jee advanced paper 1 solutions 2013

7/27/2019 JEE Advanced Paper 1 Solutions 2013

1/39

~ l f l l ' : 3 ' Q U~

Time : 3 Hours

~ - 1C o d e [ 1 J

l{ W ft l{ 3l

31'1,~ ~

&m crm:!


2/39

( PHYSICS )

PART- I PHYSICS

SECTION-1 : (Only One option corre ct type)~ - 1 : (~ ~ ~ ~ fclW


3/39

( PHYSICS )2. The image of an object , formed by a plano -convex lens at a d istance of 8 m behind the lens, is real and is

2one -th ird the si ze of the ob ject. The wave length of light inside the lens is 3 times the wavelength in free

space. The rad ius of the curved surface of the lens is :~ ~ ~ ~ ~ m ~ ~ ~ f . l > i l~ ~ 8 m~


4/39

=0 .001 em= tP1A CflT ~ c p+ ~ tP1A CflT ~ c p X ~ ~= .10 + (24 ) (0.001 )=5.124 em.

PHYSICS

4. The work done o n a part icle of mass m by a fo rce , K [ {x ' + : 2 } " 1 {x ' + : ' ) " i] K being a constant o fappropr iate dime nsio ns ), wh e n the part icle is taken from the point (a , 0 ) to the poi nt (0 , a ) a long a c ircu lar pathof rad ius a about the or ig in in the x -y pla ne is :

~ 6IC1 K[(x2 +> )312 i (x2 +> )3' 2 } K~ mm, fct+rr CflT f ~ ; f f i c p~).~ m~ ~ CflUIem(a,O)~ ~(O,a ) CfCP ~ a ~ ~ ~' l~ 'lX ~ \1fmT ~ . fGR:!CflT~ x-y Cfe1 CflT ~ ~ ~ 1 orc1 &TXTfcn


5/39

PHYSICS

(:)Sol. Y = M 1 .. . (i)

L

(:A)Y = M 2 .. .(ii)2L

6. A ray o f ligh t travelling in the direct ion ~

0J3

])s inc ide nt on a p lan e mirror. A fter re flec tion , it trav e ls a long

th e direct io n ~ ( - J31).The a ngle o f inc ide nc e is :

~ ~ ~ 1R ;m-qftm >1CPrnfc!JxurCift~ fum ~ (i + J3 1)'g I W ~~ ~ ~ fum ~ (i - J3 1)m\JIT"ff!'g I fcnxurCfi'T ~ cmur 'g :(A) 30 ( B) 45 ( C) 60 ( D) 75

Ans. (A)

Sol. Angle be tw een g iven rays is 120 so ang le of nc ide nce is 30

y

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6/39

Hindi. tr ~ fc!Jxoit~ 11Ul cpfoT120 g~ : ~ cmu-r3o o g

( PHYSICS )

y

7. Two rectangu lar blocks, having indent ical dimensions , can be arranged either in con figuration I or in configurationI I as sho wn in the figure , On of the blocks has the rma l conductivity k and the other 2k . The temperaturedifference be tween the ends a long the x-axis is the same in both the conf igurat ions. It takes 9s to transport acerta in amount of heat f rom the hot end to the co ld end in the configu ration I. The time to transport the sameamount of heat in the configuration I I s :en~ ~ ~ x :rcCBTem~ ~ fu?ITJmXen frRnm I x li B ~ fc!J


7/39

( PHYSICS )

frRm:r 2 -B ~ m ~ < l~ ~ ~m ~ > l " f c R T ~oc m q ~ f.r9 m q c#t ~ ~ \3)1SI1f >rm-g -B fWn Tfr


8/39

( PHYSICS )

10. Two non-react ive monoatom ic idea l gases have the ir atomic masses in t he ra tio 2 : 3. The ra tio of the ir part ia lpressures , when enc losed in a vessel kep t at a constan t temperature , is 4 : 3. The ra tio of the ir dens ities is:en ~ f i ! R n ~~ 4 '


9/39

S o l.

S o l.

q ____

At po int PIf resu ltant e lectr ic fie ld is zerothen

2.!. =P2

At po int QIf resu ltant e lectric fie ld is zerothen

P1 32P2 25

q ____

2.!. =P2

P1 32- -P2 25

(p1 must be negat ive )

( PHYSICS )

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10/39

( PHYSICS )

12. A ho rizonta l stretched str ing, f ixed at two ends , is vibrat ing in its fifth harmonic accord ing to the equat ion, y(x,t) = (0.01 m ) sin [(62 .8 m - 1) x] cos [(628 s- 1)t] . Assuming n = 3 .14, the correct statement( s ) is (are ) :

cn-;:fttmT 1R ~ 1\lfcWr~ ~ ~ : r r ~~ x u r. y(x , t ) = (0 .01 m ) s in [(62 .8 m- 1) x] cos [(628 s -1)t]&TXTC l ) ~tn ~ ~ I 1 C P ~mft ~ ~ ~(A ) Th e number of nodes is 5 .(B) The length of the st ring is 0 .25 m .

