jee mains maths solutions year 2010 download doubtnut … · 2018. 10. 3. · 15 jee mains maths...

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JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut Today Ques No. Question 1 JEE MAINS MATHS SOLUTIONS - 2010 Let and let where , then (1) (2) (3) (4) Watch Free Video Solution on Doubtnut 2 JEE MAINS MATHS SOLUTIONS - 2010 Let S be a non-empty subset of R. Consider the following statement: P: There is a rational number such that . Which of the following statements is the negation of the statement P ? There is no rational number such that (9) Every rational number satisfies (18) and (27) is not rational There is a rational number such that (36) Watch Free Video Solution on Doubtnut 3 JEE MAINS MATHS SOLUTIONS - 2010 Let and . Then vector satisfying and is (1) (2) (3) (4) Watch Free Video Solution on Doubtnut cos(α + β)= 4 5 s ∈( αβ)= 5 13 0≤ α , β π 4 tan2 α = 56 33 19 12 20 7 25 16 x S x >0 x S x ≤0 x S x ≤0 x S x ≤0⇒ x x S x ≤0 a = ˆ j ˆ k c = ˆ i ˆ j ˆ k b a × b + c = 0 a . b =3 2 i j +2 k ˆ i ˆ j −2 ˆ k ˆ i + j −2 ˆ k ˆ i + ˆ j −2 ˆ k

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Page 1: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

JEEMAINSMATHSSOLUTIONS

YEAR2010

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QuesNo. Question

1

JEEMAINSMATHSSOLUTIONS-2010

Let and let where , then

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

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2

JEEMAINSMATHSSOLUTIONS-2010

LetSbeanon-empty subset ofR.Consider the following statement:P:There is arationalnumber such that .Which of the following statements is thenegationofthestatementP?Thereisnorationalnumber suchthat (9)Everyrationalnumber satisfies (18) and (27)isnotrationalThereisarationalnumber suchthat (36)

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3

JEEMAINSMATHSSOLUTIONS-2010

Let and . Then vector satisfying

and is (1) (2) (3)(4)

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cos(α + β) =45

s ∈ (αβ) =5

130 ≤ α, β ≤

π

4tan2α =

5633

1912

207

2516

x ∈ S x > 0x ∈ S x ≤ 0

x ∈ S x ≤ 0 x ∈ S x ≤ 0 ⇒ xx ∈ S x ≤ 0

→a = j − k

→c = i − j − k

→b

→a ×

→b + →

c =→0 →

a

.→b = 3 2

→i −

→j + 2

→k i − j − 2k

i + j − 2k − i + j − 2k

Page 2: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

4

JEEMAINSMATHSSOLUTIONS-2010

Theequationofthetangenttothecurve ,thatisparalleltothex-axis,is

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

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5

JEEMAINSMATHSSOLUTIONS-2010

Solutionofthedifferentialequation

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6

JEEMAINSMATHSSOLUTIONS-2010

Theareaboundedbythecurves

between theordinates is (1) (2) (3)

(4)

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7

JEEMAINSMATHSSOLUTIONS-2010

IftwotangentsdrawnfromapointPtotheparabola areatrightangles,thenthelocusofPis

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y = x +4

x2

y = 1 y = 2 y = 3 y = 0

cos xdy = y(sinx

− y)dx, 0 < x <π

2

y = cosxandy= sinx

x = 0andx =3π2

4√2 + 2 4√21 4√2 + 1

4√22

y2 = 4x

Page 3: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

8

JEEMAINSMATHSSOLUTIONS-2010

Ifthevectors

aremutuallyorthogonal,then (1) (2) (3) (4)

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9

JEEMAINSMATHSSOLUTIONS-2010

Consider the following relations:R= {(x, y) | x, y are real numbers and x =wy forsomerationalnumberw};

.Then(1)neitherRnorSisanequivalencerelation(2)SisanequivalencerelationbutRisnotanequivalencerelation(3)RandSbothareequivalencerelations(4)RisanequivalencerelationbutSisnotanequivalencerelation

