue (Solution)

... x

3 + 1, y −

2

3

=

5

3,

1

3

∴ x

3 + 1 =

5

3 ⇒

x

3 =

5

3 − 1 =

2

3

Ùee x = 2

leLee y − 2

3 =

1

3 ⇒ y =

1

3 +

2

3 =

3

3 = 1

Dele: x = 2 leLee y = 1

ØeM ve 5. Ùeefo a ∗ b = (a + 3b2) Éeje heefjYeeef<ele meYeer hetCeeËkeâeW kesâ

mecegÛÛeÙe Z hej efÉDeeOeejer mebef›eâÙee * nes, lees 2 ∗ 4 %eele keâerefpeS~If * be the binary operation on the set Z of all integers

defined by a ∗ b = (a + 3b2), find 2 * 4.(C.B.S.E., Delhi, 2009)

2 Rajeev’s Model Paper (ieefCele XII)

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nue (Solution)

%eele n w, a ∗ b = a + 3b2

∴ 2 ∗ 4 = 2 + 3(4)2 [... a = 2 leLee b = 4 jKeves hej]

= 2+ 48 = 50

ØeM ve 6. meYeer MetvÙeslej JeemleefJekeâ mebKÙeeDeeW kesâ mecegÛÛeÙe ceW ceevee * Skeâ

efÉDeeOeejer mebe f›eâÙee nw pees a, b ∈ R − {0} kesâ efueS a ∗ b = ab

5 Éeje Øeoòe nw~

Ùeefo 2 ∗ (x ∗ 5) = 10, leye x keâe ceeve %eele keâere fpeS~Let * be a binary operation on the set of all non-zero

real numbers given by a ∗ b = ab

5 for all a, b ∈ R − {0}. Find

the value of x given that 2 * (x ∗ 5) = 10.(C.B.S.E., Delhi, 2014)

nue (Solution)

∀ a, b ∈ R — {0}, a * b = ab

5

∴ 2∗ (x ∗ 5) = 10

⇒ 2 ∗ 5x

5 = 10

⇒ 2∗ x = 50

2x

5 = 10, ∴ x = 25

ØeM ve 7. Ùeefo (If) f(x) = sin x3 Deewj (and) g(x) = (log x)4

, leye(then) gof Deewj fog keâe ceeve %eele keâere fpeS~ (Find the value of gof

and fog.) (U.S.E.B., 2009)

nue (Solution)

... f(x) = sin x3 Deewj g(x) = (log x)4

∴ gof(x) = g[f(x)]= g(sin x

3) = [log sin x

3]

4 = (3 log sin x)4

Deewj fog(x) = f(g(x))

= f{(log x)4}

= sin{(log x4)3

} = sin (log x)12

ØeM ve 8. ceevee efkeâ f(x) = 2x2 leLee g(x) = 3x − 4; x ∈ R. efvecveefueefKele

keâe ceeve %eele keâerefpeS :

Let f(x) = 2x2 and g(x) = 3x − 4; x ∈ R. Find thefollowing : (J.A.C., 2014)

(i) fof(x) (ii) gog(x)(iii) fog(x) (iv) fog(x)

nue (Solution)

(i) fof(x) = f[f(x)]= f(2x

2) = 2(2x2)2

= 8x4

(ii) gog(x) = g[g(x)]= g[3x − 4]

= 3(3x − 4) − 4

= 9x − 16

(iii) fog(x) = f[g(x)] = f(3x − 4)

= 2(3x − 4)2

(iv) gof(x) = g[f(x)]= g(2x

2)= 3(2x

2) − 4

= 6x2 − 4

ØeM ve 9. Ùeefo Heâueve f : R → R, f (x) = x2 + 2 Deewj g : R → R,

g(x) = x

x − 1 Éeje efoÙee ieÙee nes lees fog Deewj gof %eele keâerefpeS~

If the function f : R → R be given by f(x) = x2 + 2 and

g : R → R be given by g(x) = x

x − 1. Find fog and gof.

nue (Solution)

ÙeneB nce Øese f#ele keâjle s nQ e fkeâ f keâe hejeme = g keâe Øeevle g keâe hejeme = f keâe Øeevle ∴ fog Deewj gof oesveeW keâe Dee fmlelJe nw~

(fog) (x) = f(g(x)) = f

x

x − 1 =

x

x − 1

2

+ 2

= x

2

(x − 1)2 + 2

Deewj (gof)(x) = g(f(x)) = g(x2 + 2) =

x2 + 2

(x2 + 2) − 1

= x

2 + 2

x2 + 1

.