(C ) The maximum disp lacement of the m idpo int of he str ing its equi librium posit ion is 0 .01 m .(D) The fundamental fre qu e nc y is 100 H z .( A) ~ c#t ' f i ~5 ~ I(B) mft c#t ~ 0 .25 m I(C ) fl l A= 10

(l ) 628v = - = -- = 10 ms -1

w k 62.8

5AL = -= 0 .25

2

(C ) 2A = 0 .01 = maximum amp litude of antinode

v 10(D) f = 2e = 2 x 0.25 = 20 Hz .

Sol. (A )


11/39

( PHYSICS )

13. In the circu it shown in t he figure , the re are two pa ra lle l plate capacito rs each o f capac itance C. The s witch S 1is pressed first to fully charge the capacito r C

1and then re leased. The s witch S

2is then pressed to cha rge the

capac itor C 2 . After some time , S2 is released and then S 3 is pressed. After some t ime.~ "B~ ~ - q f t q ~B. en ~ F = r r ~~ qn;i ~ = i t n f t - ? I T"B~ C f lClft mfun C 1 r R ~"B~ S 1 CflT~ \Jflcn ~cnfcn'fimfu;r C1 Tf W l ~ m I lCflcm e n ~1 ~ t ' t> l C f l ~~ ~ ~(A) The direct ion of the magnetic fie ld is - z direct ion.(B) The direct ion of the magnet ic field is +z direction

SOnM(C) The magn itude of the magnet ic fie ld 3 0 un its.

100 nM(D) The magnitude of the magnet ic field is

30un its .

(A ) ~ < lllf?r - z fum B I

SOnM(C ) J ~ c 6 ) l lI f f ~CflT m li lur 3 0 ~ I

Ans. (A,C)

(B) l J ~ c 6 ) l llff?f +z fum B 1

100 nM( D ) ~ llf?rCflT ~ 30 ~ ~ ~I

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12/39

( PHYSICS )

S o l. Component of fina l veloc ity of partic le is in pos itive y d irect ion.Centre of c ircle is present on positive y axis. so magnet ic f ield is present in negat ive z-d irect ionAng le of dev iat ion is 30 because

V y 1tane = - = r;:;

Vx v 3

e = OB tM

MeB = Ot

B 5 ~ ~n)Hindi. c p u [ ~~ W'r em t:ICclJ ~ y ~ T "B~ern


13/39

S o l.

Hindi.

On sma llsphere

4 43 nR

3 (p )g + kx = 3 nR3 (2p g

on second sp he re (la rge )

i nR3 (3p)g = _i nR3 (2p)g+k x3 3

by equa tion (i) and (ii)

4nR3 pgx =

3k

m e ~ - q x

4 43 nR

3 (p )g + kx = 3 nR3 (2p)g

~ o m ~ - q x

i nR3 (3p)g = _i nR3 (2p)g+k x3 3

~ C j ) X U [(i) q (ii)

( PHYSICS )

.. i)

... ii)

.. i)

... ii)

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14/39

SECTION-3 : (Intege r v a lue co rrec t Typ e)~ - 3 : ~ f q ;l fF f ~ W P T ~)

( PHYSICS )

This sect ion con ta ins 5 qu est io ns . The ans we r to each quest ion is a s ingle digi t integer , ranging f rom 0 to9 (both inclusive)

~ ~ ~ ~ 5 ~ T - ~t 1 ~ < P~ o tem \R R o 9 qq ; (cn-fl vrfi;f) ~ ~ e ml:!'


15/39

v.,

+====t .- : - - - --+ v

!e

So s lope will be ( ) , and it wi ll be same for both the meta ls.

So ratio of the s lopes = 1

Hindi. KG-nax= hv - ~

eVst = h v -~

y = m x + C

v .

. .======-.-;------+ v!e

31Cl: GTC1 ( ) t d ' ~T~ m mg3ti ~ ~ ~ Fl ~

31Cl: m ~ ~ C l=1

( PHYSICS )

1 8. A bob of mass m, suspended by a st ring of length / 1 , is g iven a m inimum veloc ity requ ired to comp lete a fullcirc le in the vert ical p lane , At the highest po int, it coll ides e las t ically with another bo b of mass m suspendedby a string of length /

2, which is in itia lly at rest. Both the str ings a re mass-less and inextens ible. If the second

bob , after collis ion acqu ires the m in imum speed required to complete a full circle in the vert ica l plane, the rat io

I, .I; IS :~ m ' ; : [H'C{fmctl~ \11) fcl5~ X 'ClC1 B Tf Cffi ~ < P ~~ ~ ~ t CfoTI,T ; < P T ~ ~:

Ans. 5www.examrace.com


16/39

Sol.

Hindi.

To complete the vertica l c ircle

J9i; = ~ 5 ge 2

+ - -

~ 5 gf . 1

~ x crcr ~ C j ) ~~ ~

J9i; = ~ 5 ge 2

( PHYSICS )

19. A part ic le of mass 0.2 kg is moving in one dimens ion under a force that delivers a constant po wer 0 .5 W to thepart icle. If the initia l speed (in ms -1 } of the part icle is ze ro, the speed (in ms- 1} after 5s is :~ o.2 g ~ CJ)T cpU[ ~


17/39

( PHYSICS )2 0 . A resh ly prepa red samp le of a rad ioisotope of hatf -life 1386 s has act ivity 103 dis integrat ions pe r second . Given

that In 2 = .693, the fraction of the initia lnumber o f nuc lei (expressed in neares t integer percen tage) that willdecay in the first 80s afte r prepara tion of the sample is :

Sol.

