RelationsandFunctions

Question1

Thefunctionf:N−{1}→N;definedbyf(n)=thehighestprimefactor

ofn,is:

[27-Jan-2024Shift1]

Options:

bothone-oneandonto

one-oneonly

ontoonly

neitherone-onenoronto

Answer:D

Solution:



-------------------------------------------------------------------------------------------------

Question2

Options:

Answer:A

Solution:



-------------------------------------------------------------------------------------------------

Question3

Considerthefunctionf:[1/2,1]→Rdefinedbyf(x)=4√2x3−3√2x−1.

Considerthestatements

(I)Thecurvey=f(x)intersectsthex-axisexactlyatonepoint

(II)Thecurvey=f(x)intersectsthex-axisatx=cosπ/12

Then

[29-Jan-2024Shift1]

Options:

Only(II)iscorrect

Both(I)and(II)areincorrect

Only(I)iscorrect

Both(I)and(II)arecorrect

Answer:D

Solution:



-------------------------------------------------------------------------------------------------

Question4

[29-Jan-2024Shift1]

Options:

(0,1]

[0,3)

[0,1]

[0,1)

Answer:C

Solution:



-------------------------------------------------------------------------------------------------

Question5

Letf(x)=2x−x2,x∈R.Ifmandnarerespectivelythenumberof

pointsatwhichthecurvesy=f(x)andy=f′(x)intersectsthex-axis,

thenthevalueofm+nis

[29-Jan-2024Shift1]

Answer:5

Solution:



-------------------------------------------------------------------------------------------------

Question6

Ifthedomainofthefunctionf(x)=cos−1(2-|x|/4)+(loge(3−x))

−1is[−α,β)−{y},thenα+β+γisequalto:

[30-Jan-2024Shift1]

Options:

Answer:C

Solution:



-------------------------------------------------------------------------------------------------

Question7

LetA={1,2,3,....7}andletP(1)denotethepowersetofA.Ifthe

numberoffunctionsf:A→P(A)suchthata∈f(a),∀a∈

Aismn,mandn∈Nandmisleast,thenm+nisequalto______

[30-Jan-2024Shift1]

Answer:44

Solution:



-------------------------------------------------------------------------------------------------

Question8

Ifthedomainofthefunctionf(x)=loge is(α,β],then

thevalueof5β−4αisequalto

[30-Jan-2024Shift2]

Options:

Answer:B

Solution:



-------------------------------------------------------------------------------------------------

Question9

Letf:R→Rbeafunctiondefinedf(x)= andg(x)=

f(f(f(f(x))))then18

[30-Jan-2024Shift2]

Options:

Answer:D

Solution:



-------------------------------------------------------------------------------------------------

Question10

Options:

19/20

-4

Answer:D

Solution:



-------------------------------------------------------------------------------------------------

Question11

Ifthefunctionf:(−∞,−1]⟶(a,b]definedbyf(x)= isone-one

andonto,thenthedistanceofthepoint

P(2b+4,a+2)fromthelinex+e−3y=4is:

[31-Jan-2024Shift2]

Options:

Answer:A

Solution:



-------------------------------------------------------------------------------------------------

Question12

[1-Feb-2024Shift1]

Options:

one-onebutnotonto

neitherone-onenoronto

ontobutnotone-one

bothone-oneandonto

Answer:B

Solution:



-------------------------------------------------------------------------------------------------

Question13

Ifthedomainofthefunctionf(x)= thenα2+

β3isequalto:

[1-Feb-2024Shift2]

Options:

140

175

150

125

Answer:C

Solution:



-------------------------------------------------------------------------------------------------

Question14

LetA = {1,2,3,4, ..., 10}andB = {0,1,2,3,4}.Thenumberof

elementsintherelationR = { (a,b) ∈ A×A:2(a−b)2+3(a−b) ∈ B}is

______.

[6-Apr-2023shift1]

Answer:18

Solution:

-------------------------------------------------------------------------------------------------

Question15

LetA = {0,34,6,7,8,9,10}andRbetherelationdefinedonAsuch

thatR = {x,y) ∈ A×A:x−yisoddpositiveintegerorx −y=2}.The

minimumnumberofelementsthatmustbeaaddedtotherelationR,so

thatitisasymmetricrelation,isequalto______.

A= {1,2,3, ..... . 10}

B= {0,1,2,3,4}

R= {(a,b) ∈ A×A:2(a−b)2+3(a−b) ∈ B}

Now 2(a−b)2+3(a−b) = (a−b)(2(a−b) + 3)

⇒a=b or a −b= −2

When a =b⇒10 orderpairs

When a −b= −2⇒8 orderpairs

Total =18

[8-Apr-2023shift1]

Answer:19

Solution:

-------------------------------------------------------------------------------------------------

Question16

LetA = {1,2,3,4,5,6,7}.Thentherelation

R= {(x,y) ∈ A×A:x+y=7}is

[8-Apr-2023shift2]

Options:

A.Symmetricbutneitherreflexivenortransitive

B.Transitivebutneithersymmetricnorreflexive

C.Anequivalencerelation

D.Reflexivebutneithersymmetricnortransitive

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question17

LetA = {2,3,4}andB = {8,9,12}.Thenthenumberofelementsin

therelationR = { ((a1,b1), (a2,b2)) ∈ ( A×.B,A×B) : a1dividesb2and

a2dividesb1}is

[10-Apr-2023shift2]

Options:

A.18

B.24

A= {0,3,4,6,7,8,9,10}3,7,9→odd

R= {x−y=odd +veor x −y=2}0,4,6,8,10 →even

3C1⋅5C1=15 + (6,4), (8,6), (10,8), (9,7)

Minmorderedpairstobeaddedmustbe

:15 +4=19

R= {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

C.12

D.36

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question18

LetA = {1,3,4,6,9}andB = {2,4,5,8,10}.LetRbearelation

definedonA ×BsuchthatR = { ( (a1,b1), ( a2,b2...)):a1≤b2and

b1≤a2}.ThenthenumberofelementsinthesetRis

[11-Apr-2023shift2]

Options:

A.52

B.160

C.26

D.180

Answer:B

Solution:

a1dividesb2

Eachelementshas2choices

⇒3×2=6

a2dividesb1

Eachelementshas2choices

⇒3×2=6

Total =6×6=36

Leta1=1⇒5choicesofb2

a1=3⇒4choicesofb2

a1=4⇒4choicesofb2

a1=6⇒2choicesofb2

a1=9⇒1choicesofb2

For(a1,b2)16ways.

Similarly,b1=2⇒4choicesofa2

b1=4⇒3choicesofa2

b1=5⇒2choicesofa2

-------------------------------------------------------------------------------------------------

Question19

Thenumberoftherelations,ontheset{1,2,3}containing(1,2)and

(2,3),whicharereflexiveandtransitivebutnotsymmetric,is________.

[12-Apr-2023shift1]

Answer:3

Solution:

-------------------------------------------------------------------------------------------------

Question20

LetA = {−4, −3, −2,0,1,3,4}andR = { (a,b) ∈ A×A:b= | a| .or

2=a+1}bearelationonA.Thentheminimumnumberofelements,

thatmustbeaddedtotherelationRsothatitbecomesreflexiveand

symmetric,is________

[13-Apr-2023shift2]

Answer:7

Solution:

-------------------------------------------------------------------------------------------------

Question21

b1=8⇒1choicesofa2

RequiredelementsinR=160

A= {1,2,3}

ForReflexive(1,1)(2,2), (3,3) ∈ R

Fortransitive:(1,2)and(2,3) ∈ R⇒ (1,3) ∈ R

Notsymmetric:(2,1)and(3,2) ∉ R

R1= {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}

R2= {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)(2,1)}

R3= {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)(2,1)}

R= [ (−4,4), (−3,3), (3, −2), (0,1), (0,0), (1,1),

(4,4), (3,3) }

Forreflexive,add⇒(−2, −2), (−4, −4), (−3, −3)

Forsymmetric,add⇒(4, −4), (3, −3), (−2,3), (1,0)

LetA = {1,2,3,4}andRbearelationonthesetA ×Adefinedby

R= { ( (a,b, (c,d) : 2a +3b =4c +5d }.Thenthenumberofelementsin

Ris________

[15-Apr-2023shift1]

Answer:6

Solution:

-------------------------------------------------------------------------------------------------

Question22

Let5f (x) + 4f 1

x=1

x+3,x>0.Then18 2

∫

1f(x)dxisequalto:

[6-Apr-2023shift1]

Options:

A.10loge2−6

B.10loge2+6

C.5loge2−3

D.5loge2+3

Answer:A

Solution:

( )

A= {1,2,3,4}

R= {(a,b), (c,d)}

2a +3b =4c +5d =α let

2a = {2,4,6,8}4c = {4,8,12,16}

3b = {3,6,9,12}5d = {5,10,15,20}

2a +3b =

5 8 11 14

7 10 13 16

9 12 15 18

11 14 17 20

4c +5d

9 14 19 24

13 18...

17 22....

21 26....

Possiblevalueofα=9,13,14,14,17,18

Pairsof{(a,b), (c,d)} = 6

{ } { }

5f (x) + 4f 1

x=1

x+3...(1)

x→1

x

5f 1

x+4f (x) = x+3...(2)

(1) × 5− (2) × 4

( )

-------------------------------------------------------------------------------------------------

Question23

LetA = {x∈ ℝ : [x+3] + [x+4] ≤ 3},

B=x∈ ℝ : 3x∞

∑

r=1

10x

x−3

<3−3x ,where[t]denotesgreatest

integerfunction.Then,

[6-Apr-2023shift1]

Options:

A.A ⊂B,A≠B

B.A ∩B=φ

C.A =B

D.B ⊂C,A≠B

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question24

LetthesetsAandBdenotethedomainandrangerespectivelyofthe

functionf (x) = 1

√[x] − x,where[x]denotesthesmallestintegergreater

thanorequaltox.Thenamongthestatements:

(S1) : A∩B= (1,∞) − Nand

{ ( ) }

⇒f(x) = 5

9x −4

9x+1

3

⇒18

∫

f(x)dx =18 5

9ln 2 −4

9×3

2+1

=10 ln 2 −6

( )

A= {x∈ℝ: [x+3] + [x+4] ≤ 3}

2[x] + 7≤3

2[x] ≤ −4

[x] ≤ −2⇒x< −1...(A)

B=x∈ℝ:3x∞

∑

r=1

10x

x−3

<3−3x

3x∞

∑

r−1

10x

x−3

<3−3x

32x −310

x−3

⇒1

x−3<3−5x

⇒36−2x <33−5x

⇒6−2x <3−5x

⇒3< −3x

⇒1

10 )x< −1...(B)

A=B

{ ( ) }

( )

(S2):A ∪B= (1,∞)

[6-Apr-2023shift2]

Options:

A.only(S1)istrue

B.neither(S1)nor(S2)istrue

C.only(S2)istrue

D.both(S1)and(S2)aretrue

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question25

Letf ,g: ℕ − {1} → ℕbefunctionsdefinedbyf (a) = α,whereαisthe

maximumofthepowersofthoseprimespsuchthatpαdividesa,and

g(a) = a+1,foralla ∈ ℕ − {1}.Then,thefunctionf +gis

[27-Jul-2022-Shift-1]

Options:

A.one-onebutnotonto

B.ontobutnotone-one

f(x) = 1

√[x] − x

Ifx∈I[x] = [x](greatestintegerfunction)

Ifx∉I[x] = [x] + 1

⇒f(x) =

√[x] − xx∈I

√[x] + 1−xx∉I

⇒f(x) =

√−{x}x∈I(doesnotexist )

√1− {x}x∉I

⇒domainof f(x) = R−I

Now, f(x) = 1

√1− {x},x∉I

⇒x< {x} < 1

⇒0<1√1− {x} < 1

⇒1

√1− {x}>1

⇒Range(1,∞)

⇒A=R−I

B= (1∞)

So, A ∩B= (1,∞) − N

A∪B≠ (1,∞)

⇒S1 isonlycorrect.

{

C.bothone-oneandonto

D.neitherone-onenoronto

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question26

Ifdomainofthefunctionloge

6x2+5x +1

2x −1+cos−12x2−3x +4

3x −5is

(α,β) ∪ (γ,δ],then,18(α2+β2+γ2+δ2)isequalto

[8-Apr-2023shift2]

Answer:20

Solution:

-------------------------------------------------------------------------------------------------

Question27

( ) ( )

f,g:N− {1} → Ndefinedas

f(a) = α,whereαisthemaximumpowerofthoseprimespsuchthatpαdividesa.

g(a) = a+1

Now,

f(2) = 1,g(2) = 3⇒ (f+g)(2) = 4

f(3) = 1,g(3) = 4⇒ (f+g)(3) = 5

f(4) = 2,g(4) = 5⇒ (f+g)(4) = 7

f(5) = 1,g(5) = 6⇒ (f+g)(5) = 7

∵(f+g)(5) = (f+g)(4)

∴f+gisnotone-one

Now,∵fmin =1,gmin =3

So,theredoesnotexistanyx∈N− {1}suchthat(f+g)(x) = 1,2,3

∴f+gisnotonto

6x2+5x +1

2x −1>0

(3x +1)(2x +1)

2x −1>0

x∈−1

√2,1

√2∪5

3,∞...(B)

x<5

3...(C)

A∩B∩C≡−1

2,−1

3∪1

2,1

√2

So 18(α2+β2+γ2+δ2) = 18 1

4+1

9+1

4+1

=18 +2=20

[ ] ( )

( ) ( ]

( )

LetR = {a,b,c,d,e}andS = {1,2,3,4}.Totalnumberofonto

functionsf :R→Ssuchthatf(a) ≠ 1,isequalto_______.

[8-Apr-2023shift2]

Answer:180

Solution:

-------------------------------------------------------------------------------------------------

Question28

Ifthedomainofthefunctionf(x) = sec−12x

5x +3is[α,β)U(γ,δ],then

|3α +10(β+γ) + 21δ|isequalto________.

[10-Apr-2023shift2]

Answer:24

Solution:

( )

Totalontofunction

⌊3⌊× ⌊ 4=240

Nowwhenf(a) = 1

4+⌊4

⌊2⌊2× ⌊ x3 =24 +36 =60.

sorequiredfn=240 −60 =180

⌊

f(x) = sec−12x

5x +3

5x +3≥1⇒2x ≥5x +3

(2x)2− (5x +3)2≥0

(7x +3)(−3x −3) ≥ 0

∴domain −1,−3

5∪−3

5,−3

α= −1,β=−3

5,γ=−3

5,δ=−3

3α +10(β+γ) + 21δ = −3

−3+10 −6

5+−3

721 = −24

| |

| | | | | |

[ ) ( ]

( ) ( )

-------------------------------------------------------------------------------------------------

Question29

Thedomainofthefunctionf (x) = 1

[x]2−3[x] − 10 is(where[x]denotesthe

greatestintegerlessthanorequaltox)

[11-Apr-2023shift2]

Options:

A.(−∞, −3] ∪ [6,∞)

B.(−∞, −2) ∪ (5,∞)

C.(−∞, −3] ∪ (5,∞)

D.(−∞, −2) ∪ [6,∞)

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question30

LetA = {1,2,3,4,5}andB = {1,2,3,4,5,6}.Thenthenumberof

functionsf :A→Bsatisfyingf (1) + f(2) = f(4) − 1isequalto________.

[11-Apr-2023shift2]

Answer:360

Solution:

√

F(x) = 1

[x]2−3[x] − 10

[x]2−3[x] − 10 >0

([x] + 2)([x] − 5) > 0

[x] < −2 or [x] > 5

[x] ≤ −3 or [x] ≥ 6

x< −2 or x ≥6

x∈ (−∞, −2) ∪ [6,∞)

√

f(1) + f(2) + 1=f(4) ≤ 6

f(1) + f(2) ≤ 5

Case(i)f(1) = 1⇒f(2) = 1,2,3,4⇒4mappings

Case(ii)f(1) = 2⇒f(2) = 1,2,3⇒3mappings

Case(iii)f(1) = 3⇒f(2) = 1,2⇒2mappings

Case(iv)f(1)4⇒f(2) = 1⇒1mapping

-------------------------------------------------------------------------------------------------

Question31

LetDbethedomainofthefunctionf (x) = sin −1log3x

6+2log3x

−5x .If

therangeofthefunctiong :D→Rdefinedbyg(x) = x− [x], ( [x]isthe

greatestintegerfunction),is(α,β),thenα2+5

βisequalto

[12-Apr-2023shift1]

Options:

A.46

B.135

C.136

D.45

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question32

Forx ∈R,tworealvaluedfunctionsf(x)andg(x)aresuchthat,

g(x) = √x+1andfog(x) = x+3− √x.Thenf(0)isequalto

( ( ) )

f(5)&f(6)bothhave6mappingseach

Numberoffunctions = (4+3+2+1) × 6×6=360

6+2log3x

−5x >0&x>0&x≠1

thisgives x ∈0,1

27 .. . (1)

−1≤log3x

6+2log3x

−5x ≤1

3x ≤6+2log3x

−5x ≤1

15x2+6+2log3x≥0 6 +2log3x+5

3≥0

x∈0,1

27 .. . (2)x≥3

−23

6... . (3)

form(1),(2)&(3)

x∈3

−23

6,1

∴αissmallpositivequantity

&β=1

∴α2+5

βisjustgreaterthan135

( )

[ )

[13-Apr-2023shift1]

Options:

A.5

B.0

C.-3

D.1

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question33

Forthedifferentiablefunctionf :R− {0} → R,let

3f (x) + 2f 1

x=1

x−10,then f (3) + f′1

4isequalto

[13-Apr-2023shift1]

Options:

A.13

B. 29

C. 33

D.7

Answer:A

Solution:

( ) | ( ) |

g(x) = √x+1

fog (x) = x+3− √x

= (√x+1)2−3(√x+1) + 5

=g2(x) − 3g(x) + 5

⇒f(x) = x2−3x +5

∴f(0) = 5

But,ifweconsiderthedomainofthecompositefunctionfog(x)theninthatcasef(0)willbenotdefinedasg(x)cannot

beequaltozero.