ØeM ve 10. a ∗ b = 1 + 12b + ab, ∀ a, b ∈ Q Éeje heefjYeeef<ele Q hejSkeâ efÉDeeOeejer mebe f›eâÙee * ueW~

Consider the binary operation * on Q defined bya * b = 1 + 12b + ab ∀ a, b ∈ Q.

(i) 2 * 3 ØeeHle keâjW~ (Find 2 * 3).

(ii) efoKeeFS efkeâ * ›eâceefJeefvecesÙe veneR nw~ (Show that * is not

commutative).

nue (Solution)

ÙeneB a ∗ b = 1 + 12b + ab ∀ a, b ∈ Q ...(1)

(i) 2 ∗ 3 Øeehle keâjves n sleg (1) ce W a = 2, b = 3 jKeW2 ∗ 3 = 1 + 12 × 3 + 2 × 3

= 1 + 36 + 6 = 43

(ii) ceevee 1 , 2 ∈ Q

leye 1 ∗ 2 = 1 + 12(2) + 1(2) = 1 + 24 + 2 = 27 ...(2)

meeLe ner 2 ∗ 1 = 2 + 12(1) + 2(1) = 1 + 12 + 2 = 15 ...(3)

(2) Deewj (3) mes, 1 ∗ 2 ≠ 2 ∗ 1

∴ efÉDeeOeeje r mebÙeespeve * ›eâce-efJee fvece sÙe veneR nw~ØeM ve 11. hetCeeËkeâ I kesâ mecegÛÛeÙe hej Skeâ Heâueve ‘0’ Fme Øekeâej heefjYeeef<ele

nw efkeâ a 0 b = a − b. keäÙee ‘0’ Skeâ efÉDeeOeejer mebe f›eâÙee nw?A function ‘0’ on the set of integers I is defined by

a 0 b = a − b. Is ‘0’ a binary operation? (J.A.C., 2009)

nue (Solution)

ÛetBe fkeâ nce peevele s nQ e fkeâ oes hetCeeËkeâe W keâe Devlej ncesMee Skeâ hetCeeËkeâ neslee n w~DeLee&le d ∀ a ∈ I, b ∈ I

a − b ∈ I ⇒ a 0 b ∈ I

Dele: ‘0’ Skeâ efÉDeeOeejer mebe f›eâÙee nw~

ieefCele XII (mecyevOe Deewj Heâueve) 3

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ØeM ve 12. Ùeefo (If) f : R → R leLee (and) f(x) = 2x

2x2 + 1

, lees (then)

%eele keâj W (find) f f

12

(J.A.C., 2010)

nue (Solution)

... f(x) = 2x

2x2 + 1

(efoÙee nw)

∴ f 1

2

=

2 × 1⁄22 × 1⁄4 + 1

= 2

3

∴ f f

12

= f

23

=

2 × 2⁄32 × 4⁄9 + 1

= 1217

ØeM ve 13. keäÙee Heâueve f : R → R DeeÛÚeokeâ Heâueve nw peyeefkeâf(x) = 2x ? mekeâejCe efueKeW~

Is the function f : R → R onto function where f(x) = 2x?Give reasons.) (B.S.E.B., 2014)

nue (Solution)

ceevee men[esce sve R keâe keâesF& mJe sÛÚ DeJeÙeJe y nw~leye f(x) = y

⇒ 2x = y

⇒ x = y

2 ∈ R ([escesve)

Dele: fy

2 = 2

y

2 = y (f keâe r hee fjYee<ee mes)

DeLee&le d ∀ y ∈ R (men[escesve)

∃ y

2 ∈ R ([esce sve)

fy

2 = y

∴ f[R] = R

∴ f DeeÛÚeokeâ n w~ØeM ve 14. Ùeefo meYeer heefjcesÙe mebKÙeeDeeW kesâ mecegÛÛeÙe Q ceW Skeâ efÉÛej

mebe f›eâÙee o : Q × Q → Q, o(x, y) = x o y = x + y − xy lees efmeæ keâjW efkeâo meenÛeÙe& nw~ (B.S.E.B., 2014)

nue (Solution)

ceevee a, b, c ∈ Q

leye (a o b) o c = (a + b − ab)o c

= a + b − ab + c − (a + b − ab)c= a + b + c − ab − bc − ca + abc

leLee a o (b o c) = a o(b + c − bc)= a + b + c − bc − a(b + c − bc)= a + b + c − ab − bc − ca + abc

