PAPER-2 (B.ARCH) of JEE(MAIN) Code-X / 2. (1) 6i (2) 3i (3) 2i (4) 6. = 7, xy = 12 y 2 = 7

Size: px

Start display at page:

Download "PAPER-2 (B.ARCH) of JEE(MAIN) Code-X / 2. (1) 6i (2) 3i (3) 2i (4) 6. = 7, xy = 12 y 2 = 7"

Victoria Powell
5 years ago
Views:

1 PAPER- (B.ARCH) of JEE(MAIN) Code-X 60. If ( +i) 7 + 4i, then a value of Ans. (1) Sol. ;fn ( +i) 7 + 4i gs] rks is : dk,d eku gs % (1) 6i () i () i (4) 6 ( + i) 7 + 4i + i() i 1 7, or 9 ± Also blf, / / ( + i) ( i) i ± 6i 61. The equation of the circle, which is the mirror image of the circle, + 0, in the line, Ans. (1) is : ml o`ùk dk lehdj.k] tks o`ùk + 0 dk js[kk esa niz.k izfrfcac gs] gs : (1) () () (4) Sol. Centre of circle (1, 0), radius 1 Image of (1, 0) w.r.t. Line + 0 is 1 0 (1 0 ), Now circle with centre (, ) and radius 1 is ( ) + ( ) Hindi. o`ùk dk dsunz (1, 0), fkt;k 1 (1, 0) dk js[kk + 0 ds lkis{k izfrfcec 1 0 (1 0 ), dsunz (, ) rfkk fkt;k 1 okk o`ùk ( ) + ( ) log(sin7 cos7) lim equals : 0 sin log(sin7 cos7) lim cjkcj gs% 0 sin (1) 1 log 7 () 7 () 14 (4) 1 CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 16

2 PAPER- (B.ARCH) of JEE(MAIN) Code-X Ans. () Sol. log(sin7 cos7) 1 log(1 sin14) lim lim 0 sin 0 sin 1 log(1 sin14) sin14 14 lim sin14 14 sin 6. If the line, m, bisects the area of the region {(, ) : 0, }, then m equals ;fn js[kk m, {ksk {(, ) : 0, } dks nks cjkcj Hkkksa esa fohkkftr djrh gs] rks m dk eku gs % (1) 9 16 () 9 8 () 1 (4) 1 6 Ans. (4) D C m Sol. B A ( 4 1) As per the question given, Area of OAB Area of region OBCD Area of OACD Area of OAB or (Area of region OAB) (Area of OACD) 1 m 9 m 4 m / or m 1 6 (1 4 )d 9m CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 17

3 PAPER- (B.ARCH) of JEE(MAIN) Code-X D C m Hindi. B A ( 4 1) iz'ukuqlkj, OAB dk {kskq OBCD ifjlj dk {kskq OACD ifjlj dk {kskq OAB dk {kskq or (OAB ifjlj dk {kskq) (OACD ifjlj dk {kskq) 1 m 9 m 4 m / m 1 6 (1 4 )d ;k 8 8 9m The product of the perpendiculars drawn from the foci of the ellipse, 5 to it at the point,, is : upon the tangent 5 nh?kzo`ùk dh ukfhk;ksa ls blds fcanq 5, ij [khaph bz Li'kZ js[kk ij Mks ;s acks dk Ans. () q.kuq gs : (1) 1 () 9 () (4) 18 Sol. Propert : Product of the perpendicular drawn from focii of ellipse upon an tangent to it is square of semi minor ais. ans 9 Hindi. q.k/kez : ukfhk;ksa ls fdlh Li'kZ js[kk ij Mks ;s Ecksa dh EckbZ;ksa dk q.kuq v/kz?kqv{k ds oz ds cjkcj gksrk gsa CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 18