3f(x) + 2f 1

x=1

x−10 ×3

2f(x) + 3f 1

x=x−10 ×2

5f(x) = 3

x−2x −10

f(x) = 1

x−2x −10

[ ( ) ]

( )

-------------------------------------------------------------------------------------------------

Question34

Therangeoff (x) = 4sin −1x2

x2+1is

[13-Apr-2023shift2]

Options:

A.[0,π)

B.[0,π]

C.[0,2π)

D.[0,2π]

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question35

Ifthedomainofthefunction

f(x) = loge(4x2+11x +6) + sin −1(4x +3) + cos−110x +6

3is(α,β],then

36 |α+β|isequalto

[15-Apr-2023shift1]

Options:

A.72

B.63

C.45

( )

f′(x) = 1

5−3

x2−2

f(3) + f′1

4=1

5(1−6−10) + 1

5(−48 −2)

= | −3−10 | = 13

( )

| ( ) | | |

f(x) = 4sin −1x2

1+x2

0≤x2

1+x2<1

⇒0≤sin −1x2

1+x2<π

⇒0≤4sin −1x2

1+x2<2π

Range:[0,2π)

( )

D.54

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question36

TherelationR = {(a,b) : gcd(a,b) = 1,2a ≠b,a,b∈ ℤ}is:

[24-Jan-2023Shift1]

Options:

A.transitivebutnotreflexive

B.symmetricbutnottransitive

C.reflexivebutnotsymmetric

D.neithersymmetricnortransitive

Answer:D

Solution:

f(x) = ln(4x2+11x +6) + sin −1(4x +3)

+cos−110x +6

(i) 4x2+11x +6>0

4x2+8x +3x +6>0

(4x +3)(x+2) > 0

x∈ (−∞, −2) ∪ − 3

4,∞

(ii)4x +3∈ [−1,1]

x∈ [−1, −1∕2]

(iii) 10x +6

3∈ [−1,1]

x∈ − 9

10, − 3

x∈ − 3

4, − 1

2α= − 3

4,β= − 1

α+β= − 5

36 |α+β| = 45

( )

[ ]

( ]

Reflexive:(a,a)gcdof(a,a) = 1

WhichisnottrueforeveryaE Z.

Symmetric:

Takea=2,b=1 gcd(2,1) = 1

Also2a =4≠b

Nowwhena=1,b=2 gcd(1,2) = 1

Alsonow2a =2=b

Hencea=2b

⇒RisnotSymmetric

Transitive:

Leta=14,b=19,c=21

gcd(a,b) = 1

gcd(b,c) = 1

gcd(a,c) = 7

-------------------------------------------------------------------------------------------------

Question37

LetRbearelationdefinedonNasaRbis2a +3bisamultipleof

5,a,b∈ ℕ.ThenRis

[29-Jan-2023Shift2]

Options:

A.notreflexive

B.transitivebutnotsymmetric

C.symmetricbutnottransitive

D.anequivalencerelation

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question38

Theminimumnumberofelementsthatmustbeaddedtotherelation

R= {(a,b), (b,c)}ontheset{a,b,c}sothatitbecomessymmetricand

transitiveis:

[30-Jan-2023Shift1]

Options:

A.4

B.7

Hencenottransitive

Risneithersymmetricnortransitive.

aRa ⇒5aismultipleit5

Soreflexive

aRb ⇒2a +3b =5α,

NowbRa

2b +3a =2b +5α −3b

2⋅3

=15

2α−5

2b=5

2(3α −b)

2(2a +2b −2α)

=5(a+b−α)

Hencesymmetric

aRb ⇒2a +3b =5α .

bRc ⇒2b +3c =5β

Now 2a +5b +3c =5(α+β)

⇒2a +5b +3c =5(α+β)

⇒2a +3c =5(α+β−b)

⇒aRc

Hencerelationisequivalencerelation.

( )

C.5

D.3

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question39

LetRbearelationonN ×Ndefinedby(a,b)R(c,d)ifandonlyifad

(b−c) = bc(a−d).ThenRis

[31-Jan-2023Shift1]

Options:

A.symmetricbutneitherreflexivenortransitive

B.transitivebutneitherreflexivenorsymmetric

C.reflexiveandsymmetricbutnottransitive

D.symmetricandtransitivebutnotreflexive

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

ForSymmetric(a,b), (b,c) ∈ R

⇒(b,a), (c,b) ∈ R

ForTransitive(a,b), (b,c) ∈ R

⇒(a,c) ∈ R

Now

1.Symmetric

∴(a,c) ∈ R⇒ (c,a) ∈ R

2.Transitive

∴ (a,b), (b,a) ∈ R

⇒ (a,a) ∈ R&(b,c), (c,b) ∈ R

⇒ (b,b)&(c,c) ∈ R

∴Elementstobeadded

(b,a) (c,b) (a,c) (c,a)

, (a,a) (b,b) (c,c)

Numberofelementstobeadded = 7

{ }

(a,b)R(c,d) ⇒ ad (b−c) = bc(a−d)

Symmetric:

(c,d)R(a,b) ⇒ cb(d−a) = da(c−b)⇒

Symmetric

Reflexive:

(a,b)R(a,b) ⇒ ab(b−a) ≠ ba(a−b)⇒

Notreflexive

Transitive:(2,3)R(3,2)and(3,2)R(5,30)but

((2,3), (5,30)) ∉ R⇒Nottransitive

Question40

Amongtherelations

S= (a,b) : a,b∈ ℝ − {0}, 2+a

b>0

AndT = {(a,b) : a,b∈ ℝ, a2−b2∈Z},

[31-Jan-2023Shift2]

Options:

A.SistransitivebutTisnot

B.TissymmetricbutSisnot

C.NeitherSnorTistransitive

D.BothSandTaresymmetric

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question41

LetRbearelationonℝ,givenbyR = { (a,b) : 3a −3b + √7isan

irrationalnumber}.ThenRis

[1-Feb-2023Shift1]

Options:

A.Reflexivebutneithersymmetricnortransitive

B.Reflexiveandtransitivebutnotsymmetric

C.Reflexiveandsymmetricbutnottransitive

D.Anequivalencerelation

Answer:A

Solution:

{ }

ForrelationT=a2−b2= −I

Then,(b,a)onrelationR

⇒b2−a2= −I

∴Tissymmetric

S= (a,b) : a,b∈R− {0}, 2+a

b>0

2+a

b>0⇒a

b> −2, ⇒ b

a<−1

If(b,a) ∈ Sthen

2+b

anotnecessarilypositive

∴Sisnotsymmetric

{ }

Solution:

-------------------------------------------------------------------------------------------------

Question42

LetP(S)denotethepowersetofS = {1,2,3, ..., 10}.Definethe

relationsR1andR2onP(S)asAR1Bif(A∩Bc) ∪ (B∩A9) = ∅andAR2Bif

A∪Bc=B ∪Ac, ∀A,B∈P(S).Then:

[1-Feb-2023Shift2]

Options:

A.bothR1andR2areequivalencerelations

B.onlyR1isanequivalencerelation

C.onlyR2isanequivalencerelation

D.bothR1andR2arenotequivalencerelations

Answer:A

Solution:

Checkforreflexivity:

As3(a-a)+√7= √7whichbelongstorelationsorelationisreflexive

Checkforsymmetric:

Takea=√7

3,b=0

Now(a,b) ∈ Rbut(b,a) ∉ R

As3(b−a) + √7=0whichisrationalsorelationisnotsymmetric.

CheckforTransitivity:

Take(a,b)as √7

3,1

&(b,c)as 1,2√7

Sonow(a,b) ∈ R&(b,c) ∈ Rbut(a,c) ∉ Rwhichmeansrelationisnottransitive

( )

S= {1,2,3, ... . .10}

P(S) = powersetof S

AR,B⇒A∩→

B∪→

A∩B=φ

R1 isreflexive,symmetric

Fortransitive

A∩→

B∪→

A∩B=φ; {a} = φ= {b}A=B

B∩→

C∪→

B∩C=φ∴B=C

∴A=C equivalence.

R2≡A∪→

B=→

A∪B

R2→Reflexive,symmetric

fortransitive

( ) ( )

-------------------------------------------------------------------------------------------------

Question43

Theequationx2−4x + [x] + 3=x[x],where[x]denotesthegreatest

integerfunction,has:

[24-Jan-2023Shift1]

Options:

A.exactlytwosolutionsin(−∞,∞)

B.nosolution

C.auniquesolutionin(−∞,1)

D.auniquesolutionin(−∞,∞)

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question44

Letf (x)beafunctionsuchthatf (x+y) = f(x) ⋅ f(y)forallx,y∈N.If

f(1) = 3and n

∑

k=1f(k) = 3279,thenthevalueofnis

[24-Jan-2023Shift2]

Options:

A.6

B.8

C.7

D.9

A∪→

B=→

A∪B⇒ {a,c,d} = {b,c,d}

{a} = {b}∴A=B

B∪→

C=→

B∪C⇒B=C

∴A=C∴A∪→

C=→

A∪C∴Equivalence

x2−4x + [x] + 3=x[x]

⇒x2−4x +3=x[x] − [x]

(x−1)(x−3) = [x] . (x−1)

⇒x=1 or x −3= [x]

⇒x− [x] = 3

{x} = 3 (NotPossible)

Onlyonesolutionx=1in(−∞,∞)

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question45

Iff (x) = 22x

22x +2,x∈Rthenf 1

2023 +f2

2023 + ...... + f2022

2023 equalto

[24-Jan-2023Shift2]

Options:

A.2011

B.1010

C.2010

D.1011

Answer:D

Solution:

( ) ( ) ( )

f(x+y) = f(x) ⋅ f(y) ∀x,y∈N,f(1) = 3

f(2) = f2(1) = 32

f(3) = f(1)f(2) = 33

f(4) = 34

f(k) = 3k

∑

k=1

f(k) = 3279

f(1) + f(2) + f(3) + ......... + f(k) = 3279

3+32+33+ ........ . 3k=3279

3(3k−1)

3−1=3279

3k−1

2=1093

3k−1=2186

3k=2187

k=7

f(x) = 4x

4x+2

f(x) + f(1−x) = 4x

4x+2+41−x

41−x+2

=4x

4x+2+4

4+2(4x)

=4x

4x+2+2

2+4x

⇒f(x) + f(1−x) = 1

Now f 1

2023 +f2

2023 +f3

2023 + .......+

.......... + f 1 −3

2023 +f 1 −2

2023 +f 1 −1

2023

Nowsumoftermsequidistantfrombeginningandendis1

Sum =1+1+1+ ......... + 1 (1011times)

( ) ( ) ( )

-------------------------------------------------------------------------------------------------

Question46

Forsomea,b,c∈ ℕ,letf (x) = ax −3andg(x) = xb+c,x∈ ℝ.If

(fog )−1(x) = x−7

1∕3then(fog)(ac) + (gof )(b)isequalto________.

[25-Jan-2023Shift1]

Answer:2039

Solution:

-------------------------------------------------------------------------------------------------

Question47

Letf : ℝ → ℝbeafunctiondefinedbyf (x) =

log√m{√2(sin x −cos x) + m−2},forsomem,suchthattherangeoffis

[0,2].Thenthevalueofmis_______

[25-Jan-2023Shift2]

Options:

A.5

B.3

C.2

D.4

Answer:A

Solution:

( )

=1011

Letfog(x) = h(x)

⇒h−1(x) = x−7

⇒h(x) = fog(x) = 2x3+7

fog (x) = a(xb+c) − 3

⇒a=2,b=3,c=5

⇒fog (ac) = fog(10) = 2007

g(f(x) = (2x −3)3+5.

⇒gof(b) = gof(3) = 32

⇒sum =2039

( )

Since,

−√2≤sin x −cos x ≤ √2

-------------------------------------------------------------------------------------------------

Question48

Thenumberoffunctionsf : {1,2,3,4} → {a∈ ℤ: | a| ≤8}satisfying

f(n)+1

nf(n+1) = 1, ∀n∈ {1,2,3}is

[25-Jan-2023Shift2]

Options:

A.3

B.4

C.1

D.2

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question49

Letf(x) = 2xn+λ,λ∈ ℝ, n∈ ℕ,andf(4) = 133,f (5) = 255.Thenthesum

ofallthepositiveintegerdivisorsof(f(3) − f(2))is

[25-Jan-2023Shift2]

Options:

A.61

B.60

∴ − 2≤ √2(sin x −cos x) ≤ 2

(Assume √2(sin x −cos x) = k)

−2≤k≤2... (i)

f(x) = log√m(k+k−2)

Given,

0≤f(x) ≤ 2

0≤log√min(k+m−2) ≤ 2

1≤k+m−2≤m

−m+3≤k≤2... ..ii)

Fromeq.(i)&(ii),weget−m+3= −2

⇒m=5

f: {1,2,3,4} → {a∈ℤ:|a| ≤8}

f(n) + 1

nf(n+1) = 1, ∀n∈ {1,2,3}

f(n+1)mustbedivisiblebyn

f(4) ⇒ −6, −3,0,3,6

f(3) ⇒ −8, −6, −4, −2,0,2,4,6,8

f(2) ⇒ −8, ..............., 8

f(1) ⇒ −8, ..............., 8

f(4)

3mustbeoddsincef(3)shouldbeeventherefore2solutionpossible.

C.58

D.59

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question50

Letf :R→Rbeafunctionsuchthatf (x) = x2+2x +1

x2+1.Then

[29-Jan-2023Shift1]

Options:

A.f (x)ismany-onein(−∞, −1)

B.f (x)ismany-onein(1,∞)

C.f (x)isone-onein[1,∞)butnotin(−∞,∞)

D.f (x)isone-onein(−∞,∞)

Answer:C

Solution:

f(x) = 2xn+λ

f(4) = 133

f(5) = 255

133 =2×4n+λ...(1)

255 =2×5n+λ...(2)

(2) − (1)

122 =2(5n−4n)

⇒5n−4n=61

∴n=3&λ=5

Now,f(3) − f(2) = 2(33−23) = 38

NumberofDivisorsis1,2,19,38; &theirsumis60

-------------------------------------------------------------------------------------------------

Question51

Thedomainoff (x) = log(x+1)(x−2)

e2logex− (2x +3),x∈Ris

[29-Jan-2023Shift1]

Options:

A.ℝ − {1−3}

B.(2,∞) − {3}

C.(−1,∞) − {3}

D.ℝ − {3}

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question52

Considerafunctionf : ℕ → ℝ,satisfying

f(1) + 2f (2) + 3f (3) + ... + xf (x) = x(x+1)f(x); x≥2withf (1) = 1.Then

f(2022)+1

f(2028)isequalto

x−2>0⇒x>2

x+1>0⇒x> −1

x+1≠1⇒x≠0 and x >0

Denominator

x2−2x −3≠0

(x−3)(x+1) ≠ 0

x≠ −1,3

SoAns(2,∞) − {3}

[29-Jan-2023Shift2]

Options:

A.8200

B.8000

C.8400

D.8100

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question53

Supposef :R→ (0,∞)beadifferentiablefunctionsuchthat

5f(x+y) = f(x) ⋅ f(y), ∀x,y∈R.Iff (3) = 320,then 5

∑

n=0f(n)isequalto:

[30-Jan-2023Shift1]

Options:

A.6875

B.6575

C.6825

D.6528

Answer:C

Solution:

Givenforx≥2

f(1) + 2f (2) + .... + xf (x) = x(x+1)f(x)

replace x by x +1

⇒x(x+1)f(x) + (x+1)f(x+1)

= (x+1)(x+2)f(x+1)

⇒x

f(x+1)+1

f(x)=(x+2)

f(x)

⇒xf (x) = (x+1)f(x+1) = 1

2,x≥2

f(2) = 1

4,f(3) = 1

Nowf(2022) = 1

4044

f(2028) = 1

4056

So, 1

f(2022)+1

f(2028)=4044 +4056 =8100

Option(3)

5f(x+y) = f(x) ⋅ f(y)

-------------------------------------------------------------------------------------------------

Question54

LetS = {1,2,3,4,5,6}.Thenthenumberofonefunctionsf :S→P(S),

whereP(S)denotethepowersetofS,suchthatf (n) ⊂ f(m)wheren <m

is________.

[30-Jan-2023Shift1]

Answer:3240

Solution:

5f(0) = f(0)2⇒f(0) = 5

5f(x+1) = f(x) ⋅ f(1)

⇒f(x+1)

f(x)=f(1)

⇒f(1)

f(0)⋅f(2)

f(1)⋅f(3)

f(2)=f(1)

⇒320

5=(f(1))3

53⇒f(1) = 20

∴5f(x+1) = 20 ⋅f(x) ⇒ f(x+1) = 4f(x)

∑

n=0

f(n) = 5+5.4 +5.42+5.43+5.44+5.45

=5[46−1]

3=6825

( )

LetS= {1,2,3,4,5,6},thenthenumberofone-onefunctions,f:S⋅P(S),whereP(S)denotesthepowersetofS,such

thatf(n) < f(m)wheren<mis

n(S) = 6

P(S) = φ{1}.. . {6} {1,2} ...