Dele: (a o b) o c = a o(b o c)⇒ o meenÛeÙe & nw~

ØeM ve 15. f(x) = sin x Éeje Øeoòe Heâueve f : 0,

π2

→ R leLee

g(x) = cos x Éeje Øeoòe Heâueve g : 0,

π2

→ R hej efJeÛeej keâere fpeS~ efmeæ

keâere fpeS efkeâ f leLee g Skewâkeâer nQ, hejvleg f + g Skewâkeâer veneR nw~

Consider a function f : 0,

π2

→ R] given by f(x) = sin x

and g : 0,

π2

→ R given by g(x) = cos x. Show that f and g

are one-one, but f + g is not one-one. (U.S.E.B., 2014)

nue (Solution) :

ceevee x1, x2 ∈ 0,

π2

leLee x1 ≠ x2

leye sin x1 ≠ sin x2

⇒ f(x1) ≠ f(x2)∴ f Skewâkeâe r nw~leLee cos x1 ≠ cos x2

⇒ g(x1) ≠ g(x2)∴ g Skewâkeâe r nw~

he gve: 0, π2

∈ 0,

π2

leLee 0 ≠ π2

leye (f + g)(0) = f(0) + g(0)= sin 0 + cos 0 = 0 + 1 = 1

leLee (f + g)π2

= f

π2

+ g

π2

= sin π2

+ cos π2

= 1 + 0 = 1

∴ (f + g)(0) = (f + g)π2

∴ f + g Skewâkeâer veneR nw~➠ oerIe& GòejerÙe ØeMve (Long Answer Type Questions)

ØeM ve 1. Skeâ mecyevOe r mejue jsKeeDeeW kesâ mecegÛÛeÙe hej Fme Øekeâej efoÙeengDee nw efkeâ arb Ùeefo Deewj kesâJeue Ùeefo ‘‘a, b kesâ meceevlej nw~'' efoKeeFS efkeâr Skeâ leguÙelee mecyevOe nw~

A relation r on the set of straight lines is given by arbif ‘‘a is parallel to b.’’ Show that ‘r’ is an equivalence relation.

(J.A.C., 2009)

nue (Solution)

(a) mJeleguÙelee—ceevee L Skeâ efveÙele j sKee n w r ceW,∴ L || L, lees (L, L) ∈ r

∴ ‘r’ mJeleguÙe nw~(b) meceefcele—ceevee L1, L2 oes mejue jsKeeSB Fme Øekeâej nQ efkeâ

(L1, L2) ∈ r

∴ L1 || L2 ⇒ L2 || L1

⇒ (L2, L1) ∈ r

∴ ‘r’ mecee fcele nw~(c) meb›eâecekeâ—ceevee L1, L2, L3 lee rve mejue jsKeeS B Fme Øekeâej nQ efkeâ

(L1, L2) ∈ r Deewj (L2, L3) ∈ r

L1 || L2 Deewj L2 || L3

∴ L1 || L3

⇒ (L1, L3) ∈ r

∴ ‘r’ meb›eâecekeâ n w~Dele: r Skeâ le guÙelee mecyevOe nw~ØeM ve 2. Ùeefo Q heefjcesÙe mebKÙeeDeeW keâe mecegÛÛeÙe nw Deewj Heâueve

f : Q → Q, f(x) = 2x + 3 Éeje heefjYeeef<ele nw lees efmeæ keâerefpeS efkeâ Heâueve f

Skewâkeâer-DeeÛÚeokeâ nw~If Q is the set of rational numbers and the function

f : Q → Q, is defined by f(x) = 2x + 3; then show that thefunction f is one-one onto. (U.S.E.B., 2010)

nue (Solution)

f : Q → Q, f(x) = 2x + 3 (efoÙee nw)ceevee x1, x2 ∈ Q Fme Øekeâej nQ e fkeâ

4 Rajeev’s Model Paper (ieefCele XII)

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f(x1) = f(x2)2x1 + 3 = 2x2 + 3

⇒ 2x1 = 2x2

⇒ x1 = x2

Dele: Heâueve f Skewâkeâe r nw~he gve: ceevee e fkeâ y ∈ Q Fme Øekeâej nw e fkeâ

f(x) = y

2x + 3 = y

∴ x = y − 3

2

peneB f y − 3

2

= 2

y − 3

2

+ 3

= y − 3 + 3 = y

... men-[esce sve Q kesâ ØelÙe skeâ hee fjce sÙe DeJeÙeJe y keâe hetJe & Øeeflee fyecye

x = y − 3

2 Yeer Skeâ DeefÉleerÙe hee fjce sÙe mebKÙee n w DeLee &led Ùen [esce sve keâe Skeâ DeJeÙeJe n w~

Dele: Heâueve f-DeeÛÚeokeâ n w~ØeM ve 3. efmeæ keâere fpeS efkeâ JeemleefJekeâ mebKÙeeDeeW kesâ mecegÛÛÙe R ceW R =