4 PAPER- (B.ARCH) of JEE(MAIN) Code-X 65. An observer standing at a point P on the top of a hill near the sea-shore notices that the angle of Ans. (1) depression of a ship moving towards the hill in a straight line at a constant speed is 0. After 45 minutes, this angle becomes 45. If T (in minutes) is the total time taken b the ship to move to a point in the sea where the angle of depression from P of the ship is 60, then T is equal to :,d izs"kd leqnz ds ikl flfkr,d igkm+h dh pksvh P ij [km+k gksdj,d tgkt dk voueu dks.k 0 ikrk gsa tgkt,d leku pk ls,d lj js[kk esa igkm+h dh vksj vk jgk gsa 45 feuv ds ckn ;g dks.k 45 gks tkrk gsa ;fn leqnz esa og,d fcanq ftldk fcanq P ls tgkt dk voueu dks.k 60 gs] rd igq pus dk dq le; (feuvksa esa) T gs] rks T cjkcj gs% (1) () 45 1 () 45 1 A (4) 45 1 h Sol E D C B Let height of hill is h AB h BD h, EB h, CB ED 1 h Now speed v Now DC vt, h ( 1)h h T1 45 T 1 45 T CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 19

5 PAPER- (B.ARCH) of JEE(MAIN) Code-X A Hindi. E D C B ekuk igkmh dh ÅpkbZ h AB BD h, EB h, CB ED 1 h vr% os v vc DC vt, 1 h h ( 1)h h T 45 h T 1 45 T Two numbers are selected at random (without replacement) from the first si positive integers. If X denotes the smaller of the two numbers, then the epectation of X, is : izfke N% /kuiw.kkzdksa esa ls nks la[;k, (fcuk izfrlfkkiuk ds) ;kn`pn;k pquh bza ;fn X nksuksa esa ls NksVh la[;k dks iznf'kzr djrk gs] rks X dh izr;k'kk (Epectation) gs : (1) 5 () 14 () 1 (4) 7 Ans. (4) Sol Possible pairs are (lehko ;qxe) (1, ), (1, ), (1, 4), (1, 5), (1, 6), (, ), (, 4), (, 5), (, 6), (, 4), (, 5), (, 6), (4, 5), (4, 6), (5, 6) Hence epectation of X will be (vr% X ) (1 5) ( 4) ( ) (4 ) (5 1) CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 0

6 PAPER- (B.ARCH) of JEE(MAIN) Code-X 67. The perpendicular distance from the point (, 1, 1) on the plane passing through the point (1,, ) and containing the line, r î + ĵ + ( î + ĵ + 4 ˆk ), is : fcanq (, 1, 1) ls,d leer] tks fcanq (1,, ) ls gksdj tkrk gs rfkk ftl ij js[kk r î + ĵ + ( î + ĵ + 4 ˆk ) flfkr gs] ij Mks, ac dh ackbz gs% (1) 11 () 1 11 () 4 41 (4) 0 Ans. (4) R(1,,) r (i ˆ ˆj) (i ˆ ˆj 4k) ˆ Sol. Q( 1,0, 4) P(1,1,0) Let equation of plane through P, Q & R be a( 1) + b( ) + c(z ) 0, where ekuk P, Q o R ls tkus okk ler a( 1) + b( ) + c(z ) 0, tgk ˆi ˆj kˆ ai ˆ bj ˆ ckˆ 0 1 ˆi 6j ˆ kˆ 7 (a, b, c) (1, 6, ) equation of plane will be (ler dk lehdj.k) + 6 z 7 0 distance of (, 1, 1) from the plane, d (, 1, 1) dh ler ls nwjh, d The integral ( ) 1 d is equal to (1) 1 cot 1 + C () 1 1 tan 1 ( 1) + C () tan 1 + C (4) ( 1) (where C is a constant of integration) cot C CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 1

7 PAPER- (B.ARCH) of JEE(MAIN) Code-X lekd ( ) 1 d cjkcj gs% (1) 1 cot 1 + C () 1 1 tan 1 ( 1) + C () tan 1 + C (4) ( 1) (tgk C,d lekdu vpj gs) Ans. () Sol ( )d ( ) 1 Let ekuk + 1 t d t dt cot C (t 1)t dt [(t 1) t ].t (t 1)dt 4 [(t t 1) (t t 1) (t t 1)dt [(t t 1)(t t 1) (t 1)dt [(t t 1)(t t 1) dt dt t t 1 t t 1 4t tan t tan t tan t 4 1 t 1 ( 1) tan tan 1 t 1 ( 1) I cot c 69. The value of 1 1 is cos85º sin 55º 1 1 dk eku gs& cos85º sin 55º Ans. () (1) () () 4 (4) CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page #