{5,6} ... {1,2,3,4,5,6}

−64 elements 

case−1

f(6) = Si.e.1option,

f(5) = any5elementsubsetAofSi.e.6options,

f(4) = any4elementsubsetBofAi.e.5options,

f(3) = any3elementsubsetCofBi.e.4options,

f(2) = any2elementsubsetDofCi.e.3options,

f(1) = any1elementsubsetEofDoremptysubseti.e.3

options,

Totalfunctions = 1080

Case-2

f(6) = any5elementsubsetAofSi.e.6options,

f(5) = any4elementsubsetBofAi.e.5options,

f'(4)=any3elementsubsetCofBi.e.4options,

f(3) = any2elementsubsetDofCi.e.3options,

f'(2) = any1elementsubsetEofDi.e.2options,

f(1) = emptysubseti.e.1option

Totalfunctions = 720

Case−3

f(6) = S

f(5) = any4elementsubsetAof'Si.e.15options,

f(4) = any3elementsubsetBofAi.e.4options,

f(3) = any2elementsubsetCofBi.e.3options,

f(2) = any1elementsubsetDofCi.e.2options,

f(1) = emptysubseti.e.1option

Totalfunctions = 360

Case−4

{ }

-------------------------------------------------------------------------------------------------

Question55

Letf 1(x) = 3x +2

2x +3,x∈R−−3

Forn ≥2,definefn(x) = f10fn−1(x).

Iff5(x) = ax +b

bx +a,gcd(a,b) = 1,thena +bisequalto________.

[30-Jan-2023Shift1]

Answer:3125

Solution:

-------------------------------------------------------------------------------------------------

Question56

Therangeofthefunctionf (x) = √3−x+ √2+xis

[30-Jan-2023Shift2]

Options:

{ }

f(6) = S

f(5) = any5elementsubsetAofSi.e.6options,

f(4) = any3elementsubsetBofAi.e.10options,

f(3) = any2elementsubsetCofBi.e.3options,

f(2) = any1elementsubsetDofCi.e.2options,

f(1) = emptysubseti.e.1option

Totalfunctions = 360

Case−5

f(6) = S

f(5) = any5elementsubsetAofSi.e.6options,

f(4) = any4elementsubsetBofAi.e.5options,

f(3) = any2elementsubsetCofBi.e.6options,

f(2) = any1elementsubsetDofCi.e.2options,

f(1) = emptysubseti.e.1option

Totalfunctions = 360

Case-6

f(6) = S

f(5) = any5elementsubsetAofSi.e.6options,

f(4) = any4elementsubsetBofAi.e.5options,

f(3) = any3elementsubsetCofBi.e.4options,

f(2) = any1elementsubsetDofCi.e.3options,

f(1) = emptysubseti.e.1option

Totalfunctions = 360

∴Numberofsuchfunctions = 3240

f1(x) = 3x +2

2x +3

⇒f2(x) = 13x +12

12x +13

⇒f3(x) = 63x +62

62x +63

∴f5(x) = 1563x +1562

1562x +1563

a+b=3125

A.[√5, √10]

B.[2√2, √11]

C.[√5, √13]

D.[√2, √7]

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question57

LetA = {1,2,3,5,8,9}.Thenthenumberofpossiblefunctionsf :A→A

suchthatf(m⋅n) = f(m) ⋅ f(n)foreverym,n∈Awithm ⋅n∈Aisequal

to_______.

[30-Jan-2023Shift2]

Answer:432

Solution:

-------------------------------------------------------------------------------------------------

Question58

Ifthedomainofthefunctionf (x) = [x]

1+x2,where[x]isgreatestinteger

≤x,is[2,6),thenitsrangeis

[31-Jan-2023Shift1]

Options:

A. 5

26,2

5−9

29,27

109,18

89,9

( ] { }

y2=3−x+2+x+2√(3−x)(2+x)

=5+2 6 +x−x2

y2=5+225

4−x−1

ymax = √5+5= √10

ymin = √5

√

√( )

f(1) = 1;f(9) = f(3) × f(3)

i.e.,f(3) = 1or3

Totalfunction = 1×6×2×6×6×1=432

B. 5

26,2

C. 5

37,2

5−9

29,27

109,18

89,9

D. 5

37,2

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question59

Theabsoluteminimumvalue,ofthefunction

f(x) = x2−x+1| +[x2−x+1],where[t]denotesthegreatestinteger

function,intheinterval[−1,2],is:

[31-Jan-2023Shift2]

Options:

A. 3

B. 3

C. 1

D. 5

Answer:A

Solution:

( ]

( ] { }

( ]

f(x) = 2

1+x2x∈ [2,3)

f(x) = 3

1+x2x∈ [3,4)

f(x) = 4

1+x2x∈ [4,5)

f(x) = 5

1+x2x∈ [5,6)

Solution:

-------------------------------------------------------------------------------------------------

Question60

Letf : ℝ − {2,6} → ℝberealvaluedfunctiondefinedasf (x) = x2+2x +1

x2−8x +12.

Thenrangeoff is

[31-Jan-2023Shift2]

Options:

A. −∞, − 21

4∪ [0,∞)

B. −∞, − 21

4∪ (0,∞)

C. −∞, − 21

4∪21

4,∞

D. −∞, − 21

4∪ [1,∞)

Answer:A

Solution:

( ]

( )

( ] [ )

( ]

f(x) = | x2−x+1| +[x2−x+1]; x∈ [−1,2]

Letg(x) = x2−x+1

=x−1

2+3

4

∵ | x2−x+1|and [x2−x+2]

Bothhaveminimumvalueatx=1∕2

⇒Minimum f(x) = 3

4+0

( )

Lety=x2+2x +1

x2−8x +12

Bycrossmultiplying

yx2−8 xy +12y −x2−2x −1=0

x2(y−1) − x(8y +2) + (12y −1) = 0

Case1,y≠1

D≥0

⇒ (8y +2)2−4(y−1)(12y −1) ≥ 0

⇒y(4y +21) ≥ 0

y∈ −∞,−21

4∪ [0,∞) − {1}

Case2,y=1

x2+2x +1=x2−8x +12

10x =11

x=11

10 So, y canbe 1

Hencey∈ −∞,−21

4∪ [0,∞)

( ]

-------------------------------------------------------------------------------------------------

Question61

Letf :R− {0,1} → Rbeafunctionsuchthatf (x) + f1

1−x=1+x.

Thenf (2)isequalto:

[1-Feb-2023Shift2]

Options:

A. 9

B. 9

C. 7

D. 7

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question62

Letf : ℝ → ℝbedefinedasf (x) = x−1andg : ℝ − {1, −1} → ℝbe

definedasg(x) = x2

x2−1.Thenthefunctionfogis:

[26-Jun-2022-Shift-2]

Options:

A.one-onebutnotonto

B.ontobutnotone-one

C.bothone-oneandonto

D.neitherone-onenoronto

Answer:D

( )

f(x) + f1

1−x=1+x

x=2⇒f(2) + f(−1) = 3

x= −1⇒f(−1) + f1

2=0...(2)

x=1

2⇒f1

2+f(2) = 3

2...(3)

(1) + (3) − (2) ⇒ 2f(2) = 9

∴f(2) = 9

( )

Solution:

Letf :R→Rbeafunctiondefinedbyf (x) = 2e2x

e2x +e.

Then

100 +f2

100 +f3

100 + ... + f99

100

isequalto____

[27-Jun-2022-Shift-1]

Answer:99

Solution:

( ) ( ) ( ) ( )

Given,

f(x) = 2e2x

e2x +e

∴f(1−x) = 2e2(1−x)

e2(1−s)+e

2⋅e2

e2x

e2π +e

=2e2

e2+e2x ⋅e

=2e2

e(e+e2x)

=2e

e+e2∆

∴f(x) + f(1−x) = 2e2x

e2x +e+2e

e2x +e

=2(e2x +e)

e2x +e

=2......(1)

Now,

100 +f99

100

=f1

100 +f 1 −1

100

=2[asf(x) + f(1−x) = 2]

100 +f 1 −2

100 =2

⁞

f49

100 +f 1 −49

100 =2

∴Totalsum = 49 ×2

Remainingterm = f50

100 =f1

Putx=1

2inequation(1),weget

2+f 1 −1

2=2

⇒2f 1

2=2

⇒f1

2=1

∴Sum =49 ×2+1=99

( ) ( )

( )

Question63

Question64

LetS = {1,2,3,4,5,6,7,8,9,10}.Definef :S→Sas

f(n) = 2n ,if n =1, 2, 3, 4, 5

2n −11 ,if n =6, 7, 8, 9, 10,

Letg :S→Sbeafunctionsuchthatfog(n) = n+1 ,ifnisodd

n−1 ,ifniseven

Theng(10)g(1) + g(2) + g(3) + g(4) + g(5) )isequalto

[27-Jun-2022-Shift-2]

Answer:190

Solution:

-------------------------------------------------------------------------------------------------

Question65

Letafunctionf :N→Nbedefinedby

f(n) =

2n n =2, 4, 6, 8, ......

n−1 n =3, 7, 11, 15, ......

n+1

2n=1, 5, 9, 13, .......

then,f is

[28-Jun-2022-Shift-1]

Options:

A.one-onebutnotonto

B.ontobutnotone-one

C.neitherone-onenoronto

{

[

∵f(n) = 2n n =1234 5

2n −11 n =678910.

∴f(1) = 2,f(2) = 4, ......, f(5) = 10

and f (6) = 1,f(7) = 3,f(8) = 5, ......., f(10) = 9

Now,f(g(n)) = n+1 if n isodd

n−1 if n iseven .

f(g(10)) = 9⇒g(10) = 10

f(g(1)) = 2⇒g(1) = 1

. f (g(2)) = 1⇒g(2) = 6

∴f(g(3)) = 4⇒g(3) = 2

f(g(4)) = 3⇒g(4) = 7

f(g(5)) = 6⇒g(5) = 3

∴g(10)g(1) + g(2) + g(3) + g(4) + g(5) ) = 190

{

D.one-oneandonto

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question66

Theprobabilitythatarandomlychosenone-onefunctionfromtheset

{a,b,c,d}totheset{1,2,3,4,5}satisfiesf (a) + 2f (b)−f(c) = f(d)is

[28-Jun-2022-Shift-2]

Options:

A. 1

B. 1

C. 1

D. 1

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question67

LetS = {1,2,3,4}.Thenthenumberofelementsintheset

{f:S×S→S:f isontoandf (a,b) = f(b,a) ≥ a∀(a,b) ∈ S×S }is____

[28-Jun-2022-Shift-2]

Whenn=1,5,9,13then n+1

2willgivealloddnumbers.

Whenn=3,7,11,15......

n−1willbeevenbutnotdivisibleby4

Whenn=2,4,6,8...

Then2nwillgiveallmultiplesof4

SorangewillbeN.

Andnotwovaluesofngivesamey,sofunctionisone-oneandonto.

Numberofone-onefunctionfrom{a,b,c,d}toset{1,2,3,4,5}is5P4=120n(s).

Therequiredpossiblesetofvalue(f(a),f(b), f(c), f(d) )suchthatf(a) + 2f (b) − f(c) = f(d)are

(5,3,2,1), (5,1,2,3), (4,1,3,5), (3,1,4,5), (5,4,3,2)and(3,4,5,2)

∴n(E) = 6

∴Requiredprobability = n(E)

n(S)=6

120 =1

Answer:37

Solution:

-------------------------------------------------------------------------------------------------

Question68

Thedomainofthefunctioncos−1

2sin−11

4x2−1

πis:

[29-Jun-2022-Shift-1]

Options:

A.R − − 1

2,1

B.(−∞, −1] ∪ [1,∞) ∪ {0}

C. −∞,−1

2∪1

2,∞∪ {0}

D. −∞,−1

√2∪1

√2,∞∪ {0}

Answer:D

Solution:

(( ) )

{ }

( ) ( )

( ] [ )

Thereare16orderedpairsinS×S.Wewritealltheseorderedpairsin4setsasfollows.

A= {(1,1)}

B= {(1,4), (2,4), (3,4)(4,4), (4,3), (4,2), (4,1)}

C= {(1,3), (2,3), (3,3), (3,2), (3,1)}

D= {(1,2), (2,2), (2,1)}

AllelementsofsetBhaveimage4andonlyelementofAhasimage1.

AllelementsofsetChaveimage3or4andallelementsofsetDhaveimage2or3or4.

Wewillsolvethisquestionintwocases.

CaseI:WhennoelementofsetChasimage3.

Numberofontofunctions = 2(whenelementsofsetDhaveimages2or3)

CaseII:WhenatleastoneelementofsetChasimage3.

Numberofontofunctions = (23−1)(1+2+2) = 35

Totalnumberoffunctions = 37

−1≤

2sin −11

4x2−1

π≤1

⇒− π

2≤sin −11

4x2−1≤π

⇒−1≤1

4x2−1≤1

∴1

4x2−1+1≥0

⇒1+4x2−1

4x2−1≥0

( )

-------------------------------------------------------------------------------------------------

Question69

Letc,k∈R.Iff (x) = (c+1)x2+ (1−c2)x+2kand

f(x+y) = f(x) + f(y) − xy,forallx,y∈R,thenthevalueof

|2(f(1) + f(2) + f(3) + ...... + f(20))|isequalto____

[29-Jun-2022-Shift-1]

Answer:3395

Solution:

-------------------------------------------------------------------------------------------------

⇒4x2

4x2−1≥0

⇒4x2

(2x +1)(2x −1)≥0..... . (1)

∴x∈ −α, − 1

2∪ {0} ∪ 1

2,α

And 1

4x2−1−1≤0

⇒1−4x2+1

4x2−1≤0

⇒2−4x2

4x2−1≤0

⇒2x2−1

4x2−1≥0

⇒(√2x+1)(√2x−1)

(2x +1)(2x −1)≥0

x∈ −α, − 1

√2∪ − 1

2,1

2∪1

√2,α

From(3)and(4),weget

∴x∈ −α, − 1

√2∪1

√2,α∪ {0}

( ) ( )

( ) ( ) ( )

[ ) [ )

f(x)ispolynomial

Puty=1∕xingivenfunctionalequationweget

f x +1

x=f(x) + f1

x−1

⇒(c+1)x+1

2+ (1−c2)x+1

x+2K

= (c+1)x2+ (1−c2)x+2K + (c+1)1

x2+ (1−c2)1

x+2K −1

⇒2(c+1) = 2K −1.... . (1)

andputx=y=0weget

f(0) = 2+f(0) − 0⇒f(0) = 0⇒k=0

∴k=0 and 2c = −3⇒c= −3∕2

f(x) = − x2

2−5x

4=1

4(5x +2x2)

∑

i=1

f(i) = −2

5.20.21

2+2.20.21.41

=−1

2(2730 +5740)

= − 6790

2=3395.

( ) ( )

|||( ) |

| |

Question70

Letf (x)andg(x)betworealpolynomialsofdegree2and1respectively.

Iff (g(x)) = 8x2−2xandg(f(x)) = 4x2+6x +1,thenthevalueof

f(2) + g(2)is_____

[29-Jun-2022-Shift-2]

Answer:18

Solution:

-------------------------------------------------------------------------------------------------

Question71

Thedomainofthefunction

f(x) =

cos−1x2−5x +6

x2−9

loge(x2−3x +2)is:

[24-Jun-2022-Shift-1]

Options:

A.(−∞,1) ∪ (2,∞)

B.(2,∞)

C. −1

2,1∪ (2,∞)

D. −1

2,1∪ (2,∞) − 3,3+ √5

2,3− √5

Answer:D

Solution:

( )

[ )

[ ) { }

f(g(x) = 8x2−2x.

g(f(x) = 4x2+6x +1.

So, g(x) = 2x −1

&f(x) = 2x2+3x +1

f(2) = 8+6+1=15

Ans.18

-------------------------------------------------------------------------------------------------

Question72

Thenumberofone-onefunctionsf : {a,b,c,d} → {0,1,2, ......, 10}

suchthat2f (a) − f(b) + 3f (c) + f(d) = 0is

[24-Jun-2022-Shift-1]

Answer:31

Solution:

f(d)can'tbe9and10asif

(

) = 9or10then

(

) = 2+9=11or

(

) = 2+10 =12,whichisnotpossibleashereany

function'smaximumvaluecanbe10.

∴Nofunctionispossibleinthiscase.