~ ~ Cfrfu t ) c p ~~ I~ In 2 =0 .693 . cr.T > l '~ 80 s B fm:rfecf ~ q >fR'flrcprFc1ml ~ ';:fffi1cm*r '&.tT = No x 100= 1- e -AI) X 1QQ,. At X 1QQ=5 X 1Q-'1 X 80 X 1QQ= 4

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18/39

PART - II : CHEMISTRY

SECTION - 1 : (Onl y One o ption co rre ct Type)~ ~ - 1 : (~ ~ ~ fctcm;q ~ )

CHEMISTRY

This s e ction co nta ins 10 multiple ch o ice questions. Eac h quest ion has four c ho ice s (A), (B), (C) and (D) outof which ONLY ONE is co rrec t.

21. The standa rd e ntha lpies of formation of C0 2(g), H p (t) and g lucose( s ) at 25C are -400 kJ/mo l, - 300 kJ/mo land - 1300 k J /mo l, respec tive ly. The sta ndard e ntha lpy of combus tion p er g ram of glucose at 25C is

C0 2(g ), H p (l)mn wrn (oRr) c#t ~ + r r ~~ ~ ~25C w+ro: -400 J ~ .- 300 J ~ ~ ~ - 1300kJ /BtC1 'g I >lfc:rm11 wrn


19/39

CHEMISTRY

23 . The co m pound tha t does NO T libe ra te C0 2, on t rea tme nt with aqueous sodium b icarbo nate so lution , is(A) Ben zoic ac id (B) Ben ze nesu lphon ic acid

(C ) Sa licylic acid (D) Ca rbo lic ac id (P he nol)

l R r c i J~( C ) ~ ~ f f i ~ C f 5~ (D)


20/39

CHEMISTRY

P + Q ~ R + S-B P * r 75 % e m ~P C!ft 5 0% ~ -B~ Tf"'[ ~ * r gc;r=n-B~ ~ 1 O * r frtfiF=r~ . ~~ ~ fu?r B ~ ~ ~ I ~ ~ < n* r 'fl1'ffil ~ ~ I

[O]o[Q]

(A) 2 (B) 3 (C) 0 (D) 1Ans. (D)Sol. Fo r P, if t50 % = x

the n ~ 5% = 2xThis happens o nly in f irst order react ion .so , order with respect to P is 1.

Fo r Q, the g rap h s hows tha t co ncentra tion o f Q dec reases linear ly with time. So rate , with respect to Q ,remains constan t. Hence , it is ze ro order wrt Q .So , overall order is 0 + 1 =1 ~ Ans. is D

Sol. P ~


21/39

Ans . (A)S o . The given ar rangeme nt is octa hedra lvoid arrangemen t.

~ ~ ~ T ~ ftfcffic#t C l T C R ~'g Ir,\ 2 0 .414 x25 0

r,\ 2 103.5 pm .

& r,\ < 183 pm

CHEMISTRY

So, we ha ve to c hoose from 104 pm and 125 pm . As no othe r informat ion is given, we consider e xac t fit, andhenc e 104 pm is considered as ans we r.~ : 1r1 104 pm Tim 125 pm -B~ ~ m ~ C!JW~ ~ ' C f n~ ~ ~ . ~ : 1r1 104 pm emi3 m l1l"rcmrFcP~ ~ ' f ! ~~ qrc;rr l1[q 'g I

28 . Upon treatment with ammo niaca l H2S , the meta l ion tha t prec ipitates as a su lfide is~ ~ H2S m ~~ m -qx fGrn ~ n- g~ CiJT 3fql\)1l01 ~ ~ ~ ~ "Bmmt ~ 'g(A) Fe (lll) (B)AI (Ill) (C)Mg(ll) (D)Zn(ll)

Ans . (D)S o l. Ammoniaca l H2S is group reage nt of fourt h group ca tionic rad ica ls. Fe

3 &AP will prec ipitate Fe (OH )3 andAI(OH)3 respect ive ly. On ly Zinc willform white prec ipitate o f ZnS .r i ~ H2S, llllft


22/39

CHEMISTRY

S o l. Common ore o fAg - Ag2S , Cu - CuFeS 2 ,Pb - PbS, Sn - Sn0 2 , Mg - KCI. MgCI2 6Hp , AI- A 1p 3 xHpSo ans we r is (A)

S o l. Ag ~ m + r r~ ~ - Ag2S , Cu CfiT ~ r r= r r~ ~ - CuFeS 2,

31.

Ans .S o l.

S o l.

32 .

Ans .S o l.

S o l.

33 .

Ans .