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10

JEEMAINSMATHSSOLUTIONS-2010

Let bedefinedby

.Iffhasalocalminimumat ,thenapossiblevalueofkis(1)0(2) (3)

→a = i − j + 2

→k ,

→b = 2 i + 4j

+ kand→c = λi + j

+ mk

(λ, μ) = (2, 3) (2, 3) (3, 2) (3, 2)

S

= {( ,

)m,n,pandqareintegerssuchthatn,q ≠ 0andqm=pn}

m

np

q

f : R → Rf(x) = {k − 2x,

if x ≤ − 12x+ 3, fx ≻ 1}

x = 1 −12

−1

Page 4: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

(4)1

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11

JEEMAINSMATHSSOLUTIONS-2010

Thenumberof3´3non-singularmatrices,withfourentriesas1andallotherentriesas0,is(1)5(2)6(3)atleast7(4)lessthan4

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12

JEEMAINSMATHSSOLUTIONS-2010

Fournumbersarechosenatrandom(withoutreplacement)fromtheset{1,2,3,.....,20}. Statement-1: The probability that the chosen numberswhen arranged in some

orderwill formanAPIs .Statement-2: If the fourchosennumbers fromanAP,

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

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13

JEEMAINSMATHSSOLUTIONS-2010

Statement-1:ThepointA(3,1,6)isthemirrorimageofthepointB(1,3,4)intheplane.Statement-2:Theplanex bisectsthelinesegmentjoining

A(3,1,6)andB(1,3,4).(1)Statement-1istrue,Statement-2istrue;Statement-2isnot the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 isfalse(3)Statement-1isfalse,Statement-2istrue(4)Statement-1istrue,Statement-2istrue;Statement-2isthecorrectexplanationforStatement-1

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JEEMAINSMATHSSOLUTIONS-2010

2−1

185

xy + z = 5 xy + z = 5

Page 5: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

14

Let

Statement-1: Statement-2:

. (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correctexplanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3)Statement-1isfalse,Statement-2istrue(4)Statement-1istrue,Statement-2istrue;Statement-2isthecorrectexplanationforStatement-1

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15

JEEMAINSMATHSSOLUTIONS-2010

Let A be a matrix with non-zero entries and let , where I isidentitymatrix.DefineTr(A)=sumofdiagonalelementsofAand|A|=determinantofmatrix A. Statement-1: Statement-2: (1) Statement-1 is true,Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2)Statement-1istrue,Statement-2isfalse(3)Statement-1isfalse,Statement-2istrue(4)Statement-1istrue,Statement-2istrue;Statement-2isthecorrectexplanationforStatement-1

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16

JEEMAINSMATHSSOLUTIONS-2010

Let f:RRbeacontinuous functiondefinedby .Statement-1:

forsome .Statement-2: forall . (1)

Statement-1istrue,Statement-2istrue;Statement-2isnotthecorrectexplanationfor

S1 =10

∑j= 1

j(j − 1)10Cj , S2

=10

∑j = 1

j10Ci(andS)3

=10

∑j = 1

j210Cj.

S3 = 55 × 29

S1 = 90 × 28andS2

= 10 × 28

2 × 2 A2 = I 2 × 2

T r(A) = 0 |A| = 1

f(x) =1

ex + 2e−x

f(c) = ,13

c ∈ R 0 < f(x) ≤ ,1

2√2x ∈ R

Page 6: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false,Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is thecorrectexplanationforStatement-1

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17

JEEMAINSMATHSSOLUTIONS-2010

Foraregularpolygon,letrandRbetheradiioftheinscribedandthecircumscribedcircles. A false statement among the following is There is a regular polygon with

(17) There is a regular polygon with (30) There is a regular

polygonwith (47)Thereisaregularpolygonwith (60)

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18

JEEMAINSMATHSSOLUTIONS-2010

If and aretherootsoftheequation ,then (1)(2)1(3)2(4)