{(a, b) : a ≤ b2} Éeje heefjYeeef<ele mecyevOe R, ve lees mJeleguÙe nw, ve meceefcele nw

Deewj ve ner meb›eâecekeâ nw~Show that the relation R in the set R of real numbers,

defined as R = {(a, b) : a ≤ b2} is neither reflexive, norsymmetric nor transitive. (B.S.E.R., 2014)

nue (Solution) :

e foÙee n w R = {(a, b) : a ≤ b2; a, b ∈ R}

(1) mJeleguÙe : nce osKele s nQ e fkeâ

1

2 ≤

1

2

2

DemelÙe nw

∴ 12

, 12

≤/ R

∴ R mJele guÙe veneR nw~(2) meceefcele : nce osKele s nQ e fkeâ

− 1 ≤ 32

hejvleg 3 ≤/ (− 1)2

∴ (− 1, 3) ∈ R hejvle g (3, − 1) ∉ R

∴ R mecee fcele veneR nw~(3) meb›eâecekeâ : nce osKele s nQ e fkeâ 2 ≤ (− 3)2

Deewj − 3 ≤ (1)2 hejvle g 2 ≤ (1)2

∴ (2, − 3) ∈ R

Deewj (− 3, 1) ∈ R

hejvleg (2, 1) ∉ R

∴ R meb›eâecekeâ veneR nw~Dele: Øeoòe mecyevOe R ve lee s mJele guÙe nw, ve meceefcele n w Dee wj ve ner

me b›eâecekeâ n w~ØeM ve 4. leerve Heâueve f : N → N, g : N → N leLee h : N → R hej

efJeÛeej keâerefpeS peneB f(x) = 2x, g(y) = 3y + 4 leLee h(z) = sin z, ∀ x, y

leLee z ∈ N~ efoKeeFS efkeâ ho(gof) = (hog)of~Consider three functions f : N → N, g : N → N and

h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z,∀ x, y and z ∈ N. Show that ho(gof) = (hog)of.

(U.S.E.B., 2013)

nue (Solution)

[ho(gof)](x) = h[(gof)(x)] = h[gof(x)}]

= h[g(2x)]= h[3(2x) + y]

= h(6x + 4)= sin(6x + 4) ...(1)

he gve: [(hog)of](x) = (hog)[f(x)]= (hog)(2x)= h[g(2x)]= h[3(2x + 4)]= h(6x + 4)= sin(6x + 4) ...(2)

(1) Deewj (2) mes nce heeles n Q[ho(gof)(x) = [(hog)of](x)

⇒ ho(gof) = (hog)of

2 Øeefleueesce ef$ekeâesCeefceleerÙe Heâueve[INVERSE TRIGONOMETRIC FUNCTIONS]

Yeeie (De) : Jemlegefve… ØeM ve[Objective Type Questions]

➠ yengefJekeâuHeerÙe ØeM ve (Multiple Choice Questions)

efveoxM e—efvecve ØeMvee W keâe Skeâ mener Gòej nw~ efoS ieS efJekeâuHee W ce W mes menere fJekeâuHe hej efveMeeve ueieeÙe W— ieefCele XII (Øeefleueesce ef$ekeâesCeefceleerÙe Heâueve)

Instruction : In the following questions there is onecorrect answer. In each question you have to mark thatcorrect option from the given options :

1. tan−1

x

y

− tan

−1 x − y

x + y

yejeyej nw (is equal to) :

(a) π3

(b) π4

(c) π2

(d) − 3 π4

2. cos−1

1

2

+ 2 sin

−1 1

2

keâe ceeve nw (The value of

cos−1

12

+ 2 sin

−1 12

is equal to) :

(a) π4

(b) π6

(c) 2π3

(d) 5π6

3. 2 tan− 1 1

3 + tan− 1

1

7 = (B.S.E.B., 2010)

(a) tan− 1

44

29(b)

π2

(c) 0 (d) π4

ieefCele XII (Øeefleueesce ef$ekeâesCeefceleerÙe Heâueve) 5

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4. cos− 1

cos

7π6

=

(a) 7π6

(b) π3

(c) π6

(d) 5π6

5. tan−1

√ 3 − sec−1

(−2) =

[B.S.E.B., 2011; C.B.S.E., 2012 (All India)]

(a) π (b) − π3

(c) π3

(d) 2π3

6. Ùee fo (If) sin−1

2x

1 + x2

+ sin

−1

2y

1 + y2

= 2 tan

−1 a, leye a keâe ceeve

n w (then a is equal to) :

(a) x − y

1 + xy(b)

y

1 + xy

(c) y

1 − xy(d)

x + y

1 − xy.