Pb CfiT ~ ~ - PbS, Sn CfiT m ~ ~ - Sn0 2 , Mg CflT ~ - KCI. MgCI2 6HpAICfil~ ~ - A ip 3 . x H p~ : \3'eH ( A ) ~ I

SECTION- 2: (Onl y o r m ore o pt io ns co rrect Type)~ ~ - 2 : (-.:rP m ~ ~ f c t q ; - ~~ " ; ()

This sec tion conta ins 5 mul t ip le c ho ice qu es t ions. Each question has fou r cho ices (A), (B), (C) and (D) ou tofwhich ONE o r MORE are correct.

Benzene and naph thalene form an idea l so lution at room tempera ture. For th is process, t he true stateme nt(s )is (are )(A) to G is posi tive (B) toSsystemis posi tive

(C) toSsurroundings= 0 (D) toH = 0~ ~ ~ " ; : [mmxur m ~ m- ~ ;mcm ~ OATC)'g 1 >rWl'f ~ ~ -mftC f l~ ~ ('g)(A) oG ~ ~ I (B) toS{f.'l r e m ~l'Cfil'l! ( i lg & l > & ~ ~} . Ma2b2 (CI< f i"' C"lcl)OQ)(D) [Pt (NH3MN0 3)]CIC1m [P t ( N H3)3C I] Br ~ '


23/39

CHEMISTRY

S o l. As ester hydrolys is is first order with respect to [H].~

l "~ cmfe


24/39

35.

Ans.

Sol.

Among P , Q , R and S , the a roma tic compound (s ) is /are(A) p (B) Q (C ) R

C l

)d AICI, p

0Na H

Q

-or(NH.)2CO,

R100 -115 c

0 0 HCIh

s

P , Q , R ~ SB ~


25/39

SECTION- 3 : (Inte ger va lue co rr ec t Ty pe)~ ~ - 3 : ( ~ < P1 { A ~ "WPJ";()

CHEMISTRY

This sect ion contains 5 qu es t io ns . The a nswer to eac h ques tion is a s ingl e digi t integ e r, ra nging f rom 0 to9 (both inc lusive ) .

36 . The to ta l number of lone -pa irs of e lectro ns in melamine is~ " ; : [1R ~ ~ - ; : : i j~ ~ ~ T l ' l )c#t 'WI m:c:m'g

Ans . 6S o l. Structure of melamine is as follo ws

Tota l no . of lone pa irs of e lectro n is '6'.S o l. ~ c#t "fR'q.'f[~ 'g :

H).JyNI(NH 2

:NyN:NH2

37 . The tota l number of carboxylic ac id groups in the p roduct P is

~ P -B c m o i f ~~ ~ c#t cge1~ 'g

Ans .

S o l.

c(;qo 1. H,o , b. p2. 030 0 3. H,0 2

2

1.H,o ' n - r - Y O O H - 2 ~ 0, ' ~~ C O O H ~

No. of - COOH group is '2'.- COOH c#t m:c:m2 g I

O/ H20 2 H O O C ~H O O C ~

38 . The atomic masses of He and Ne are 4 and 20 a.m .u. , respect ive ly. The value of the de B rog lie wavele ngth o fHe gas at - 73 oc is "M" times tha t of the de B rog lie wave length of Neat 727 oc. M isH e~ x N e~ lR+ITUJ ~ " ; : [w + r n :4 ~ x2 0a.m.u. g I, He i ff i Cift-73 C lR ~ cRTf ~ Ne c#t 727 C lR~ m ~ ffiTf~ ~ "M":fiT 'g I M


26/39

S o l.

3 9.

An s .

S o l.

40 .

An s .S o l.

S o l.

CHEMISTRY

h

'A= J2m(KE) KE ac: T

AHe mNeKENe 20 x 1000ANe - mHeKEHe

= = 5.4 x 200

EDTA4- is e thy1enediam ine tetraace ta te ion. The tota l numbe r of - C ~ Obond a ng les in [Co(EDTA)j1- comp lexion is :E D T A 4 -~ ' SI~ q 4 )1 e ~l ~ ' < f l e c :~ ~ I ~ C 1~ [Co(EDTA)f- B - C ~ O; m o mcp)uft '$1 tgC1 " ' < ' i ~~8

A te trapept ide h a s - COOH g roup o n a lanine. Th is p roduces g lyc ine (G iy), val ine (Va l), phe nyl a lan ine (Phe)and a lan ine (Ala) , on comp lete hydrolys is . Fo r th is tetrapep tide , the numbe r of poss ible sequences (p rimarystructures ) with - NH 2 group attached to a chira l ce nte r is :~ ~ ~ B " ; : [1R - COOH -gq fcM+rr;=r~ I ~ ~ \JfC1~ &RT " ; : [(Giy) , ( V al) , ~~ (Phe )d'm ~ (Aia ))ITC{lmet 'g 1 ~ t R ' r ~*r fi'l:rrFcrn~ 3 i t( m ~~ 3 i t )*r "


27/39

PART - Ill MATHEMATICS

SECTION- IStraight Objective Type

( MATHEMATICS )

This sect ion conta ins 10 questions. Each quest ion has 4 cho ices (A), (B), (C) and (D) for its ans wer , out ofwhich ONLY ONEis cor rect.