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19

JEEMAINSMATHSSOLUTIONS-2010

Thenumberofcomplexnumberszsuchthat equals(1)1(2)2(3) (4)0

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20

JEEMAINSMATHSSOLUTIONS-2010

AlineABinthree-dimensionalspacemakesangles and withthepositivex-axis and the positive y-axis respectively. If AB makes an acute angle q with thepositivez-axis,thenqequals

=r

R

1

√2=

r

R

23

=r

R

√32

=r

R

12

α β x2-x + 1 = 0 α2009 + β2009 = 12

|z1| = |z + 1| = |zi|∞

45o 120o

Page 7: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

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21

JEEMAINSMATHSSOLUTIONS-2010

The line L given by passes through the point (13, 32). The line K is

paralleltoLandhastheequation ThenthedistancebetweenLandKis

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

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22

JEEMAINSMATHSSOLUTIONS-2010

A person is to count 4500 currency notes. Let an denote the number of notes hecountsinthenthminute.If

and are inA.P.withcommondifference2, then the time takenbyhim to count all notes is (1) 34 minutes (2) 125 minutes (3) 135 minutes (4) 24minutes

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23

JEEMAINSMATHSSOLUTIONS-2010

Let be a positive increasing function with . Then

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

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JEEMAINSMATHSSOLUTIONS-2010

+ = 1x

5y

b+ = 1

x

c

y

3

√1717

√15

23

√17

23

√15

a1 = a2 = . . . . . .= a10 = 150

a10, a11, . . . . . .

f : R → R limx→ ∞

= 1f(3x)

f(x)

limx→ ∞

=f(2x)

f(x)23

32

( ) = (1 )

Page 8: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

24Let p(x) be a function defined on R such that , for all

and .Then equals(1)21(2)41(3)42

(4)

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25

JEEMAINSMATHSSOLUTIONS-2010

Let beadifferentiablefunctionwith

.Let

.Then (1) (2)0(3) (4)4

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26

JEEMAINSMATHSSOLUTIONS-2010

Therearetwourns.UrnAhas3distinctredballsandurnBhas9distinctblueballs.Fromeachurn twoballsare takenoutat randomand then transferred to theother.The number of ways in which this can be done is (1) 36 (2)66(3)108(4)3

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27

JEEMAINSMATHSSOLUTIONS-2010

Considerthesystemoflinearequations:

Thesystemhas(1)exactly3solutions(2)auniquesolution(3)nosolution(4)infinitenumberofsolutions

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p' (x) = p' (1x)

x ∈ [0, 1], p(0) = 1 p(1) = 41 ∫ 1

0p(x)dx

√41

f : (1, 1) → Rf(0) = − 1 and

f ' (0) = 1

g(x)

= [f(2f(x) + 2)]2

g' (0) = 4 2

x1 + 2x2 + x3 = 3 2x1 + 3x2 + x3 = 33x1 + 5x2 + 2x3

= 1

Page 9: JEE MAINS MATHS SOLUTIONS YEAR 2010 Download Doubtnut … · 2018. 10. 3. · 15 JEE MAINS MATHS SOLUTIONS - 2010 Let A be a matrix with non-zero entries and let , where I is identity

28

JEEMAINSMATHSSOLUTIONS-2010

Anurncontainsnineballsofwhich threeare red, fourareblueand twoaregreen.Three balls are drawn at randomwithout replacement from the urn. The probability

thatthethreeballshavedifferentcolouris(1) (2) (3) (4)

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JEEMAINSMATHSSOLUTIONS-2010

For two data sets, each of size 5, the variances are given to be 4 and 5 and thecorresponding means are given to be 2 and 4, respectively. The variance of the

combineddatasetis(1) (2)2(3) (4)

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JEEMAINSMATHSSOLUTIONS-2010

Thecircle

intersects the line at two distinct points if (1) (2)(3) (4)

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27

121

223

13

112

132

52

x2 + y2 = 4x + 8y+ 5

3x4y = m 35 < m < 1515 < m < 65 35 < m < 85 85 < m < 35