∴Totalpossiblefunctionswhenf(c)=0andf(a)=1,2,3and4are=7+5+3+2=17

CaseII:

(1)Whenf(c)=1andf(a)=0then

3×1+2×0+f(d)=f(b)

⇒3+f(d)=f(b)

∴Possiblevalueoff(d)=2,3,4,5,6,7

∴Totalpossiblefunctionsinthiscase=6

(2)Whenf(c)=1andf(a)=2then

3×1+2×2+f(d)=f(b)

⇒7+f(d)=f(b)

∴Possiblevalueoff(d)=0,3

∴Totalpossiblefunctionsinthiscase=2

(3)Whenf(c)=1andf(a)=3then

3×1+2×3+f(d)=f(b)

⇒9+f(d)=f(b)

∴Possiblevalueoff(d)=0

∴Totalpossiblefunctionsinthiscase=1

∴Totalpossiblefunctionswhenf(c)=1andf(a)=0,2and3are=6+2+1=9

CaseIII:

(1)Whenf(c)=2andf(a)=0then

3×2+2×0+f(d)=f(b)

⇒6+f(d)=f(b)

∴Possiblevaluesoff(d)=1,3,4

∴Totalpossiblefunctionsinthiscase=3

(2)Whenf(c)=2andf(a)=1then,

3×2+2×1+f(d)=f(b)

⇒8+f(d)=f(b)

∴Possiblevaluesoff(d)=0

∴Totalpossiblefunctioninthiscase=1

∴Totalpossiblefunctionswhenf(c)=2andf(a)=0,1are=3+1=4

CaseIV:

(1)Whenf(c)=3andf(a)=0then

3×3+2×0+f(d)=f(b)

⇒9+f(d)=f(b)

∴Possiblevaluesoff(d)=1

∴Totalone-onefunctionsfromfourcases

=17+9+4+1=31

-------------------------------------------------------------------------------------------------

Question73

LetR1andR2berelationsontheset{1,2, ......., 50}suchthat

R1= { (p,pn) : p.isaprimeandn ≥0isaninteger}and

R2= { (p,pn) : p.isaprimeandn =0or1}

Then,thenumberofelementsinR1−R2is___

[28-Jun-2022-Shift-1]

Answer:8

Solution:

R1−R2= {(2,22), (2,23), (2,24), (2,25), (3,32), (3,33), (5,52), (7,72)}

Sonumberofelements = 8

Question74

LetR1= {(a,b) ∈ N×N: | a−b| ≤13}and

R2= {(a,b) ∈ N×N: | a−b| ≠13}.ThenonN :

[28-Jun-2022-Shift-2]

Options:

A.BothR1andR2areequivalencerelations

B.NeitherR1norR2isanequivalencerelation

C.R1isanequivalencerelationbutR2isnot

D.R2isanequivalencerelationbutR1isnot

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question75

TheprobabilitythatarelationRfrom{x,y}to{x,y}isbothsymmetric

andtransitive,isequalto

[29-Jun-2022-Shift-2]

Options:

A. 5

B. 9

R1= {(a,b) ∈ N×N:|a−b| ≤13}and

R2= {(a,b) ∈ N×N:|a−b| ≠13}

InR1:∵ | 2−11 | = 9≤13

∴ (2,11) ∈ R1and(11,19) ∈ R1but(2,19) ∉ R1

∴R1isnottransitive

HenceR1isnotequivalence

InR2: (13,3) ∈ R2and(3,26) ∈ R2but(13,26) ∉ R2(∵ | 13 −26 | = 13)

∴R2isnottransitive

HenceR2isnotequivalence.

C. 11

D. 13

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question76

Thenumberofbijectivefunctions

f: {1,3,5,7, ..., 99} → {2,4,6,8, .. . 100},suchthat

f(3) ≥ f(9) ≥ f(15) ≥ f(21) ≥ ... . f(99),is

[25-Jul-2022-Shift-2]

Options:

A.50P17

B.50P33

C.33! × 17!

D. 50!

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question77

Letf (x)beaquadraticpolynomialwithleadingcoefficient1suchthat

f(0) = p,p≠0,andf (1) = 1

3.Iftheequationsf (x) = 0and

f∘f∘f∘f(x) = 0haveacommonrealroot,thenf (−3)isequalto___

[25-Jul-2022-Shift-2]

Totalno.ofrelations = 22×2=16

Fav.relation = φ, {(x,x)}, {(y,y)}, {(x,x)(y,y)}

{(x,x), (y,y), (x,y)(y,x)}

Prob. =5

Asfunctionisone-oneandonto,outof50elementsofdomainset17elementsarefollowingrestriction

f(3) > f(9) > f(15)...... > f(99)

Sonumberofways = 50C17 ⋅1.33!

=50P33

Answer:25

Solution:

-------------------------------------------------------------------------------------------------

Question78

Letf :R→Rbeacontinuousfunctionsuchthatf (3x) − f(x) = x.If

f(8) = 7,thenf (14)isequalto:

[26-Jul-2022-Shift-1]

Options:

A.4

B.10

C.11

D.16

Answer:B

Solution:

Letf(x) = (x−α)(x−β)

Itisgiventhatf(0) = p⇒αβ =p

andf(1) = 1

3⇒ (1−α)(1−β) = 1

Now,letusassumethat,αisthecommonrootoff(x) = 0andfofofof(x) = 0

fofofof(x) = 0

⇒fofof (0) = 0

⇒fof(p) = 0

So,f(p)iseitherαorβ.

(p−α)(p−β) = α

(αβ −α)(αβ −β) = α⇒ (β−1)(α−1)β=1(∵α≠0)

So,β=3

(1−α)(1−3) = 1

3

α=7

f(x) = x−7

6(x−3)

f(−3) = −3−7

6(3−3) = 25

( )

f(3x) − f(x) = x...... (1)

x→x

f(x) − fx

3=x

3..... . (2)

Againx→x

3−fx

9=x

32......

Similarly

( )

( ) ( )

-------------------------------------------------------------------------------------------------

Question79

Thedomainofthefunction

f(x) = sin−1[2x2−3] + log2log 1

(x2−5x +5),where[t]isthegreatest

integerfunction,is:

[27-Jul-2022-Shift-2]

Options:

A. −5

2,5− √5

B. 5− √5

2,5+ √5

C. 1,5− √5

D. 1,5+ √5

Answer:C

Solution:

( )

(√)

( )

[ )

3n−2−fx

3n−1=x

3n−1... . (n)

Addingalltheseandapplyingn→∞

lim

n→∞

f(3x) − fx

3n−1=x 1 +1

3+1

32+ ...

f(3x) − f(0) = 3x

Puttingx=8

f(8) − f(0) = 4

⇒f(0) = 3

Puttingx=14

f(14) − 3=7⇒f(14) = 0

( ) ( )

( ( ) ) ( )

−1≤2x2−3<2

or2≤2x2<5

or1≤x2<5

x∈ − 5

2, −1∪1,5

log 1

(x2−5x +5) > 0

0<x2−5x +5<1

x2−5x +5>0&x2−5x +4<0

x∈ −∞,5− √5

2∪5+ √5

2,∞

&x∈ (−∞,1) ∪ (4,∞)

Takingintersection

x∈1,5− √5

(√] [ √)

( ) ( )

( )

-------------------------------------------------------------------------------------------------

Question80

Thenumberoffunctionsf ,fromthesetA = {x∈N:x2−10x +9≤0}

tothesetB = {n2:n∈N}suchthatf (x) ≤ (x−3)2+1,foreveryx ∈A,

is_________.

[27-Jul-2022-Shift-2]

Answer:1440

Solution:

-------------------------------------------------------------------------------------------------

Question81

Consideringonlytheprincipalvaluesoftheinversetrigonometric

functions,thedomainofthefunctionf (x) = cos−1x2−4x +2

x2+3is:

[28-Jul-2022-Shift-1]

Options:

A. −∞,1

B. −1

4,∞

C.(−1∕3,∞)

D. −∞,1

Answer:B

Solution:

( )

( ]

[ )

( ]

A= {x∈N,x2−10x +9≤0}

= {1,2,3, ..., 9}

B= {1,4,9,16, ......}

f(x) ≤ (x−3)2+1

f(1) ≤ 5,f(2) ≤ 2, ......... . f(9) ≤ 37

x=1has2choices

x=2has1choice

x=3has1choice

x=4has1choice

x=5has2choices

x=6has3choices

x=7has4choices

x=8has5choices

x=9has6choices

∴Totalfunctions = 2×1×1×1×2×3×4×5×6=1440

Solution:

-------------------------------------------------------------------------------------------------

Question82

Letα,βandγbethreepositiverealnumbers.Let

f(x) = αx5+βx3+γx,x∈Randg :R→R.besuchthatg(f(x)) = xforall

x∈R.Ifa1,a2,a3, ..., anbeinarithmeticprogressionwithmeanzero,

thenthevalueoff g 1

∑

i=1f(ai)isequalto:

[28-Jul-2022-Shift-1]

Options:

A.0

B.3

C.9

D.27

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question83

Thenumberofelementsintheset

S=x∈ ℝ : 2 cos x2+x

6=4x+4−xis:

[29-Jul-2022-Shift-2]

( ( ) )

{ ( ) }

−1≤x2−4x +2

x2+3≤1

⇒−x2−3≤x2−4x +2≤x2+3

⇒2x2−4x +5≥0 −4x ≤1

x∈R&x≥ − 1

Sodomainis − 1

4,∞

[ )

f g 1

∑

i=1

f(ai)

a1+a2+a3+ ...... + an

n=0

∴Firstandlastterm,secondandsecondlastandsoonareequalinmagnitudebutoppositeinsign.

f(x) = αx5+βx3+γx

∑

i=1

f(ai) = α(a1

5+a2

5+a3

5+ ... + an

5) + β(a1

3+a2

3+ ... + an

3) + γ(a1+a2+ ... + an)

=0α +0β +0γ

∴f g 1

∑

i=1

f(ai) = 1

∑

i=1

f(ai) = 0

( ( ) )

Options:

A.1

B.3

C.0

D.infınite

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question84

Thedomainofthefunctionf (x) = sin−1x2−3x +2

x2+2x +7is:

[29-Jul-2022-Shift-2]

Options:

A.[1,∞)

B.[−1,2]

C.[−1,∞)

D.(−∞,2]

Answer:C

Solution:

( )

2 cos x2+x

6=4x+4−x

L.H.S≤2.&R.H.S.≥2

HenceL.H.S = 2&R.H.S = 2

2 cos x2+x

6=2 4x+4−x=2

Checkx=0Possiblehenceonlyonesolution.

( )

f(x) = sin −1x2−3x +2

x2+2x +7

−1≤x2−3x +2

x2+2x +7≤1

x2−3x +2

x2+2x +7≤1

x2−3x +2≤x2+2x +7

5x ≥ −5

x≥ −1

And x2−3x +2

x2+2x +7≥ −1

x2−3x +2≥ −x2−2x −7

2x2−x+9≥0

( )

-------------------------------------------------------------------------------------------------

Question85

Thetotalnumberoffunctions,

f: {1,2,3,4} → {1,2,3,4,5,6}suchthatf (1) + f(2) = f(3),isequalto

[25-Jul-2022-Shift-1]

Options:

A.60

B.90

C.108

D.126

Answer:B

Solution:

x∈R

(i)∩(ii)

Domain∈[−1,∞)

Given,f(1) + f(2) = f(3)

Itmeansf(1), f(2)andf(3)aredependentoneachother.Butthereisnoconditiononf(4),sof(4)canbe

f(4) = 1,2,3,4,5,6.

Forf(1), f(2)andwehavetofindhowmanyfunctionspossiblewhichwillsatisfytheconditionf(1) + f(2) = f(3)

Case1:

Whenf(3) = 2thenpossiblevaluesoff(1)andf(2)whichsatisfyf(1) + f(2) = f(3)isf(1) = 1andf(2) = 1.

Andf(4)canbe = 1,2,3,4,5,6

∴Totalpossiblefunctions = 1×6=6

Case2:

Whenf(3) = 3thenpossiblevalues

(1)f(1) = 1andf(2) = 2

(2)f(1) = 2andf(2) = 1

Andf(4)canbe = 1,2,3,4,5,6.

∴Totalfunctions = 2×6=12

Case3:

Whenf(3) = 4then

(1)f(1) = 1andf(2) = 3

(2)f(1) = 2andf(2) = 2

(3)f(1) = 3andf(2) = 1

Andf(4)canbe = 1,2,3,4,5,6

∴Totalfunctions = 3×6=18

Case4:

Whenf(3) = 5then

(1)f(1) = 1andf(4) = 4

(2)f(1) = 2andf(4) = 3

(3)f(1) = 3andf(4) = 2

(4)f(1) = 4andf(4) = 1

Andf(4)canbe = 1,2,3,4,5and6

∴Totalfunctions = 4×6=24

Case5:

Whenf(3) = 6then

(1)f(1) = 1andf(2) = 5

(2)f(1) = 2andf(2) = 4

(3)f(1) = 3andf(2) = 3

(4)f(1) = 4andf(2) = 2

(5)f(1) = 5andf(2) = 1

Andf(4)canbe = 1,2,3,4,5and6

-------------------------------------------------------------------------------------------------

Question86

Letf :N→Rbeafunctionsuchthatf (x+y) = 2f (x)f(y)fornatural

numbersxandy.Iff(1)=2,thenthevalueofαforwhich

∑

k=1f(α+k) = 512

3(220 −1)

holds,is:

[25-Jun-2022-Shift-1]

Options:

A.2

B.3

C.4

D.6

Answer:C

Solution:

∴Totalpossiblefunctions = 5×6=30

∴Totalfunctionsfromthose5casesweget = 6+12 +18 +24 +30 = 90

Question87

Letf :R→Rbeafunctiondefinedbyf (x) = 2 1 −x25

2(2+x25)

50.If

Answer:2

Solution:

( ( ) )

thefunctiong(x) = f(f(f(x)) + f(f(x)),thenthegreatestintegerless

thanorequaltog(1)is___

[25-Jun-2022-Shift-1]

Letf (x) = x−1

x+1,x∈R− {0, −1,1}.Iff n+1(x) = f(fn(x))foralln ∈N,

thenf 6(6) + f7(7)isequalto:

[26-Jun-2022-Shift-1]

Options:

A. 7

B.−3

C. 7

D.−11

Answer:B

Solution:

Question88

Question89

Therangeofthefunction,

f(x) = log√53+cos 3π

4+x +cos π

4+x+cos π

4−x −cos 3π

4−x

[2021,01Sep.Shift-II]

( ( ) ( ) ( ) ( ) )

Options:

A.(0, √5)

B.[−2,2]

C. 1

√5, √5

D.[0,2]

Answer:D

Solution:

[ ]

Solution:

f(x) = log√53+cos 3π

4+x+cos π

4+x

+cos π

4−x−cos 3π

4−x

=log√5(3− √2sin x + √2cos x)

∵−2≤ −√2sin x + √2cos x ≤2

⇒1≤3− √2sin x + √2cos x ≤5

⇒log√51≤log√5(3− √2sin x + √2cos x)

⇒0≤f(x) ≤ 2

⇒f(x) ∈ [0,2]

( ( ) ( )

( ) ( ) )

Question90

Thedomainofthefunctionf (x) = sin−13x2+x−1

(x−1)2+cos−1x−1

x+1is

[2021,31Aug.Shift-II]

( ) ( )

Options:

A. 0,1

B.[−2,0] ∪ 1

4,1

C. 1

4,1

2∪ {0}

D. 0,1

Answer:C

Solution:

[ ]

f(x) = sin−13x2+x−1

(x−1)2+cos−1x−1

x+1

−1≤x−1

x+1≤1⇒ −1−1≤x−1

x+1−1≤1−1

⇒ − 2≤−2

x+1≤0⇒0≤1

x+1≤1

⇒1≤x+1<∞

⇒0≤x<∞

⇒x∈ [0,∞)

and−1≤3x2+x−1

(x−1)2≤1

⇒ − (x−1)2≤3x2+x−1≤ (x−1)2,x≠1

⇒ − (x2−2x +1) ≤ 3x2+x−1

and3x2+x−1≤x2−2x +1

⇒4x2−x≥0

and2x2+3x −2≤0

⇒x(4x −1) ≥ 0

and(x+2)(2x −1) ≤ 0

⇒x∈ (−∞,0] ∪ 1

4,∞

andx∈ −2,1

⇒x∈ (−2,0] ∪ 1

4,1

DomainoffinEq.(i)∩Eq.(ii)

∴x∈ {0} ∪ 1

4,1

( ) ( )

[ )

[ ]

[2021,31Aug.Shift-11]

Question91

Letf :N→N beafunctionsuchthatf (m+n) = f(m) + f(n)forevery

m,n ∈N .Iff (6) = 18,thenf (2) ⋅ f(3)isequalto

Options:

A.6

B.54

C.18

D.36

Answer:B

Solution:

Question92

Thedomainofthefunctioncosec−11+x

xis

[2021,26Aug.Shift-II]

Options:

A. −1, − 1

2∪ (0,∞)

B. −1

2,0∪ [1,∞)

C. −1

2,∞− {0}

D. −1

2,∞− {0}

Answer:D

Solution:

( )

( ]

[ )

( )

[ )

f(m+n) = f(m) + f(n), m,n∈N

∴f(3+3) = f(3) + f(3)

⇒f(6) = 2f (3) = 18 [∵f(6) = 18]

Also f (3) = f(2+1) = f(2) + f(1)

=f(1+1) + f(1)

f(3) = f(1) + f(1) + f(1)

⇒9=3f (1) ⇒ f(1) = 3

∴f(2) = f(1+1) = f(1) + f(1) = 3+3=6

Hence, f (2) ⋅ f(3) = 6⋅9=54

f(x) = cosec−11+x

1+x

x≥1

Clearly,x≠0

r|1+x|2≥ | x|2

( ) | |

1+x2+2x ≥x2

2x +1≥0

x≥ − 1

So,

x∈ − 1

2,∞− {0}

[ ]

Question93

WhichofthefollowingisnotcorrectforrelationRonthesetofreal

numbers?