~ - I~ ' C )q f . : lc s o~~ ~ ~ B 1 0 ~ 'g I ~ ~ - $4 fctcp(Yll(A), (B), (C)cr m (D) l G r~ ~ ~ ~ ~ 'g I

23(A) 25

Sol. (B)23

25(B) 23

cot z : cot - 1 (1+2 + 4 +6 + .... . .+ 2n )0 = 1

cot L:cor 1(1 + n(n + 1))

(n + 1)- ncot L: tan -

1

1+ n (n + 1)

23cot Z : (tan - 1(n + 1) - tan- 1 n)

0 = 1

cot (tan -124 - tan -11)

(- 1 24 -1)

cot tan 1+ 24

cot cot - = - - 1 25 ) 2523 23

(C ) 2324

24(D) 23

42. Let PR = 3 + ] - 2k and s o = i - 3]- 4k determin e diagona ls of a para llelogram PQRS and PT = i + 2]+ 3k

be another vecto r. Then the volume of the paralle le piped determ ined by the vectors PT , PQ and P5 is

11AJfct5ffi = 3 i + }- 2k mrr s o = i - 3}- 4k ~ ' W f R R ~P O R S- $ ~ ~ ~ ' g mFiT = i + 2} + 3k

(A) 5 (B) 20 (C) 10 (D) 30

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28/39

Sol. (C)- -R = PO+PS

- - -O =P O - PS- - .

PS =PR - SO

2

- -PO = PR + SO2v = I f O PS PT] I

v = I f R+SO , PR - SO , PT] I

V = ~ I f R, SO , Pr ] l

3 1 - 2

_!_1 - 3 - 42 1 2 3

12 (- 3 - 7 - 10) = 10

( MATHEMATICS )

R

143. Let comp lex numbers a. and - lies on circle s (x - Xo)2 + (y - y0)2 = r2 and (x - Xo)2 + (y - y0 )2 = 4r2,a.

Sol.

respec tively. If Zo = x0 + iy0 satis fie s t he equa tion 21 ol2= r2+ 2, the n la. l =

l=IT9Tfcn ~ ~~ ~ a. mn : w+ro: Cfti (x - x0)2 + (y - y0)2= r2Tim (x - x0 ) 2 + (y - y0 ) 2= 4r2 ~ f t ~e 1~ 1 I a. 12- Zoa - a. Zo + 1 ol2 = r2

1 - Zoa - Zoa. + 1 ol21 .l2= 4r21a.l2=> (la.21- 1) + 1 ol2 (1 - l a.l2) = r2 (1 - 4a. 2)

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29/39

(la l2 - 1) - ~2

) = r2(1 - 4 1a l2 )

lal 2 - 1 = - 2 + 8 la f

1la l = J7

( MATHEMATICS )

44. Fo r a> b > c > 0, the d istance between (1, 1) a nd the po int of intersect io n of the lines ax + by + c = 0 a nd bx

+ ay + c = 0 is less t han 2J2 . The n

a > b > c > 0 ~ ~ (1, 1 ) ~T WT3it ax+ by+c =O q- bx + ay + c = O ~ ~ ~ ~ ~ c#t ~ 2J2~ C j ) l 1 l ( ' f 6 1

(A) a + b - c > 0 (B) a - b + c < 0 (C ) a - b + c > 0 (D) a+ b - c < 0

So l.(A)(a - b )x + (b - a )y = 0~ X = y

(-c -c)~ Po int of intersec tion a + b , a + b

Now ( 1+ - c - )2

+ ( 1+ - c - )2

< 2J2a+b a + b

J 2 ( a+b + c ) < 2J 2~ a + b

~ a + b- c > O

45.X+ 2 y + 1 Z

Perpend icu lar a re d rawn from poin ts o n the line -2

- = --=-1 = 3 to the p lane x + y + z = 3 . The feet of

pe rpend icu lars lie o n the line

(A) ~ = y - 1 = z - 25 8 - 13

(B) ~ = y - 1 = z - 22 3 - 5

So l. (D)

. . X+2 y + 1 ZAny pom t on hne -- = - = - = A,

2 - 1 3Le t any two poin ts on th is line are

x y - 1 z - 2(C ) - = -= -3 - 7

A(- 2, - 1, 0 ), B (0 , - 2, 3) Put (A-= 0 , 1)Le t foot of perpe ndicu lar from A ( - 2 , - 1, 0) on pla ne is (a , 13, y)

~ a + 2 = 13+1 = 13-0 = iJ (say )1 1 1

Also , a + 13 + Y = 3~ iJ - 2 + i J - 1 + i J= 3 ~ i J = 2~ M(O, 1, 2)

( D) ~ = y - 1 = z -22 - 7 5

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30/39

(2 - 4 11 )

S im ilar ly foot of perpe ndicu la r from B (O, - 2 , 3) on p la ne is N 3' 3 '3

x - 0 y -1 z - 2So , equation o f MN is - 2- = _ 7 = - 5- .