7. Ùee fo (If) sin−1

(1 − x) − 2 sin−1

x = π2

; leye x keâe ceeve nw (then x

is equal to) : (B.S.E.B., 2012)

(a) 0 (b) 0, − 1

2

(c) 0, 1

2(d) FveceW mes keâe sF& veneR

8. Ùee fo tan− 1

x + tan− 1

y = π4

, xy < 1 nw, lee s x + y + xy keâe ceeve n w (If

tan− 1

x + tan− 1

y = π4

, xy < 1, then the value of x + y + xy

is) :(a) − 1 (b) 1 (c) 0

(d) Fvece W mes keâe sF& veneR (None of these)

9. tan2 tan

− 1 15

keâe ceeve nw (The value of tan

2 tan

− 1 15

is) :

(a) 12

5(b)

2

5(c)

5

12(d)

7

12

10. tan− 1

(1) + cos− 1

−

12

+ sin

− 1−

12

keâe ce gKÙe ceeve n w (The

principal value of tan− 1

(1) + cos− 1

−

1

2

+ sin

− 1−

1

2

is) :

(a) π4

(b) π2

(c) π6

(d) 3π4

11. sin− 1

1

√ 2

keâe ce gKÙe ceeve n w (The principal value of

sin− 1

1√ 2

is) :

(a) π4

(b) π2

(c) π3

(d) π6

12. sinπ3

− sin− 1

−

12

keâe ceeve nw :

The value of sinπ3

− sin− 1−

12

is :

(a) − 1 (b) 12

(c) − 12

(d) 1

13. tan− 1

(√ 3 ) − cot− 1(− √ 3 ) keâe ceeve nw :

The value of tan− 1(√ 3 ) − cot− 1(− √ 3 ) is :

(a) π (b) − π2

(c) 0 (d) 2√ 3

14. cosπ2

+ sin− 1

1

3

keâe ceeve n w :

The value of cos π2

+ sin− 11

3

is :

(a) 1

3(b) −

1

3(c) 1 (d) 0

15. sinsin

− 1 12

+ cos− 1

12

keâe ceeve nw :

The value of sin sin− 1

12

+ cos− 1 12

is :

(a) π2

(b) 1 (c) 0 (d) 12

16. cot− 1

tan

π7

keâe ceeve nw :

The value of cot− 1tan

π7

is :

(a) 5π14

(b) 9π14

(c) π14

(d) 3π14

17. sin (cot− 1

x) kesâ e fueS yee rpee rÙe JÙebpekeâ nw :

Algebraic expression for sin (cot− 1 x) is :

(a) 1

1 + x2(b)

1

√1 + x2(c)

x

√1 + x2

(d) Gheje skeäle ce W keâe sF& veneR (None of these)

18. sec2 (tan

− 1 4) + cosec

2 (cot

− 1 2) keâe ceeve n w :

The value of sec2 (tan− 1 4) + cosec2 (cot− 1 2) is :

(a) 12 (b) 18 (c) 22 (d) 25

19. Ùee fo (If) sin− 1

x + sin− 1

y = π3

, lees (then the value of)

cos− 1

x + cos− 1

y keâe ceeve nw (is) :

(a) π6

(b) π3

(c) 2π3

(d) π

20. Ùee fo (If) tan− 1

x + tan− 1

y + tan− 1

z = π2

, lees xy + yz + zx keâe

ceeve n w :

If tan− 1 x + tan− 1 y + tan− 1 z = π2

, then the value of

xy + yz + zx is :

(a) − 1 (b) 1 (c) 0

(d) Fvece W mes keâe sF& veneR (None of these)

[Gòej—1. (b), 2. (c), 3. (d), 4. (d), 5. (b), 6. (d), 7. (a), 8. (b),

9. (c), 10. (d), 11. (a), 12. (d), 13. (b), 14. (b), 15.

(b), 16. (a), 17. (b), 18. (c), 19. (c), 20. (b)]

Yeeie-ye : iewj-Jemlegefve… ØeM ve [Non-Objective Type Questions]

➠ ueIeg GòejerÙe ØeMve (Short Answer Type Questions)

ØeM ve 1. e fm eæ keâere fpeS (Prove that) :

12

tan− 1

x = cos− 1

1 + √ 1 + x2

2√ 1 + x2

1

2

(B.S.E.R., 2013)

6 Rajeev’s Model Paper (ieefCele XII)

Mathematics (E-Model Paper) Class XIIth

Publisher : Sbpd Publications ISBN : 9789351679950 Author : SBPD EditorialBoard

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