[2021,31Aug.Shift-1]

Options:

A.(x,y) ∈ R⇔0< | x|−|y| ≤1isneithertransitivenorsymmetric.

B.(x,y) ∈ R⇔0< | x−y| ≤1issymmetricandtransitive.

C.(x,y) ∈ R⇔x|−|y| ≤1isreflexivebutnotsymmetric.

D.(x,y) ∈ R⇔x−y| ≤1isreflexiveandsymmetric.

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question94

LetN bethesetofnaturalnumbersandarelationRonN bedefinedby

R= { (x,y) ∈ N×N:x3−3x2y−xy2+3y3=0}.ThentherelationRis

[2021,27JulyShift-11]

Options:

A.symmetricbutneitherreflexivenortransitive.

B.reflexivebutneithersymmetricnortransitive.

C.reflexiveandsymmetric,butnottransitive.

D.anequivalencerelation.

Answer:B

Solution:

Accordingtothequestion,let'sconsideroption(b)(2,3)and(3,4)satisfy0< | x−y|leq 1but(2,4)doesnotsatisfyit.

Solution:

-------------------------------------------------------------------------------------------------

Question95

DefinearelationRoveraclassofn ×nrealmatricesAandBas"ARB,

ifthereexistsanon-singularmatrixPsuchthatPAP−1=B′.Thenwhich

ofthefollowingistrue?

[2021,18MarchShift-II]

Options:

A.Rissymmetric,transitivebutnotreflexive.

B.Risreflexive,symmetricbutnottransitive.

C.Risanequivalencerelation.

D.Risreflexive,transitivebutnotsymmetric.

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Given,relationRonNisdefinedby

R= {(x,y) ∈ N×N:x3−3x2−xy2+3y3=0}

x3−3x2y−xy2+3y3=0

⇒x3−xy2−3x2y+3y3=0

⇒x(x2−y2) − 3y(x2−y2) = 0

⇒ (x−3y)(x2−y2) = 0

⇒ (x−3y)(x−y)(x+y) = 0

Now, x−x=0

⇒x=x, ∀(x,x) ∈ N×N

So,Risareflexiverelation.

Butnotsymmetricandtransitiverelationbecause,

(3,1)satisfiesbut(1,3)doesnot.Also,(3,1)and

(1, −1)satisfiesbut(3, −1)doesnot.

Hence,relationRisreflexivebutneithersymmetricnortransitive.

Forreflexiverelation,∀(A,A) ∈ RformatrixP.

⇒A=PAP−1istrueforP=1

So,Risreflexiverelation.

Forsymmetricrelation,

Let(A,B) ∈ RformatrixP.

⇒A=PBP−1Afterpre-multiplybyP−1andpost-multiplybyP1

weget

P−1AP =B

So,(B,A) ∈ RformatrixP−1.

So,Risasymmetricrelation.

Fortransitiverelation,

LetARBandBRC

So,A=PBP−1andB=PCP−1

Now, A=P(PCP−1)P−1

⇒A= (P)2C(P−1)2⇒A= (P)2⋅C⋅ (P2)−1

∴(A,C) ∈ RformatrixP2.

∴Ristransitiverelation.

Hence,Risanequivalencerelation.

Question96

LetA = {2,3,4,5, ...., 30}and'′′beanequivalencerelationonA ×A,

definedby(a,b) ∼ (c,d),ifandonlyifad =bc.Then,thenumberof

orderedpairs,whichsatisfythisequivalencerelationwithorderedpair

(4,3)isequalto

[2021,16MarchShift-II]

Options:

A.5

B.6

C.8

D.7

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question97

LetR = { (P,O)∣, PandQareatthesamedistancefromtheorigin\}be

arelation,thentheequivalenceclassof(1, −1)istheset

[2021,26Feb.Shift-1]

Options:

A.S = {(x,y) ∣ x2+y2=4}

B.S = {(x,y) ∣ x2+y2=1}

C.S = {(x,y) ∣ x2+y2= √2}

D.S = {(x,y) ∣ x2+y2=2}

Answer:D

Solution:

A= {2,3,4,5, ..., 30}

a=bc

∴ (a,b)R(4,3)

⇒3a =4b

⇒a=4

bmustbeamultipleof3,bcanbe

(3,6,9, .. . 30).

Also,amustbelessthanorequalto30.

(a,b) = (4,3), (8,6), (12,9), (16,12), (20,15)

(24,18), (28,21)

⇒7orderedpairs

( )

Solution:

-------------------------------------------------------------------------------------------------

Question98

Let{S=1,2,3,4,5,6,7}.Thenthenumberofpossiblefunctions

f:S→Ssuchthatf (m⋅n) = f(m) ⋅ f(n)foreverym,n∈Sandm ⋅n∈S

isequalto............

[2021,27JulyShift-I]

Answer:490

Solution:

-------------------------------------------------------------------------------------------------

LetP(a,b)andQ(c,d)areanytwopoints.

Given,OP =00

i.e. a2+b2=c2+d2

Squaringonbothsides,

R= {((a,b), (c,d)) : a2+b2=c2+d2}

R(x,y), S(1, −1), OR =OS(equivalenceclass)

ThisgivesOR =x2+y2andOS = √2

l⇒x2+y2= √2

⇒x2+y2=2(Squaringonbothsides)

∴S= {(x,y) : x2+y2=2}

√ √

√

S= {1,2,3,4,5,6,7}

f:S→S

f(m⋅n) = f(m)f(n)

m,n∈S⇒m,n∈S

If mn ∈S⇒mn ≤7

So, (1⋅1,1⋅2, ..., 1⋅7) ≤ 7

(2⋅2,2⋅3) ≤ 7

Whenm=1,f(n) = f(1) ⋅ f(n) ⇒ f(1) = 1

Whenm=n=2,

f(4) = f(2)f(2) = f(2) = 1⇒f(4) = 1 or

f(2) = 2⇒f(4) = 4.

When,m=2,n=3

f(6) = f(2)f(3)

When, f (2) = 1

f(3) = 1 to 7

When, f (2) = 2

f(3) = 1 or 2 or 3.

Andf(5), f(7)cantakeanyvalue(1-7)[ ∵f(5) = f(1) ⋅ f(5) ≤ 7andf(7) = f(1) ⋅ f(7) ≤ 7}Thepossiblecombinationis

llf(1) = 1 f (1) = 1

f(2) = 1 f (2) = 2

f(3) = (1−7)f(3) = (1−3)

f(4) = 1 f (4) = 4

f(5) = (1−7)f(5) = (1−7)

f(6) = f(3)f(6) = f(3)

f(7) = (1−7)f(7) = (1−7)

So,total = (1×1×7×1×7×1×7)

+(1×1×3×1×7×1×7)

=490

{

Question99

If[x]bethegreatestintegerlessthanorequaltox,then 100

∑

n=8

(−1)nn

2is

equalto

[25July2021,Shift-III]

Options:

A.0

B.4

C.-2

D.2

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question100

Ifthedomainofthefunctionf (x) = cos−1x2−x+1

sin−12x −1

istheinterval(α,β),

thenα +βisequalto

[2021,22JulyShift-II]

Options:

A. 3

B.2

C. 1

D.1

Answer:A

Solution:

[ ]

√

√( )

Wehave,

100

∑

n=8

(−1)nn

2( ∵[x]isthegreatestintegerfunction)

Substitutethevaluesofn

= [4] + [−4.5] + [5] + [−5.5]

+ ... + [−49.5] + [50]

=4−5+5−6+ ... − 50 +50

[ ]

Solution:

-------------------------------------------------------------------------------------------------

Question101

Let[x]denotethegreatestinteger≤x,wherex ∈R.Ifthedomainofthe

realvaluedfunction

f(x) = |[x]| − 2

|[x]| − 3is(−∞,a) ∪ [b,c)

∪[u,∞), a<b<c,thenthevalueof

a+b+cis

[2021,20JulyShift-I]

Options:

A.8

B.1

C.−2

D.−3

Answer:C

Solution:

√

f(x) = cos−1x2−x+1

sin−12x −1

⇒x∈R,x(x−1) ≤ 0

x2−x+1≥0andx2−x+1≤1

0≤x≤1⋅⋅⋅⋅⋅⋅ ⋅ (i)⇒ 0<sin−12x −1

2<π

⇒0<2x −1

2<1

⇒1

2<x<3

2⋅⋅⋅⋅⋅⋅ ⋅ (ii)

(A) ∩ (B) = x∈1

2,1

∴α+β=3

√

√( )

( )

( ]

f(x) = |[x]| − 2

|[x]| − 3

|[x]| − 2

|[x]| − 3≥0

Let|[x]| = t

√

-------------------------------------------------------------------------------------------------

Question102

Therealvaluedfunctionf (x) = cosec−1x

√x− [x],where[x]denotesthegreatest

integerlessthanorequaltox,isdefinedforallxbelongingto

[2021,18MarchShift-I]

Options:

A.allrealsexceptintegers

B.allnon-integersexcepttheinterval[−1,1]

C.allintegersexcept0,−1,1

D.allrealsexcepttheinterval[−1,1]

Answer:B

Solution:

l| [x] | = 3⇒x∈ [−3, −2) ∪ [3,4)

Domainof x = [−∞, −3) ∪ [−2,3) ∪ [4,∞)

a= −3

b= −2

c=3

∴a+b+c= −3+ (−2) + 3= −2

Given,f(x) = cosec−1x

√x− [x]

⇒f(x) = cosec−1x

√{x}

Forf(x)tobedefined,

|x| ≥ 1

{x} > 0. ⇒ x≤ −1 or x ≥1

x≠1 integers .

i.e.x∈ (−∞, −1] ∪ [1,∞) − {integers}

{ {

-------------------------------------------------------------------------------------------------

Question103

Ifthefunctionsaredefinedasf (x) = √xandg(x) = √1−x,thenwhatis

thecommondomainofthefollowingfunctions?

f+g,f−g,f∕g,g∕f,g−f ,where

(f±g)(x) = f(x) ± g(x), (f∕g)(x) = f(x)

g(x)

[2021,18March,Shift-1]

Options:

A.0 ≤x≤1

B.0 ≤x<1

C.0 <x<1

D.0 <x≤1

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question104

Afunctionf (x)isgivenbyf (x) = 5x

5x+5,thenthesumoftheseries

20 +f2

20 +f3

20 + ... + f39

isequalto

[2021,25Feb.Shift-II]

Options:

A. 29

( ) ( ) ( ) ( )

i.e.allnon-integersexcepttheinterval[−1,1]

(here,−1and1areincludedinexceptcase,becauseof−1and1areintegers).

Given,f(x) = √xandg(x) = √1−x

∴Domainoff(x) = D1isx≥0

i.e.D1:x∈ (0,∞)

anddomainofg(x) = D2is1−x≥0⇒x≤1

i.e.D2:x∈ (−∞1]

As,weknowthat,thedomainoff+g1f−g,g−fwillbeD1∩D2aswellasthedomainfor f

gisD1∩D2exceptall

thosevalue(s)ofx,suchthatg(x) = 0.

Similarly,for g

fisD1∩D2butf(x) ≠ 0.

Hence,commondomainfor(f+g), (f−g), f

g,g

fand(g−f)willbe0<x<1

( ) ( )

B. 49

C. 39

D. 19

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question105

Ifa +α=1,b+β=2andaf (x) + αf 1

x=bx +β

x,x≠0,thenthevalue

ofexpression f(x) + f1

x+1

is

[2021,24Feb.Shift-II]

Answer:2

Solution:

( )

Given,f(x) = 5x

5x+5,then,

f(2−x) = 52−x

52−x+5

5x+5

Thisgives,f(x) + f(2−x) = 5x+5

5x+5=1⇒f1

20 +f 2 −1

20 =f1

20 +f39

20 =1

Similarly,

cf 2

20 +f38

20 =1 andsoon,

∴f1

20 +f2

20 + ... + f38

20 +f39

=1+1+ ... + 1+f20

=19 +f(1) = 19 +1

2=39

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( )

Given,a+α=1

b+β=2

∵a⋅f(x) + α⋅f1

x=bx +β

x⋅⋅⋅⋅⋅⋅ ⋅ (i)

Replacexby 1

x,

af 1

x+af (x) = b

x+βx

( )

-------------------------------------------------------------------------------------------------

Question106

Letf (x) = sin−1xand

g(x) = x2−x−2

2x2−x−6

Ifg(2) = lim

x→2

g(x),thenthedomainofthefunctionfogis

[2021,26Feb.Shift-II]

Options:

A.(−∞, −2] ∪ − 3

2,∞

B.(−∞, −2] ∪ [−1,∞)

C.(−∞, −2] ∪ − 4

3,∞

D.(−∞, −1] ∪ [2,∞)

Answer:C

Solution:

[ )

AddingEqs.(i)and(ii),weget

&(a+α)f(x) + f1

x=x+1

x(b+β) ⋅⋅⋅⋅⋅⋅ ⋅ (ii)

⇒

f(x) + f1

x+1

=b+β

a+α=2

1=2

[ ( ) ] ( )

( )

Given,g(x) = x2−x−2

2x2−x−6,f(x) = sin−1x

f(g(x)) = sin−1(g(x))

f∘g(x) = sin−1x2−x−2

2x2−x−6

Forthedomainoff og(x),

|g(x)| ≤ 1

[ ∵Domainoff(x)is[−1,1]

⇒x2−x−2

2x2−x−6≤1

⇒(x+1)(x−2)

(2x +3)(x−2)≤1

⇒x+1

2x +3≤1

⇒ − 1≤x+1

2x +3≤1

⇒x+1

2x +3

2≤1

⇒ (x+1)2≤ (2x +3)2

⇒3x2+10x +8≥0

( )

| |

( )

-------------------------------------------------------------------------------------------------

Question107

Letg :N→N bedefinedas

g(3n +1) = 3n +2

g(3n +2) = 3n +3,

g(3n +3) = 3n +1,foralln ≥0.

Thenwhichofthefollowingstatementsistrue?

[2021,25JulyShift-1]

Options:

A.Thereexistsanontofunctionf :N→N suchthatfog =f

B.Thereexistsaone-onefunctionf :N→N suchthatf og =f

C.gogog =g

D.Thereexistsafunctionf :N→N suchthatgof =f

Answer:A

Solution:

⇒ (3x +y)(x+2) ≥ 0

Thisimplies,

x∈ (−∞, −2] ∪ − 4

3,∞

[ )

g(3n +1) = 3n +2

g(3n +2) = 3n +3

g(3n +3) = 3n +1,foralln≥0

g:N→N

g(1) = 2,g(4) = 5,g(7) = 8

g(2) = 3,g(5) = 6,g(8) = 9

g(3) = 1,g(6) = 4,g(9) = 7

⇒f[g(1)] = f(1)

⇒f(2) = f(1)

Clearly,itisnotaone-onefunction.

Now,f[g(2)] = f(2)

f(3) = f(2)

And,f[g(3)] = f(3)

f(1) = f(3)

Similarly,f[g(4)] = f(4)

f(5) = f(4)

And,soon

l f (1) = f(2) = f(3)

f(4) = f(5) = f(6)

Now,therecanbeapossibilitysuchthat

So,f(x)canbeontofunction.

Whenf(1) = f(2) = f(3) = 1

f(4) = f(5) = f(6) = 2

andsoon.

Question108

Considerfunctionf :A→Bandg :B⟶C(A,B,C⊂eqR)suchthat

(gof )−1exists,then

[2021,25JulyShift-II]

Options:

A.fandgbothareone-one

B.f andgbothareonto

C.f isone-oneandgisonto

D.f isontoandgisone-one

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question109

LetA = {0,1,2,3,4,5,6,7}.Then,thenumberofbijectivefunctions

F:A→Asuchthatf (1) + f(2) = 3−f(3)isequalto..........

[2021,22JulyShift-III]

Answer:720

Solution:

Givenfunctions,f:A→Bandg:B→C(A,B,C⊂eqR)

∴(gof)−1exists⇒gofisabijectivefunction.

⇒'f'mustbe'one-one'and'g'mustbe'onto'function.

f(1) + f(2) = 3−f(3)

A= {0,1,2,3,4,5,6,7}

f:A→A

So,f(1) + f(2) + f(3) = 3

0+1+2=3istheonlypossibility.

So,f(0)canbeeither0or1or2.

Similarly,f(1)andf(2)canbe0,1and2.

and{3,4,5,6,7} ⟶ {3,4,5,6,7}

Theyhave5!choices.

And{0,1,2}

Theyhave3!choices.