3 3 3

( MATHEMATICS )

46 . Fo ur persons independe ntly so lve a certa in p roblem cor rectly w ith p robabi lities ~ ~ . Then th eprobabili ty th at the prob lem is so lved co rrec tly by at least o ne of them is

'CfR urFcm fCiei?IC"llll fclmt ~ 'f!1'R'm em~ m ~~ ~ ~ m ~ T-etcP ~ m t C'f6T 'f!1'R'm ~ ~ ~Cj)1=[ ~ Cfll=f ~ Cllfci-"C1'~ T iTcP ~ - F c P ~\ J I T~ c#t ~ m ~

235 21 3 253(A) 256 (B) 256 (C) 256 (D) 256

Sol. (A)P (prob lem so lved by a t leas t o ne ) = 1 - P (problem is not so lved by by a ll)

= 1 - P( A ) P( B ) P(c ) P (5 )

= 1 - ( ) ()( ) ( ) = 1 - 2 ~ ~= ~ ~ ~47. The area enc losed by the curves y = sinx + cosx a nd y = lcosx - s inx l ove r the interva l [ 0, ] is

~ [0, ] -qx qwl y = s inx + COSX ( f ~Ty = lcosx- s inx l &TXT m% ~ ~(A) 4 (J 2 - 1) (B) 2J2(J2 -1 ) (C) 2 (J 2 + 1) (D) 2J2(J2 +1)

Sol. (B)G iven y = s in x + cos x x E [0 , n/2]

dy .dx = COS X- Sin X

[c o s x - s in x x E [O n/4]

y = lcosx - s in x l = sinx - cosx xe[n /4 , n / 2]+

0 n/ 4 n/ 2

n /4 n /2

req uired area = Jl(s in x + cosx ) - (cosx - s in x ~d x+ J 12cosx ldx0 n / 4

n / 4 n /2

= J 12s in x ldx + J 12cosx Idx0 n /4 .J2

= 2 (- c o s x) ~1 4 + 2(s in x) ~ ~~

= 2 [- ~ +1+1- ~ ]y = h I os X + ~ )I

= 2 ( 2 - ~ )- - - - + - - - - - ~ ~ ~ ~ ~ - - - - ~ - - - -

n/ 4 n/ 2

= 2 (2 - h ) = 4 - 2 h = 2 h (h - 1) www.examrace.com


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( MATHEMATICS )

48 . A curve passes through t he po int ( 1, i). et the s lope of the curve at each poin t (x, y) be ~ + s e c(~) ,x > 0. Then th e equation of the curve is

S o l.

em 'f!BtCj)XU['g

(A) s i n(~) = logx + ~ (B) c o s e c(~) = logx + 2

(C) sec (2: ) = log x + 2 (D ) c o s(

2: ) = 1o g x+~

(A)Given s lope at (x, y) is

dy y- = - + sec(y /x)dx x

let y_ = t => y = xtX

dtt + x dx = t + sec (t)

J ot t dt = J dxs in t = .en x + cs in(y/x) = .en x + c

=>dy dt- = t + xdx dx

This curve passes through (1, rc/6)s in(n/6) = .en(1 ) + c => c = 1/2s in(y/x) = en x + 1/2

49 . Let f: [ , 1] R (the set of all rea l numbers ) be a pos itive , non-constan t and d iffere ntiab le func tion such t ha t

f'(x) < 2 f(x) and f( ) = 1. Then the va lue of J f( x)dx lies in t he interva l1/ 2

f'( x) < 2 ( x) T I ~Tf(~) = 1 t Cf6T J f( x)dx em l1 R ~ ~ "B~1/ 2

(A) (2e-1 ,2e ) (B) ( e - 1 , 2 e - 1) (e - 1 )(C ) - 2 ,e - 1 (

e - 1)(D) 0 ,-2

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( MATHEMATICS )

S o l. (D)

50 .

f' (x) - 2f(x) < 0

ddx (e- 2x f(x)) < 0

=> e- 2x f(x) is decreas ing => x > 1/2

1 1

e- 2x f(x) < 1/e => f (x)< e2n--1 ::::>0 < Jf (x )dx < J (e2

x-1)dx

1/ 2 1/ 2

The number of poin ts in (- ao, ao), fo r wh ich x2- x s inx - cosx = 0 , is

(- ao, ao) B ~ ~ Clft ~ . ~ ~ x2- x s inx - cosx = O,(A) 6 (B) 4 (C) 2

1e - 1

0< J f (x )dx < -

2-

11 2

(D) O

S o l. (C)x2 = x s inx + cos xf(x) = x2

g(x ) = x sin x + cos xg' (x) = sin x + x cos x - s in xg ' (X) = X COS X

3rt/2

- 3 rt/2 . ....... ........ ....... ....... . .. .

O nly two so lut ion.

SECTION- IIMultiple Correc t Answer Type

This sec tion conta ins 5 ques tions. Each ques tion has 4 cho ices (A), (B), (C) and (D) for its ans we r, out ofw hich ONE OR MORE is /are correct.