Numberofbijectivefunctions

=3! × 5! = 720

-------------------------------------------------------------------------------------------------

Question110

Letf :R−α

6⟶Rbedefinedby

f(x) = 5x +3

6x −α

Then,thevalueofαforwhich(fof)(x) = x,forallx ∈R−α

6is

[2021,20JulyShift-II]

Options:

A.Nosuchαexists

B.5

C.8

D.6

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question111

Letf :R− {3} → R− {1}bedefinedbyf (x) = x−2

x−3.Letg :R→Rbegiven

asg(x) = 2x −3.Then,thesumofallthevaluesofxforwhich

f−1(x) + g−1(x) = 13

2isequalto

[2021,18MarchShift-II]

{ }

f(x) = 5x +3

6x −α

Now,f∘f(x) = f5x +3

6x −α

55x +3

6x −α+3

65x +3

6x −α−α

=5(5x +3) + 3(6x −α)

6(5x +3) − α(6x −2)

=5(5x +3) + 3(6x −α)

6(5x +3) − α(6x −2)

Given,fof(x) = x

⇒5(5x +3) + 3(6x −α)

6(5x +3) − α(6x −α)=x

⇒25x +15 +18x −3α

=30x2+18x −6αx2+α2x

⇒x2(30 −6α) − x(α2−25) + 3α −15 =0

Comparingcoefficients,

30 −6x =0

⇒6α =30

⇒α=5

( )

Options:

A.7

B.2

C.5

D.3

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question112

Theinverseofy =5log xis

[2021,17MarchShift-I]

Options:

A.x =5log y

B.x =ylog 5

C.x =y

log 5

D.x =5

log y

Answer:C

Given,f(x) = x−2

x−3

g(x) = 2x −3

Lety=f(x) = x−2

x−3

⇒xy −3y =x−2⇒ xy −x=3y −2

⇒x(y−1) = 3y −2

⇒x=3y −2

y−1

⇒f−1(y) = 3y −2

y−1

⇒f−1(x) = 3x −2

x−1

Similarly, g−1(x) = x+3

Given,f−1(x) + g−1(x) = 13

⇒3x −2

x−1+x+3

2=13

⇒x2+8x −7=13(x−1)

⇒x2−5x +6=0

⇒ (x−2)(x−3) = 0

⇒x=2,3

∴Sum = 2+3=5

Solution:

-------------------------------------------------------------------------------------------------

Question113

LetA = {1,2,3, ..., 10}andf :A→Abedefinedasdefinedas

f(x) = x+1 if x isodd

xif x iseven .

Then,thenumberofpossiblefunctionsg :A→A,suchthatgof =f is

[2021,26Feb.Shift-II]

Options:

A.105

B.10C5

C.55

D.5!

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question114

Letf ,g:N→N ,suchthatf (n+1) = f(n) + f(1) ∀n∈N andgbeany

{

y=5log x

Takinglogonbothsides,

⇒log y =log x ⋅log 5

⇒1

log 5 =log x

log y

log 5 =logyx

x=y

log 5

f(x) = x+1 xisodd

xxiseven. .

Given,g:A→Asuchthat,

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

Whenxiseven,then

g(x) = x

Whenxisodd,then

g(x+1) = x+1

Thisimplies,

g(x) = x1x iseven.

⇒Ifxisodd,theng(x)cantakeanyvalueinsetA.

So,numberofg(x) = 105

{

arbitraryfunction.Whichofthefollowingstatementsisnottrue?

[2021,25Feb.Shift-1]

Options:

A.iff ogisone-one,thengisone-one.

B.iff isonto,thenf (n) = n, ∀n∈N .

C.f isone-one.

D.ifgisonto,thenfogisone-one.

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question115

Letxdenotethetotalnumberofone-onefunctionsfromasetAwith3

elementstoasetBwith5elementsandydenotethetotal

numberofone-onefunctionsfromthesetAtothesetA ×B.Then,

[2021,25Feb.Shift-II]

Options:

A.2y =91x

B.2y =273x

C.y =91x

D.y =273x

Answer:A

Solution:

Given,f(n+1) = f(n) + f(1), ∀n∈N

⇒f(n+1) − f(n) = f(1)

ItisanAPwithcommondifference = f(1)

Also,generalterm

l l =Tn=f(1) + (n−)f(1) = nf (1)

⇒f(n) = nf (1)

Clearly,f(n)isone-one.

Forfogtobeone-one,gmustbeone-one.

Forftobeonto,f(n)shouldtakeallthevaluesofnaturalnumbers.

As,f(x)isincreasing,f(1) = 1

⇒f(n) = n

Ifgismany-one,thenf ogismanyone.

So,ifgisonto,thenfogisone-one.

x= { f:A→B,fisone-one}

y= { g:A→A×B,gisoneone}

-------------------------------------------------------------------------------------------------

Question116

Letf :R→Rbedefinedasf (x) = 2x −1andg :R− {1} → Rbedefined

asg(x) =

x−1

x−1.Thenthecompositionfunctionf (g(x))is:

24Feb2021Shift1

Options:

A.ontobutnotone-one

B.bothone-oneandonto

C.one-onebutnotonto

D.neitherone-onenoronto

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question117

LetR1andR2betworelationsdefinedasfollows:

R1= {(a,b) ∈ R2:a2+b2∈Q}andR2= {(a,b) ∈ R2:a2+b2∉Q},

whereQisthesetofallrationalnumbers.Then:

[Sep.03,2020(II)]

Options:

A.NeitherR1norR2istransitive.

NumberofelementsinA=3i.e.|A| = 3

Similarly,|B| = 5

Then,|A×B|= |A|×|B| = 3×5=15

Now,numberofone-onefunctionfromAtoBwillbe

l5P3=5!

(5−3)! =5!

2!=5×4×3=60

∴x=60

Now,numberofone-onefunctionfromA

l to A ×B willbe =15P3=15!

(15 −3)! =15!

12!

=15 ×14 ×13 =2730

∴y=2730l c ∴y=2730

Thus, 2 × (2730) = 91 × (60)

⇒2y =91x

f(g(x)) = 2g(x) − 1=22x −1

2(x−1)−1 = x

x−1=1+1

x−1

Rangeoff(g(x)) = mathbb R − {1}

Rangeoff(g(x))isnotonto

&f(g(x))isone-one

So,f(g(x))isone-onebutnotonto.

( )

B.R2istransitivebutR1isnottransitive.

C.R1istransitivebutR2isnottransitive.

D.R1andR2arebothtransitive.

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question118

Thedomainofthefunctionf (x) = sin−1|x| + 5

x2+1is(−∞, −a] ∪ [a,∞].

Thenaisequalto:

[Sep.02,2020(I)]

Options:

A. √17

B. √17 −1

C. 1+ √17

D. √17

2+1

Answer:C

Solution:

( )

(a)ForR1leta=1+ √2,b=1− √2,c=81∕4

aR1b⇒a2+b2= (1+ √2)2+ (1− √2)2=6∈Q

bR1c⇒b2+c2= (1− √2)2+ (81∕4)2=3∈Q

aR1c⇒a2+c2= (1+ √2)2+ (81∕4)2=3+4√2∉Q

∴R1isnottransitive.

ForR2leta=1+ √2,b= √2,c=1− √2

aR2b⇒a2+b2= (1+ √2)2+ (√2)2=5+2√2∉Q

bR2c⇒b2+c2= (√2)2+ (1− √2)2=5−2√2∉Q

aR2c⇒a2+c2= (1+ √2)2+ (1− √2)2=6∈Q

∴R2isnottransitive.

∵f(x) = sin−1|x| + 5

x2+1

∴−1≤|x| + 5

x2+1≤1

⇒ | x| +5≤x2+1

[∵x2+1≠0]

⇒x2− | x| −4≥0

( )

-------------------------------------------------------------------------------------------------

Question119

IfR = {(x,y) : x,y∈Z,x2+3y2≤8}isarelationonthesetofintegers

Z ,thenthedomainofR−1is:

[Sep.02,2020(I)]

Options:

A.{−2, −1,1,2}

B.{0,1}

C.{−2, −1,0,1,2}

D.{−1,0,1}

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question120

Let[t]denotethegreatestinteger≤t.Thentheequationin

x, [x]2+2[x+2] − 7=0has:

[Sep.04,2020(I)]

Options:

A.exactlytwosolutions

B.exactlyfourintegralsolutions

C.nointegralsolution

D.infinitelymanysolutions

Answer:D

Solution:

⇒ |x| − 1− √17

2|x| − 1+ √17

2≥0

⇒x∈ −∞, − 1+ √17

2∪1+ √17

2,∞

∴a=1+ √17

( ) ( )

( ) [ )

Since,R= {(x,y) : x,y∈Z,x2+3y2≤8}

∴R= {(1,1), (2,1), (1, −1), (0,1), (1,0)}

⇒DR−1= {−1,0,1}

-------------------------------------------------------------------------------------------------

Question121

Letf (x)beaquadraticpolynomialsuchthatf (−1) + f(2) = 0.Ifoneof

therootsoff (x) = 0is3,thenitsotherrootliesin:

[Sep.02,2020(II)]

Options:

A.(-1,0)

B.(1,3)

C.(-3,-1)

D.(0,1)

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question122

Letf (1,3) → Rbeafunctiondefinedbyf (x) = x[x]

1+x2where[x]denotes

thegreatestinteger≤x.Thentherangeoff is:

[Jan.8,2020(II)]

Options:

A. 2

5,3

5∪3

4,4

B. 2

5,1

2∪3

5,4

C. 2

5,4

( ) ( )

( )

Thegivenequation[x]2+2[x] + 4−7=0

⇒[x]2+2[x] − 3=0

⇒[x]2+3[x] − [x] − 3=0

⇒([x] + 3)([x] − 1) = 0⇒ [x] = 1or-3⇒x∈ [−3, −2) ∪ [1,2)

∴Theequationhasinfinitelymanysolutions.

Letf(x) = ax2+bx +c

Given:f(−1) + f(2) = 0

a−b+c+4a +2b +c=0

⇒5a +b+2c =0......(i)

andf(3) = 0⇒9a +3b +c=0......(ii)

Fromequations(i)and(ii),

1−6=b

18 −5=c

15 −9⇒a

−5=b

13 =c

6

Productofroots,αβ =c

a=−6

5andα=3

⇒β=−2

5∈ (−1,0)

D. 3

5,4

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question123

LetA = {a,b,c}andB = {1,2,3,4}.Thenthenumberofelementsin

thesetC = { f:A→B|2∈f(A)andf isnotone-one}is_______.

[NASep.05,2020(II)]

Answer:19

Solution:

-------------------------------------------------------------------------------------------------

Question124

Theinversefunctionoff (x) = 82x −8−2x

82x +8−2x,x∈ (−1,1),is_______.

[Jan.8,2020(I)]

( )

 f (x)

x2+1

x∈ (1,2)

x2+1

x∈ [2,3).



f′(x)

1−x2

1+x2

x∈ (1,2)

1−2x2

1+x2

x∈ [2,3).



∴f(x)isadecreasingfunction

∴y∈2

5,1

2∪6

10,4

5

⇒y∈2

5,1

2∪3

5,4

{

( ) ( ]

Thedesiredfunctionswillcontaineitheroneelementortwoelementsinitscodomainofwhich'2'alwaysbelongsto f (A).

∴ThesetBcanbe{2}, {1,2}, {2,3}, {2,4}

Totalnumberoffunctions = 1+ (23−2)3=19

Options:

A.1

4loge

1+x

1−x

B.1

4(log8e)loge

1−x

1+x

C.1

4loge

1−x

1+x

D.1

4(log8e)loge

1+x

1−x

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question125

Ifg(x) = x2+x−1and(gof )(x) = 4x2−10x +5,thenf 5

4isequalto:

[Jan.7,2020(I)]

Options:

A.3

B.−1

C.1

D.−3

Answer:B

Solution:

( )

y=82x −8−2x

82x +8−2x

1+y

1−y=82x

8−2x ⇒84x =1+y

1−y

⇒4x =log8

1+y

1−y

⇒x=1

4log8

1+y

1−y

∴f−1(x) = 1

4log8

1+x

1−x

( )

(gof )(x) = g(f(x)) = f2(x) + f(x) − 1

g f 5

4=45

2−10 .5

4+5 = −5

4

[∵g(f(x)) = 4x2−10x +5]

( ( ) ) ( )

-------------------------------------------------------------------------------------------------

Question126

Forasuitablychosenrealconstanta,letafunction,f :R− {−a} → Rbe

definedbyf (x) = a−x

a+x.Furthersupposethatforanyrealnumberx ≠ −a

andf (x) ≠ −a,(fof)(x) = x.Thenf −1

2isequalto:

[Sep.06,2020(II)]

Options:

A.1

B.−1

C.-3

D.3

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question127

Letf :R→Rbedefinedbyf (x) = x

1+x2,x∈R.Thentherangeoff is:

[Jan.11,2019(I)]

Options:

A. −1

2,1

( )

[ ]

g f 5

4=f25

4+f5

4−1

−5

4=f25

4+f5

4−1

f25

4+f5

4+1

4=0

4+1

2=0

4= −1

( ( ) ) ( ) ( )

( ) ( )

( ( ) )

( )

f(f(x)) =

a−a−x

a+x

a+a−x

a+x

=x

⇒a−ax

1+x=f(x)⇒a(1−x)

1+x=a−x

a+x⇒a=1

∴f(x) = 1−x

1+x⇒f−1

2=3

( )

B.R − [−1,1]

C.R − − 1

2,1

D.(−1,1) − {0}

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question128

Thedomainofthedefinitionofthefunctionf (x) = 1

4−x2+log10(x3−x)is:

[April.09,2019(II)]

Options:

A.(-1,0)∪(1,2) ∪ (3,∞)

B.(-2,-1)∪(−1,0) ∪ (2,∞)

C.(-1,0)∪(1,2) ∪ (2,∞)

D.(1,2)∪(2,∞)

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

[ ]

f(x) = x

1+x2,x∈R

Let,y=x

1+x2

⇒yx2−x+y=0⇒x=1±1−4y2

2

⇒1−4y2≥0

⇒1≥4y2

⇒ | y≤1

2

⇒ − 1

2≤y≤1

2

⇒Therangeoffis − 1

2,1

√

[ ]

Todeterminedomain,denominator≠0andx3−x>0

i.e.,4−x2≠0x ≠ ±2......(1)

andx(x−1)(x+1) > 0

x∈ (−1,0) ∪ (1,∞)........(2)

Hencedomainisintersectionof(1)&(2).

i.e.,x∈ (−1,0) ∪ (1,2) ∪ (2,∞)

Question129

Iff(x) = loge

1−x

1+x,x<1,thenf 2x

1+x2isequalto

[April8,2019(I)]

Options:

A.2f (x)

B.2f (x2)

C.(f(x))2

D.−2f (x)

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question130

Letf (x) = ax(a>0)bewrittenasf (x) = f1(x) + f2(x),wheref 1(x)isan

evenfunctionandf 2(x)isanoddfunction.Thenf 1(x+y) + f1(x−y)

equals:

[April.08,2019(II)]

Options:

A.2f 1(x)f1(y)

B.2f 1(x+y)f1(x−y)

C.2f 1(x)f2(y)

D.2f 1(x+y)f2(x−y)

Answer:A

Solution:

( ) | | ( )

f(x) = log 1−x

1+x,x<∣

f2x

1+x2=log

1−2x

1+x2

1+2x

1+2x2



=log 1+x2−2x

1+x2+2x =log 1−x

1+x

2

=2 log 1−x

1+x=2f (x)

( ) | |

( ) ( )

( )

Solution:

-------------------------------------------------------------------------------------------------

Question131

Letafunctionf : (0,∞) → (0,∞)bedefinedbyf (x) = 1−1

x.Thenf is:

[Jan.11,2019(II)]

Options:

A.notinjectivebutitissurjective

B.injectiveonly

C.neitherinjectivenorsurjective

D.(Bonus)

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question132

Thenumberoffunctionsffrom{1,2,3,...,20}onto{1,2,3,....,20}

suchthatf(k)isamultipleof3,wheneverkisamultipleof4is:

[Jan.11,2019(II)]

Options:

A.65× (15)!

B.5! × 6!

C.(15)! × 6!

| |

Givenfunctioncanbewrittenas

f(x) = ax=ax+a−x

2+ax−a−x

2

wheref1(x) = ax+a−x

2isevenfunction

f2(x) = ax−a−x

2isoddfunction

⇒f1(x+y) + f1(x−y)

=ax+y+a−x−y

2+ax−y+a−x+y

2

2[ax(ay+a−y) + a−x(ay+a−y)]

=(ax+a−x)(ay+a−y)

2=2f 1(x) ⋅ f1(y)

( ) ( )

f: (0,∞) → (0,∞)

f(x) = 1−1

xisnotafunction

∵f(1) = 0and1∈domainbut0∉co-domain

Hence,f(x)isnotafunction.

| |

D.56×15

Answer:C

Solution:

Question133

LetNbethesetofnaturalnumbersandtwofunctionsfandgbedefined

asf,g:N→Nsuchthatf (n) =

n+1

2if n is odd

2if n is even



andg(n) = n− (−1)n.Thenf ogis:

[Jan.10,2019(II)]

Options:

A.ontobutnotone-one.

B.one-onebutnotonto.

C.bothone-oneandonto.

D.neitherone-onenoronto.

Answer:A

Solution:

{

Domainandcodomain = {1,2,3, ....., 20}.

Therearefivemultipleof4as4,8,12,16and20.

andthereare6multipleof3as3,6,9,12,15,18.

Since,wheneverkismultipleof4thenf(k)ismultipleof3thentotalnumberofarrangement = 6c5×5! = 6!

Remaining15elementscanbearrangedin15!ways.

Since,foreveryinput,thereisanoutput

⇒functionf(k)inonto

∴Totalnumberofarrangement = 15!.6!

f(n) =

n+1

2,if n is odd

2,if n is even



g(n) =

2,n=1

1,n=2

4,n=3

3,n=4

6,n=5

5,n=6



Then,

{

-------------------------------------------------------------------------------------------------

Question134

LetA = { x∈R:xisnotapositiveinteger}.Defineafunctionf :A→R

asf (x) = 2x

x−1,thenf is:

[Jan.09,2019(II)]

Options:

A.notinjective

B.neitherinjectivenorsurjective

C.surjectivebutnotinjective

D.injectivebutnotsurjective

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question135

Forx ∈0,3

2,l etf (x) = √x,g(x) = tan xandh(x) = 1−x2

1+x2

Ifϕ(x) = ((hof )og)(x),thenϕ π

3isequalto

[April12,2019(I)]

( )

f(g(n)) =

g(n) + 1

2,if g(n)is odd

g(n)

2,if g(n)is even

f(g(n)) =

1,n=1

1,n=2

2,n=3

2,n=4

3,n=5

3,n=6

: :



⇒fogisontobutnotone-one

{

AsA= { x ∈R:xisnotapositiveinteger}

Afunctionf:A→Rgivenbyf(x) = 2x

x−1

f(x1) = f(x2) ⇔ x1=x2

So,fisone-one.

Asf(x) ≠ 2foranyx∈A⇒fisnotonto.

Hencefisinjectivebutnotsurjective.

Options:

A.tan π

B.tan 11π

C.tan 7π

D.tan 5π

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question136

Letf (x) = x2,x∈R.ForanyA ⊆R,defineg(A) = {x∈R:f(x) ∈ A}.If

S= [0,4],thenwhichoneofthefollowingstatementsisnottrue?

[April10,2019(I)]

Options:

A.g(f(S)) ≠ S

B.f (g(S)) = S

C.g(f(S)) = g(S)

D.f (g(S)) ≠ f(S)

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

∵ϕ(x) = ( (hof)og)(x)

∵ϕπ

3=h f g π

3=h(f(√3)) = h(31∕4)

=1− √3

1+ √3= −1

2(1+3−2√3) = √3−2= −(−√3+2)

= −tan 15° = tan(180° − 15°) = tan π −π

12 =tan 11π

( ) ( ( ( ) ) )

( )

f(x) = x2;x∈R

g(A) = {x∈R:f(x) ∈ A}S= [0,4]

g(S) = {x∈R:f(x) ∈ S}

= {x∈R:0≤x2≤4} = {x∈R: −2≤x≤2}

∴g(S) ≠ S∴f(g(S)) ≠ f(S)

g(f(S)) = {x∈R:f(x) ∈ f(S)}

= {x∈R:x2∈S2} = {x∈R:0≤x2≤16}

= {x∈R: −4≤x≤4}

∴g(f(S)) ≠ g(S)

∴g(f(S)) = g(S)isincorrect.

Question137

Forx ∈R− {0,1}, let f 1(x)=1

x,f2(x) = 1−xandf 3(x) = 1

1−xbethree

givenfunctions.Ifafunction,J (x)satisfies(f2o Jof1)(x) = f3(x)thenJ (x)

isequalto:

[Jan.09,2019(I)]

Options:

A.f 3(x)

B.1

xf3(x)

C.f 2(x)

D.f 1(x)

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question138

LetN denotethesetofallnaturalnumbers.Definetwobinaryrelations

onN asR1= {(x,y) ∈ N×N:2x +y=10}and

R2= {(x,y) ∈ N×N:x+2y =10}.Then

[OnlineApril16,2018]

Options:

A.Both R1andR2aretransitiverelations

B.BothR1andR2aresymmetricrelations

C.RangeofR2is{1,2,3,4}

Thegivenrelationis

(f2oJ of 1)(x) = f3(x) = 1

1−x

⇒(f2J)(f1(x)) = 1

1−x

⇒(f2oJ )1

x=1

1−1

x−1

∵f1(x) = 1

x

⇒(f2oJ )(x) = x

x−1

xis replaced by x 

⇒f2(J(x)) = x

x−1

⇒1−J(x) = x

x−1[∵f2(x) = 1−x]

∴J(x) = 1−x

x−1=1

1−x = f3(x)

( ) [ ]

[ ]

D.RangeofR1is{2,4,8}

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question139

Considerthefollowingtwobinaryrelationsontheset

A= {a,b,c} : R1= { (c,a)(b,b), (a,c),(c,c), (b,c), (a,a) }and

R2= { (a,b), (b,a), (c,c),(c,a), (a,a), (b,b), (a,c).Then

[OnlineApril15,2018]

Options:

A.R2issymmetricbutitisnottransitive

B.BothR1andR2aretransitive

C.BothR1andR2arenotsymmetric

D.R1isnotsymmetricbutitistransitive

Answer:A

Solution:

Here,R1= {(x,y) ∈ N×N:2x +y=10}and

R2= {(x,y) ∈ N×N:x+2y =10}

ForR1;2x +y=10andx,y∈N

So,possiblevaluesforxandyare:

x=1,y=8i.e.(1,8);

x=2,y=6i.e.(2,6);

x=3,y=4i.e.(3,4)andx=4,y=2i.e.(4,2).

R1= {(1,8), (2,6), (3,4), (4,2)}

Therefore,RangeofR1is{2,4,6,8}

R1isnotsymmetric

Also,R1isnottransitivebecause(3,4), (4,2) ∈ R1but(3,2)∉R1

Thus,optionsA,BandDareincorrect.

ForR2;x+2y =10andx,y∈N

So,possiblevaluesforxandyare:x=8,y=1i.e.(8,1);

x=6,y=2i.e.(6,2);

x=4,y=3i.e.(4,3)and

x=2,y=4i.e.(2,4)

R2= {(8,1), (6,2), (4,3), (2,4)}

Therefore,RangeofR2is{1,2,3,4}

R2isnotsymmetricandtransitive.

BothR1andR2aresymmetricas

Forany(x,y) ∈ R1,wehave

(y,x) ∈ R1andsimilarlyforR2

Now,forR2, (b,a) ∈ R2, (a,c) ∈ R2but(b,c) ∉ R2

Similarly,forR1, (b,c) ∈ R1, (c,a) ∈ R1but(b,a) ∉ R1

-------------------------------------------------------------------------------------------------

Question140

Letf :A→Bbeafunctiondefinedasf (x) = x−1

x−2,whereA =R− {2}and

B=R− {1}.Thenf is

[OnlineApril15,2018]

Options:

A.invertibleandf −1(y) = 2y +1

y−1

B.invertibleandf −1(y) = 3y −1

y−1

C.noinvertible

D.invertibleandf −1(y) = 2y −1

y−1

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question141

Thefunctionf :R→ −1

2,1

2definedasf (x) = x

1+x2,is:

[2017]

Options:

A.neitherinjectivenorsurjective

B.invertible

C.injectivebutnotsurjective

D.surjectivebutnotinjective

Answer:D

Solution:

[ ]

Therefore,neitherR1norR2istransitive.

Supposey=f(x)

⇒y=x−1

x−2

⇒yx −2y =x−1

⇒(y−1)x=2y −1

⇒x=f−1(y) = 2y −1

y−1

Asthefunctionisinvertibleonthegivendomainanditsinversecanbeobtainedasabove.

Solution:

-------------------------------------------------------------------------------------------------

Question142

Thefunctionf :N→N definedbyf (x) = x−5x

5,whereN issetof

naturalnumbersand[x]denotesthegreatestintegerlessthanorequal

tox,is:

[OnlineApril9,2017]

Options:

A.one-oneandonto.

B.one-onebutnotonto.

C.ontobutnotone-one.

D.neitherone-onenoronto.

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question143

Letf (x) = 210 .x+1andg(x) = 310 .x−1.If(f og)(x) = x,thenxisequal

to:

[OnlineApril8,2017]

[ ]

Wehavef:R→ −1

2,1

2,

f(x) = x

1+x2∀x∈R

⇒f′(x) = (1+x2) . 1−x.2x

(1+x2)2 = −(x+1)(x−1)

(1+x2)2

signoff′(x)

⇒f′(x)changessignindifferentintervals.

∴NotinjectiveNowy=x

1+x2

⇒y+yx2=x

⇒yx2−x+y=0

Fory≠0,D=1−4y2≥0

⇒y∈−1

2,1

2− {0}

Fory=0⇒x=0

∴Rangeis −1

2,1

2

⇒Surjectivebutnotinjective

[ ]

f(1) = 1−5[1∕5] = 1

f(6) = 6−5[6∕5] = 1→Manyone

f(10) = 10–5(2) = 0whichisnotinco–domain.

Neitherone–onenoronto.

}

Options:

A. 310 −1

310 −2−10

B. 210 −1

210 −3−10

C. 1−3−10

210 −3−10

D. 1−2−10

310 −2−10

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question144

Forx ∈R,x≠0,letf 0(x) = 1

1−xandf n+1(x) = f0(fn(x))n=0,1,2, .....

Thenthevalueoff 100(3) + f1

3+f2

2isequalto:

[OnlineApril9,2016]

Options:

A.8

B.4

C.5

D.1

Answer:C

Solution:

( ) ( )

f(g(x)) = x

⇒f(310x−1) = 210(310 .x−1) + 1=x

⇒210(310x−1) + 1=x

⇒x(610 −1) = 210 −1

⇒x=210 −1

610 −1=1−2−10

310 −2−10

f1(x) = f0+1(x) = f0(f0(x)) = 1

1−1

1−x

=x−1

x

f2(x) = f1+1(x) = f0(f1(x)) = 1

1−x−1

=x

-------------------------------------------------------------------------------------------------

Question145

LetA = {x1,x2, ....., x7}andB = {y1,y2,y3}betwosetscontaining

sevenandthreedistinctelementsrespectively.Thenthetotalnumberof

functionsf :A→Bthatareonto,ifthereexistexactlythreeelementsx

inAsuchthatf (x) = y2,isequalto

(OnlineApril11,2015)

Options:

A.14.7C3

B.16.7C3

C.14.7C2

D.12. .7C2

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question146

Letf :R→Rbedefinedbyf (x) = |x| − 1

|x| + 1thenfis:

[OnlineApril19,2014]

Options:

A.bothone-oneandonto

f3(x) = f2+1(x) = f0(f2(x)) = f0(x) = 1

1−x

f4(x) = f3+1(x) = f0(f3(x)) = x−1

x

∴f0=f3=f6= .... = 1

1−x

f1=f4=f7= ...... = x−1

x

f2=f5=f8= ........ = x

f100(3) = 3−1

3=2

3f1

3 =

3−1

= −1

2

2=3

2

∴f100(3) + f1

3+f2

2=5

( )

( ) ( )

Numberofontofunctionsuchthatexactlythreeelementsinx∈Asuchthatf(x) = 1

2isequalto

=7C3, {24−2} = 14 .7C3

B.one-onebutnotonto

C.ontobutnotone-one

D.neitherone-onenoronto.

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question147

LetPbetherelationdefinedonthesetofallrealnumberssuchthat

P= {(a,b) : sec2a−tan2b=1}.ThenPis:

[OnlineApril9,2014]

Options:

A.reflexiveandsymmetricbutnottransitive.

B.reflexiveandtransitivebutnotsymmetric.

C.symmetricandtransitivebutnotreflexive.

D.anequivalencerelation.

Answer:D

Solution:

f(x) = |x| − 1

|x| + 1

forone-onefunctioniff(x1) = f(x2)then

x1mustbeequaltox2

Letf(x1) = f(x2)

|x1| − 1

|x1| + 1=|x2| − 1

|x2| + 1|x1||x2|+|x1|−|x2| −1= | x1||x2|−|x1|+|x2| −1

⇒ | x1|−|x2|=|x2|−|x1|

2|x1| = 2|x2|

|x1|= |x2|

x1=x2,x1= −x2

herex1hastwovaluesthereforefunctionismanyonefunction.

Foronto:f(x) = |x| − 1

|x| + 1

foreveryvalueoff(x)thereisavalueofxindomainset.

Iff(x)isnegativethenx=0

forallpositivevalueoff(x),domaincontainatleastoneelement.Hencef(x)isontofunction.

P= {(a,b) : sec2a−tan2b=1}

Forreflexive:

sec2a−tan2a=1(true∀a)

Forsymmetric:

sec2b−tan2a=1

L.H.S

1+tan2b− (sec2a−1) = 1+tan2b−sec2a+1

= −(sec2a−tan2b) + 2

-------------------------------------------------------------------------------------------------

Question148

Letf (n) = 1

3+3n

100 n,where[n]denotesthegreatestintegerlessthan

orequalton.Then 56

∑

n=1f(n)isequalto:

[OnlineApril19,2014]

Options:

A.56

B.689

C.1287

D.1399

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question149

Letf beanoddfunctiondefinedonthesetofrealnumberssuchthat

forx ≥0,f(x) = 3 sin x +4 cos xThenf (x)atx = − 11π

6isequalto:

[OnlineApril11,2014]

Options:

A. 3

2+2√3

[ ]

= −1+2=1

So,RelationissymmetricFortransitive:

ifsec2a−tan2b=1andsec2b−tan2c=1

sec2a−tan2c = (1+tan2b) − (sec2b−1)

= −sec2b+tan2b+2

= −1+2=1

So,Relationistransitive.

Hence,RelationPisanequivalencerelation

Letf(n) = 1

3+3n

100 n

where[n]isgreatestintegerfunction,

=0.33 +3n

100 n

Forn=1,2, ..., 22,wegetf(n) = 0andforn=23,24, ..., 55,wegetf(n) = 1×nForn=56,f(n) = 2×n

So,

∑

n=1

f(n) = 1(23) + 1(24) + ... + 1(55) + 2(56)

= (23 +24 + ... + 55) + 112

=33

2[46 +32] + 112

=33

2(78) + 112 =1399

[ ]

B.−3

2+2√3

C. 3

2−2√3

D.−3

2−2√3

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question150

Ifgistheinverseofafunctionf andf ′(x) = 1

1+x5,theng′(x)isequalto:

[2014]

Options:

A. 1

1+ {g(x)}5

B.1 + {g(x)}5

C.1 +x5

D.5x4

Answer:B

Solution:

Givenfbeanoddfunction

f(x) = 3 sin x +4 cos x

Now,f−11π

6=3 sin −11π

6+4 cos −11π

f−11π

6=3 sin −2π +π

6+4 cos −2π +π

f−11π

6=3 sin −2π −π

6+4 cos −2π −π

Foroddfunctions

sin(−θ) = −sin θ

and cos(−θ) = cos θ

∴f−11π

6= −3 sin 2π −π

6−4 cos 2π −π

⇒f−11π

6= +3 sin π

6−4 cos π

⇒f−11π

6=3×1

2−4×√3

orf−11π

6=3

2−2√3

( ) ( ) ( )

( ) { ( ) } { ( ) }

{ }

( ) ( ) ( )

( ) ( )

( )

Sincef(x)andg(x)areinverseofeachother

∴g′(f(x)) = 1

f′(x)

-------------------------------------------------------------------------------------------------

Question151

LetR = { (x,y) : x,y ∈N.andx2−4xy +3y2=0},whereN isthesetof

allnaturalnumbers.ThentherelationRis:

[OnlineApril23,2013]

Options:

A.reflexivebutneithersymmetricnortransitive.

B.symmetricandtransitive.

C.reflexiveandsymmetric,

D.reflexiveandtransitive.

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question152

LetR = { (3,3)(5,5), (9,9), (12,12), (5,12),(3,9), (3,12), (3,5) }bea

relationonthesetA = {3,5,9,12}.Then,Ris:

[OnlineApril22,2013]

Options:

A.reflexive,symmetricbutnottransitive.

B.symmetric,transitivebutnotreflexive.

C.anequivalencerelation.

D.reflexive,transitivebutnotsymmetric.

Answer:D

⇒g′(f(x)) = 1+x5 ∵f′(x) = 1

1+x5

Herex=g(y)

∴g′(y) = 1+ [g(y)]5

⇒g′(x) = 1+ (g(x))5

( )

R= { (x,y) : x,y∈Nandx2−4xy +3y2=0}

Now,x2−4xy +3y2=0

⇒(x−y)(x−3y) = 0

∴x=yorx=3y

∴R= { (1,1), (3,1), (2,2), (6,2),(3,3)(9,3), ...... }

Since(1,1), (2,2), (3,3), ......arepresentintherelation,thereforeRisreflexive.

Since(3,1)isanelementofRbut(1,3)isnottheelementofR,thereforeRisnotsymmetric

Here(3,1)∈Rand(1,1)∈R⇒ (3,1) ∈ R(6,2)∈Rand(2,2)∈R⇒ (6,2) ∈ R

Forallsuch(a,b) ∈ Rand(b,c) ∈ R

⇒(a,c) ∈ R

HenceRistransitive.

Solution:

-------------------------------------------------------------------------------------------------

Question153

LetA = {1,2,3,4}andR :A→Abetherelationdefinedby

R= {(1,1), (2,3), (3,4), (4,2)}.Thecorrectstatementis:

[OnlineApril9,2013]

Options:

A.Rdoesnothaveaninverse.

B.Risnotaonetoonefunction.

C.Risanontofunction.

D.Risnotafunction.

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question154

IfP(S)denotesthesetofallsubsetsofagivensetS,thenthenumberof

one-to-onefunctionsfromthesetS={1,2,3}tothesetP(S)is

[OnlineMay19,2012]

Options:

A.24

B.8

C.336

D.320

Answer:C

LetR= { (3,3), (5,5), (9,9), (12,12),(5,12), (3,9), (3,12),(3,5)\}bearelationonset

A= {3,5,9,12}

Clearly,everyelementofAisrelatedtoitself.

Therefore,itisareflexive.