~ ~ - I I~ ~ fctcm;q :qcJ)R

~ ~ ~ B 5 'g I ~ ~ \3'ffl ~ ~ 4 fctcPc,q (A), (B), (C) mn (D) t fGA-B~ ~ m ~ 3l#)q; mft~ I

51. A recta ngu la r s heet of fixed pe rimeter with s ides hav ing their lengt hs in th e rat io 8 : 15 is co nverted into a nopen rec tangu lar box by fo lding afte r remov ing squares of equa l area from all four corners . If he tota l area ofremoved squares is 100 , the resu lt ing box has ma x imum vo lume. T he leng th s of the sides of th e rec tangu lar

shee t a re(A) 24 (B) 32 (C)45 (D) 60~ ~ ~ c#t 3ll'!l C11ctJI'


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Sol. (A ,C)

a

Let = 8x , b = 15 x:. Vo lum e = (8x - 2a ) ( 15x - 2a ) (a ) = 4a 3 - 46 a 2x + 120 ax 2

dV- = 6a 2 - 46ax + 60 x2da

(dV )- - oda a tx =5 -

5x =3 a nd -

6

So, at x = 3 gi ves max ima

S5 . . .

o, at x = 6 g1ves m 1mma.

dVda = 0 w hen a = 5 gi ven ( :. 4a 2 = 100 g iven fo r maximum vo lume )

a t a = 5

dVby - = 0 => 6x2 - 23x + 15 = 0

da

x =3 o r 5/6So by x = 3 (fo r max volum e )

8x = 24 , 15x = 45 Ans. (A , C)

4n k(k+1)

52. Let S n = ~ ) - 1)_ 2 _ k2 . Then S0

can tak e va lue (s )2

(A) 1056 (8 ) 1088 (C ) 1120

4n k(k+1)

11AT fcp sn= 2) -1)_ 2_ k2 , Cf6l sn l1 A ~ 'f!Clml ~ :2

(A) 1056 (8 ) 1088 (C ) 1120

(D) 1332

(D) 1332

( MATHEMATICS )

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S o l. (A D)

4n k (k+1 )sn = z ) - 1)_ 2_ k2

k= 1

= 12 - 22 + 32 + 4 2 - 52 - 62 + 72 + 8 2 + ......= 32 - 1) + (4 2- 22) + (72- 52) + (82- 62) ... .=2 [4 + 6 + 12 +1 4 + 20+ 22 + ... .. .

2n te rms

=2[(4 + 12 + 20 .... .. + (6 + 14 + 22 . .... .. . ]n terms n terms

= 2[% (4 x 2+ (n -1 )8 )%(2 x 6+ (n+ 1)8 )]

=2[n (4 + 4n - 4 ) + n(6 + 4n - 4 )]= (4n 2 + (4 n + 2 )n)= (8n 2 + 2 n)=4 n(4n + 1)(A) 1056 = 32 x 33 n = 8(B) 1088 = 32 x 34 n = 8

(C) 1120 = 32 x 35 n = 8(D) 1332 = 36 x 37 n = 8

53 . A line 1 pass ing throug h the or igin is perpe ndicu lar to the lines

1 : (3 + t ) i + (- 1 + 2 t) }+ (4 + 2 t) k - 00 < t < 00

12 : (3 + 2t ) i + (3 + 2t ) } + (2 + s )k - 00 < s < 00

( MATHEMATICS )

Then , the coordi nate (s ) o f th e po int(s ) on l2 a t a d ista nce of J17 from th e po int of ntersec tion of land /1 is (are )

S o l.

(775)

(A) 3'3 '3 (B) (- 1, ,- 1, 0)~ ~ n I , \1ll ~ ~ ~ t ~ n 3 i i

(C) ( 1, 1, 1 )

1 : (3 + t ) i + (- 1 + 2 t) }+ (4 + 2 t) k - 00 < t < 00

12 : (3 + 2t ) i + (3 + 2t ) } + (2 + s )k - 00 < s < 00

(778)

(D) 9 '9 '9

1R ~ ~ 1 CfOT,12 1R f t~ m~ ( ~ 3 i i )~ ~ C J ) ,\1ll WT3ii 1 mn 1 ~ ~ ~ ~ J17 Clft~ 1R 'g ('g),f.i.:;=r~ ('g) :

(775)(A) 3'3 '3(B,D)Let equa tion of line .e is

(B) (- 1, ,- 1, 0)

x - 0 y - 0 z - 0 . -- = -- = -- = k a b cThis line is perpend icu lar to g ive n line .e, and .e2 Hence a + 2b + 2c = 0

2a + 2b + c = 0

(C) ( 1, 1, 1 ) (778)(D) 9 '9 '9

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He nce equa tion o f isX Y Z

- 2 = 3 = - 2 = k 1, k2-1, -1,

fo r 1 fo r 2Now A(- 2 k1, 3k 1 - 2k 1) B(- 2k 2, 3k2, - 2k 2)Po int A sa t isfied 1

- 2 k1 + 3 k1j - 2k 1k

= (3 + t )i

+ (- 1 + 2t ) 1 + (4 + 2t )k3 + t = - 2 k1 ...... . 1 )

- 1 + 2t = 3k 1 ...... . 2 )4 + 2t = - 2k 1 ...... . 3 )

(2 ) & (3) - 5 = 5k 1=> k1 = - 1 =>A (2 , - 3, 2 )Let a ny po in t on 2 (3 + 2S , 3 + 2S , 2 + S )

Given ~ (1 + 2S )2 + (6 + 2S )2 + (S )2 = J179S 2 + 28S + 37 = 179S 2 + 28S + 20 = 09S 2 + 18S + 1 OS + 2 0 = 09S (S + 2 ) + 10 (S + 2 ) = 0s = - 2 , - 10 /9