Now,Risnotsymmetrybecause3isrelatedto5but5isnotrelatedto3.

AlsoRistransitiverelationbecauseitsatisfiesthepropertythatifaRbandbRcthenaRc.

Domain = {1,2,3,4}

Range = {1,2,3,4}

∴Domain=Range

HencetherelationRisontofunction.

Solution:

-------------------------------------------------------------------------------------------------

Question155

IfA = { x∈z+:x<10andxisamultipleof3or4 },wherez+istheset

ofpositiveintegers,thenthetotalnumberofsymmetricrelationsonA

is.

[OnlineMay12,2012]

Options:

A.25

B.215

C.210

D.220

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question156

LetS= {1,2,3} → n(S) = 3

Now,P(S) = setofallsubsetsofS

totalno.ofsubsets = 23=8

∴n[P(S)] = 8

Now,numberofone-to-onefunctionsfromS→P(S)is8P3=8!

5!=8×7×6=336

ArelationonasetAissaidtobesymmetriciff(a,b)in A ⇒ (b,a) ∈ A, ∀a,b∈A

HereA= {3,4,6,8,9}

NumberoforderpairsofA×A=5×5=25

Divide25orderpairsofA times Ain3partsasfollows:

Part–A:(3,3),(4,4),(6,6),(8,8),(9,9)

Part–B:(3,4),(3,6),(3,8),(3,9),(4,6),(4,8),(4,9),(6,8),(6,9),(8,9)

Part–C:(4,3),(6,3),(8,3),(9,3),(6,4),(8,4),(9,4),(8,6),(9,6),(9,8)

Inpart–A,bothcomponentsofeachorderpairaresame.

Inpart–B,bothcomponentsaredifferentbutnottwosuchorderpairsarepresentinwhichfirstcomponentofoneorder

pairisthesecondcomponentofanotherorderpairandvice-versa.

Inpart–C,onlyreverseoftheorderpairsofpart–Barepresenti.e.,if(a,b)ispresentinpart–B,then(b,a)willbe

presentinpart–C

Forexample(3,4)ispresentinpart–Band(4,3)presentinpart–C.

NumberoforderpairinA,BandCare5,10and10respectively.

InanysymmetricrelationonsetA,ifanyorderpairofpart–Bispresentthenitsreverseorderpairofpart–Cwillmust

bealsopresent.

HencenumberofsymmetricrelationonsetAisequaltothenumberofallrelationsonasetD,whichcontainsallthe

orderpairsofpart–Aandpart–B.

Nown(D) = n(A) + n(B) = 5+10 =15

HencenumberofallrelationsonsetD= (2)15

⇒NumberofsymmetricrelationsonsetD= (2)15

Therangeofthefunctionf (x) = x

1+ | x|,x∈R,is

[OnlineMay7,2012]

Options:

A.R

B.(-1,1)

C.R − {0}

D.[-1,1]

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question157

LetAandBbenonemptysetsinRandf :A→Bisabijectivefunction.

Statement1:f isanontofunction.

Statement2:Thereexistsafunctiong :B→Asuchthatfog =IB

[OnlineMay26,2012]

Options:

A.Statement1istrue,Statement2isfalse.

B.Statement1istrue,Statement2istrue;Statement2isacorrectexplanationforStatement

1.

C.Statement1isfalse,Statement2istrue.

D.Statement1istrue,Statement2istrue,Statement2isnotthecorrectexplanationfor

Statement1.

Answer:D

Solution:

f(x) = x

1+ | x|,x∈R

Ifx>0, | x=x⇒f(x) = x

1+x

whichisnotdefinedforx= −1

Ifx<0, | x= −x⇒f(x) = x

1−xwhichisnotdefinedforx=1

Thusf(x)definedforallvaluesofRexcept1and-1

Hence,range = (−1,1)

LetAandBbenon-emptysetsinR.

Letf:A→Bisbijectivefunction.

Clearlystatement-1istruewhichsaysthatfisanontofunction.

Statement-2isalsotruestatementbutitisnotthecorrectexplanationforstatement-1

-------------------------------------------------------------------------------------------------

Question158

LetRbethesetofrealnumbers.

Statement-1:A = { (x,y) ∈ R×R:y−xisaninteger}isanequivalence

relationonR.

Statement-2:B = { (x,y) ∈ R×R:x=αyforsomerationalnumberα }

isanequivalencerelationonR.

[2011]

Options:

A.Statement-1istrue,Statement-2istrue;Statement-2isnotacorrectexplanationfor

Statement-1.

B.Statement-1istrue,Statement-2isfalse.

C.Statement-1isfalse,Statement-2istrue.

D.Statement-1istrue,Statement-2istrue;Statement-2isacorrectexplanationforStatement-

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question159

Thedomainofthefunctionf (x) = 1

√|x| − xis

[2011]

Options:

A.(0,∞)

B.(−∞,0)

C.(−∞,∞) − {0}

D.(−∞,∞)

Answer:B

Solution:

∵x−x=0∈I( ∴Risreflexive)

Let(x,y) ∈ Rasx−yandy−x∈I( ∵Rissymmetric)

Nowx−y∈Iandy−z∈I⇒x−z∈I

So,Ristransative.

HenceRisequivalence.

Similarlyasx=αyforα=1.Bisreflexivesymmetricandtransative.HenceBisequivalence.

Bothrelationsareequivalencebutnotthecorrectexplanation.

Solution:

-------------------------------------------------------------------------------------------------

Question160

Letf beafunctiondefinedby

f(x) = (x−1)2+1, (x≥1)

Statement-1:Theset{x:f(x) = f−1(x)} = {1,2}

Statement-2:f isabijectionandf −1(x) = 1+ √x−1,x≥1

[2011RS]

Options:

A.Statement-1istrue,Statement-2istrue;Statement-2isacorrectexplanationforStatement-

1.

B.Statement-1istrue,Statement-2istrue;Statement-2isNOTacorrectexplanationfor

Statement-1.

C.Statement-1istrue,Statement-2isfalse.

D.Statement-1isfalse,Statement-2istrue.

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question161

Considerthefollowingrelations:

R= { (x,y) | x,yarerealnumbersandx =wyforsomerationalnumber

w}; S=m

n,p

qm,n,p.andqareintegerssuchthatn,q≠0and

qm =pn }

.Then

[2010]

Options:

{ ( ) |

f(x) = 1

√|x| − x,f(x)isdefineif|x| − x>0

⇒ | x| >x, ⇒x<0

Hencedomainoff(x)is(−∞,0)

Givenfisabijectivefunction

∴f: [1,∞) → [1,∞)

f(x) = (x−1)2+1,x≥1

Lety= (x−1)2+1⇒(x−1)2=y−1

⇒x=1± √y−1⇒f−1(y) = 1± √y−1

⇒f−1(x) = 1+ √x−1{∴x≥1}

Hencestatement-2iscorrect

Nowf(x) = f−1(x)

⇒f(x) = x⇒ (x−1)2+1=x

⇒x2−3x +2=0⇒x=1,2

Hencestatement-1iscorrect

A.NeitherRnorSisanequivalencerelation

B.SisanequivalencerelationbutRisnotanequivalencerelation

C.RandSbothareequivalencerelations

D.RisanequivalencerelationbutSisnotanequivalencerelation

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question162

Letf (x) = (x+1)2−1,x≥ −1

Statement-1:Theset{x:f(x) = f−1(x) = {0, −1}.

Statement-2:f isabijection.

[2009]

Options:

A.Statement-1istrue,Statement-2istrue.Statement-2isnotacorrectexplanationfor

Statement-1.

B.Statement-1istrue,Statement-2isfalse.

C.Statement-1isfalse,Statement-2istrue.

D.Statement-1istrue,Statement-2istrue.Statement-2isacorrectexplanationforStatement-

Answer:D

Solution:

LetxRy.

⇒x=wy ⇒y=x

w

⇒(y,x) ∉ R

Risnotsymmetric

LetS:m

nSp

q

⇒qm =pn ⇒p

q=m

n

∵m

n=m

n∴reflexive

n=p

q⇒p

q=m

n∴symmetric

Letm

nSp

q,p

qSr

s

⇒qm =pn,ps =rq

⇒p

q=m

n=r

s

⇒ms =rntransitive

.Sisanequivalencerelation.

Giventhatf(x) = (x+1)2−1,x≥ −1

-------------------------------------------------------------------------------------------------

Question163

LetRbetherealline.ConsiderthefollowingsubsetsoftheplaneR ×R

:

S= { (x,y) : y=x+1and0 <x<2}

T= { (x,y) : x−yisaninteger}

Whichoneofthefollowingistrue?

[2008]

Options:

A.NeitherSnorT isanequivalencerelationonR

B.BothSandT areequivalencerelationonR

C.SisanequivalencerelationonRbutT isnot

D.T isanequivalencerelationonRbutSisnot

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question164

ClearlyDf= [−1,∞)butco-domainisnotgiven.Thereforef(x)isonto.

Letf(x1) = f(x2)

⇒(x1+1)2−1= (x2+1)2−1

⇒x1=x2

∴f(x)isone-one,hencef(x)isbijection

∵(x+1)beingsomething+ve, ∀x> −1

∴f−1(x)willexist.Let(x+1)2−1=y

⇒x+1= √y+1(+vesquarerootasx+1≥0)

⇒x= −1+ √y+1

⇒f−1(x) = √x+1−1

Thenf(x) = f−1(x)

⇒(x+1)2−1= √x+1−1

⇒(x+1)2= √x+1⇒(x+1)4= (x+1)

⇒(x+1)[(x+1)3−1] = 0⇒x= −1,0

∴Thestatement-1andstatement-2botharetrue.

Giventhat

S= { (x,y) : y=x+1and0<x<2}

∵x≠x+1foranyx ∈ (0,2)

⇒(x,x) ∉ S

So,Sisnotreflexive.

Hence,Sinnotanequivalencerelation.

GivenT= {x,y) : x−yisaninteger}

∵x−x=0isaninteger,∀x∈R

So,Tisreflexive.

Let(x,y) ∈ T⇒x−yisanintegertheny−xisalsoaninteger⇒(y,x) ∈ R

∴Tissymmetric

Ifx−yisanintegerandy−zisanintegerthen(x−y) + (y−z) = x−zisalsoaninteger.

∴Tistransitive

HenceTisanequivalencerelation.

Letf :N→Y beafunctiondefinedasf (x) = 4x +3where

Y= { y∈N:y=4x +3forsomex ∈N}.

Showthatf isinvertibleanditsinverseis

[2008]

Options:

A.g(y) = 3y +4

B.g(y) = 4+y+3

C.g(y) = y+3

D.g(y) = y−3

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question165

LetWdenotethewordsintheEnglishdictionary.DefinetherelationR

byR = (x,y) ∈ W×W|thewordsxandyhaveatleastoneletterin

common.}ThenRis

[2006]

Options:

A.notreflexive,symmetricandtransitive

B.relexive,symmetricandnottransitive

C.reflexive,symmetricandtransitive

D.reflexive,notsymmetricandtransitive

Answer:B

Solution:

Clearlyf(x) = 4x +3isoneoneandonto,soitisinvertible.

Letf(x) = 4x +3=y

⇒x=y−3

4

∴g(y) = y−3

Clearly(x,x) ∈ R, ∀x∈W

∵Allletterarecommoninsomeword.SoRisreflexive.

Let(x,y) ∈ R,then(y,x) ∈ Rasxandyhaveatleastoneletterincommon.So,Rissymmetric.

ButRisnottransitiveforexample

Letx=BOY ,y= TOYandz= THREE

-------------------------------------------------------------------------------------------------

Question166

Arealvaluedfunctionf (x)satisfiesthefunctionalequation

f(x−y) = f(x)f(y) − f(a−x)f(a+y)whereaisagivenconstantand

f(0) = 0,f(2a −x)isequalto

[2005]

Options:

A.−f(x)

B.f (x)

C.f (a) + f(a−x)

D.f (−x)

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question167

LetR={(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)}bea

relationontheset

A={3,6,9,12}.Therelationis

[2005]

Options:

A.reflexiveandtransitiveonly

B.reflexiveonly

C.anequivalencerelation

D.reflexiveandsymmetriconly

Answer:A

Solution:

then(x,y) ∈ R(O,Yarecommon)and(y,z) ∈ R(Tiscommon)but(x,z) ∉ R.(asnoletteriscommon)

Giventhatf(0) = 0andput

x=0,y=0

f(0) = f2(0) − f2(a)

⇒f2(a) = 0⇒f(a) = 0

f(2a −x) = f(a− (x−a))

=f(a)f(x−a) − f(0)f(x)

=f(a)f(x−a) − f(x) = −f(x)

⇒f(2a −x) = −f(x)

Solution:

-------------------------------------------------------------------------------------------------

Question168

Letf : (−1,1) → B,beafunctiondefinedbyf (x) = tan−12x

1−x2,thenf is

bothone−oneandontowhenBistheinterval

[2005]

Options:

A. 0,π

B. 0,π

C. −π

2,π

D. −π

2,π

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question169

Thegraphofthefunctiony =f(x)issymmetricalaboutthelinex =2,

then

[2004]

Options:

A.f (x) = −f(−x)

B.f (2+x) = f(2−x)

( )

[ )

[ ]

( )

Risreflexiveandtransitiveonly.

Here(3,3), (6,6), (9,9), (12,12) ∈ R[So,reflexive]

(3,6), (6,12), (3,12) ∈ R[So,transitive]

∵(3,6) ∈ Rbut(6,3)∉R[So,non-symmetric]

Givenf(x) = tan−12x

1−x2=2tan−1x

forx∈ (−1,1)

Ifx∈ (−1,1) ⇒ tan−1x∈−π

4,π

4

⇒2tan−1x∈−π

2,π

2

Clearly,rangeoff(x) = −π

2,π

2

Forftobeonto,codomain=range

∴Co-domainoffunction = B= −π

2,π

( )

C.f (x) = f(−x)

D.f (x+2) = f(x−2)

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question170

LetR={(1,3),(4,2),(2,4),(2,3),(3,1)}bearelationonthesetA={1,

2,3,4}..TherelationRis

[2004]

Options:

A.reflexive

B.transitive

C.notsymmetric

D.afunction

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question171

(b)Giventhatagraphsymmetrical.withrespecttolinex=2asshowninthefigure.

Fromthefigure

f(x1) = f(x2), where x1=2−x and x2=2+x

∴f(2−x) = f(2+x)

∵(1,1) ∉ R⇒Risnotreflexive

∵(2,3) ∈ Rbut(3,2)∉R

∴Risnotsymmetric

∵(4,2)and(2,4)∈Rbut(4,4)∉R

⇒Risnottransitive

Iff :R→S,definedbyf (x) = sin x − √3cos x +1,isonto,thenthe

intervalofSis

[2004]

Options:

A.[-1,3]

B.[-1,1]

C.[0,1]

D.[0,3]

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question172

Domainofdefinitionofthefunctionf (x) = 3

4−x2+log10(x3−x),is

[2003]

Options:

A.(-1,0)∪(1,2) ∪ (2,∞)

B.(a,2)

C.(-1,0)∪(a,2)

D.(1,2)∪(2,∞)

Answer:A

Solution:

Giventhatf(x)isonto

∴rangeoff(x) = codomain=S

Now,f(x) = sin x − √3 cos x +1

=2 sin x −π

3+1

weknowthat−1≤sin x −π

3≤1

−1≤2 sin x −π

3+1≤3∴f(x) ∈ [−1,3] = S

( )

f(x) = 3

4−x2+log10(x3−x)

4−x2≠0;x3−x>0

x≠ ±√4and−1<x<0or1<x<∞

-------------------------------------------------------------------------------------------------

Question173

Iff :R→Rsatisfiesf (x+y) = f(x) + f(y),forallx,

y∈Randf (1) = 7,then n

∑

r=1f(r)is

[2003]

Options:

A. 7n(n+1)

B. 7n

C. 7(n+1)

D.7n + (n+1)

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question174

Afunctionffromthesetofnaturalnumberstointegersdefinedby

f(n) =

n−1

2,when n is odd

−n

2,when n is even

is

[2003]

Options:

A.neitherone-onenoronto

{

∴D= (−1,0) ∪ (1,∞) − {√4}

D= (−1,0) ∪ (1,2) ∪ (2,∞)

f(x+y) = f(x) + f(y)

∵f(1) = 7

f(2) = f(1+1) = f(1) + f(1) = 14

f(3) = f(1+2) = f(1) + f(2) = 21

___________

∴

∑

r=1

f(r) = 7(1+2+3..... + n)

=7n(n+1)

B.one-onebutnotonto

C.ontobutnotone-one

D.one-oneandontoboth.

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Wehavef:N→I

Letxandyaretwoevennaturalnumbers,andf(x) = f(y) ⇒ −x

2=−y

2⇒x=y

∴f(n)isone-oneforevennaturalnumber.

Letxandyaretwooddnaturalnumbersandf(x) = f(y) ⇒ x−1

2=y−1

2⇒x=y

∴f(n)isone-oneforoddnaturalnumber.

Hencefisone-one.

Lety=n−1

2⇒2y +1=n

Thisshowsthatnisalwaysoddnumberfory∈I......(i)

andy=−n

2⇒ −2y =n

Thisshowsthatnisalwaysevennumberfory∈I......(ii)

From(i)and(ii)

Rangeoff=I= codomain

∴fisonto.

Hencefisoneoneandontoboth.