He nce (- 1, - 1, 0 ) , (7 /9 , 7/9 , 8 /9 )Ans. (B) & D)

54. Let f(x) = x s in n x, x > 0. The n fo r all na tura l num ber s n, f' (x) van is he s at

( MATHEMATICS )

(A) a un ique po int in t he interva l ( n.n + ) (B) a u nique poin t in the interva l ( n ~ .n+ 1)

(C ) a un iqu e po int in th e interva l (n, n +1 ) (D) tw o poin ts in the inte rva l (n , n +1)

l1AT fcl5f(X) = Xs in 1tX, X > 0, Ci6l ~ t : r l ~ ) n ~ f'(X) "QX ~ ~ tmn ~ :

( C )~ (n , n +1)-B ~ + n ? r~ ~ -qx ( D ) ~ C 1(n , n +1 ) - B e n ~ ~-qx

Sol. (B,C)f(x) = x sin nx , x > 0f ' (x) = sin n x + n x cos n x = 0

tan n x = - n x

7/2

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( MATHEMATICS )

55 . Fo r 3x3 matr ices M a nd N , which o f th e fo llowing s ta temen t(s ) is (are ) NOT correct ?

Sol.

(A) NT M N is symmetric or ske w s ymmetr ic , according as M is symmetr ic or ske w symmet ric

(B) M N - N M is s ymmetr ic for a ll symmetric matrices M a nd N(C ) M N is s ymetr ic fo r a ll s ymmetr ic matr ices M a nd N(D) (ad j M) (ad j N) = ad j(MN) fo r a ll invertib le mat rices M a nd N

3x3 Mr ~ T

N~

f.i.:;=rBcPf

";:f> 1 C P ~~

-fflt 'g ('g) 7(A) M ~


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S o l. Po int of intersec tion of ta nge nts at P a nd Q is R (2 sece , 0)

Area of P Q R= ~ 2.J3 sine (2 sec e - 2 cos e)

sin3

e [ 1 1 J=> ~ = 2.J3 cose ; whe re cos e e 4 '2

y

x=h=2cose

Nowd~ = 2 .J3[cose .3si n2 ecose- s in3 e(-s in e)] >

0de cos 2 e

As e i. T => cos -!- ~ T

8 = a = ,-;:; (1 -1 / 4f '

2= r;:; 3.J3 = 36

m '2 2v3 112 4v 3 8 8Occues a t cos = 1 2

_ _ (1- 1/ 16f ' 2 _ 15 [15 2.J3 .15 J3.JS~ m i 1- ~1 - 2.J3 114 - 8 .J3 4.4 .4 = 16

Occues cos = 1/445

=> ~ 1= 8 J5

( MATHEMATICS )

57 . The coeff icients of th ree consecutive terms of (1 + xr 5 are in the ra tio 5: 10 : 14 . Then n =

(1 + xr 5 ~ ~ Wl'IFRf ~ ~ TTICfl5 : 10 : 14 ~ C lB t C'f6T n =

S o l. ns c . nsc . n+SC = 5 . 10 . 14r-1 r r+ 1

n+S c 10---= ___,r_ = -n+5c 5

r - 1

(n + 5) - r + 1 =2

r

n + 6 = 3r

123r= - (r+ 1)5r = 4n + 6 = 12

&

&(n+5)-( r + 1) + 1 = !_

r + 1 5

n + 6 12-- - -r + 1 5

n =6

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( MATHEMATICS )

58 . Cons ider the set o f e ight vectors V = ~ ~+ b}+ ck:a ,b,cE {- 11}}.Three non-cop lana r vectors ca n be chose n

from V in 2P ways. The n p is

; m o ~e m ~V = ~ ~+ b}+ ck:a ,b,CE {- 1,1}. ~ I ~ cTt"9 31'


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( MATHEMATICS )

59 . Of t he three independent events E 1, E2 and E 3 , the pro bability t ha t on ly E 1 occurs is a, n ly ~ occu rs is 13andon ly E3 occurs is y . Let t he probability p tha t none of eve nts E 1 , E2 or E3 occurs sat isfy the equat ions (a - 213)p = al3 and (13- 3y) p = 213y . Allthe g iven pro babilities are assumed to lie in the inter val (0, 1 .

Probab ility of occu rrence of E1Then Probabili ty of occu rre nce of E 3 =

"fft"9~

tTc-=rr311E1, E2 Tim E3 B~ c r c 1

E1~

c#t mfuclmr at~ c r c 1

E2~

c#tmfucpm 13~

TimE3 ~ c#tmftrcimTy ~I I T ~fclJ tTc-=rr311E1, E2 =>=>=>

Alterna tive

=>

=>

n2 + n - 4k = 245 0n2 + n - 245 0 = 4k(n + 50)(n - 49 ) = 4kn > 49

To sat isfy t his equation n s hou ld be of the form of (4p + 1) o r (4p + 2) tak ing n = 504k = 1 00

k =25k - 20 = 5Now if we take n = 53k = 103n < kso no t poss ibleHence n ;:::53 willnot be poss ible .

jee advanced paper 1 solutions 2013

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