8 PAPER- (B.ARCH) of JEE(MAIN) Code-X Sol. 1 1 cos85 sin cos(70 15) sin(70 15) 1 1 sin15 ( cos15 ) cos ( / 4) cos15 sin15 (cos(0 15 )) sin15cos15 sin0 70. Let a 1, a, a, a 4, a 5 be a G.P. of positive real numbers such that the A.M. of a and a 4 is 117 and Ans. (1) Sol. the G.M. of a and a 4 is 108. Then the A.M. of a 1 and a 5 is ;fn a 1, a, a, a 4 rfkk a 5,d /kukred oklrfod la[;kvksa dh,slh q.kksùkj Js<h gs fd a rfkk a 4 dk lekurj ek/; 117 gs rfkk a rfkk a 4 dk q.kksùkj ek/; (G.M.) 108 gs] rks a 1 rfkk a 5 dk lekurj ek/; gs& (1) () 108 () 117 (4) a 1, a, a, a 4, a 5 are in G.P. (a 1, a, a, a 4, a 5 q.kksùkj Js.kh es gs) a + a (1) a a 4 (108)...() Let ekuk a a 1 r, a a 1 r, a 4 a 1 r, a 5 a 1 r 4, a 1 r (1 + r ) 4...() a 1 r 4 (108) a 1 r (4) ()/(4) (1 r ) r r 1r or ;k (r )(r ) 0 r or a 1 r 108 a or ;k a 1 48 a 5 a 1 r AM of a 1 & a 5 dk l-ek CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page #

9 PAPER- (B.ARCH) of JEE(MAIN) Code-X 71. The integral 5 / 4 d is equal to 1 tan / 4 lekd 5 / 4 d cjkcj gs& 1 tan / 4 (1) 18 () () 1 (4) 6 Ans. () 5 4 I d tan 5 4 d I 1 1 tan Adding above two integrals 5 4 I d 6 4 or ;k I 1 7. There vector a, b 4a.b b.c c.a is equal to rhu lfn'k a, b eku gs& d cot and c are such that a 1 rfkk c,sls gas fd a 1 nksuksa lekduksa dk ;ks djus ij, b, b, c 4, c 4 and a b c 0. Then the value of rfkk a b c 0 gs] rks 4a.b b.c c.a dk (1) 7 () 68 () 6 (4) 4 Ans. () Sol. c a b a. b 4 + b. c+ c.a 4 a. b + b. a b + a.( a b 4 a. b a.b b a a. b a.b (4) (1) 15 a.b a b c a b c CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 4

10 or;k 1 4 a.b 4 a.b 11 a.b a 15.b PAPER- (B.ARCH) of JEE(MAIN) Code-X For all real numbers, and z, the determinant z z 1 z z z 1 1 z 1 is equal to (1) ( z)(z ) () zero () ( )( z)(z ) (4) ( z)( z) z, rfkk z ds lhkh oklrfod la[;kvksa ds f, lkjf.kd z 1 z z z 1 1 z 1 cjkcj gs& Ans. () Sol. (1) ( z)(z ) () 'kwu; () ( )( z)(z ) (4) ( z)( z) z z 1 z z z 1 1 z 1 ( z) R R R1 z 1 z( z) 1 ( z) z 1 z 1 R R R1 1 ( z) ( z) 1 1 ( ) ( z) (z ) z 1 1 C C C ( z) z 0 R R R If 1 and are the two values of such that the roots and of the quadratic equation, ( 4 1 ) satisf 0, then is equal to 5 1 ;fn 1 rfkk, ds,sls nks eku gs fd f}?kkrh lehdj.k ( ) ds ew,,sls gs fd 4 0 gs] rks 1 cjkcj gs& 5 1 (1) 488 () 56 () 51 (4) 504 Ans. (1) CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 5

11 PAPER- (B.ARCH) of JEE(MAIN) Code-X Sol. + (1 ) / or ;k or Now, If the sum of the first 15 terms of the series is 15k, then k is equal to Ans. () ;fn Js.kh ds izfke 15 inksa dk ;ks 15k gs] rks k cjkcj gs& (1) 16 () 1 () 81 (4) 119 S n T n S n ,, + T n T n or;k T n r term n 1 T n + [8+(n )] (n 1)(n ) + + n (n ) or ;k T n n n + S n T n S n n(n 1)(n 1) 6 n(n 1) 4 + n CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 6

12 PAPER- (B.ARCH) of JEE(MAIN) Code-X S n S k k Water is running into an underground right circular conical reservoir, which is 10m deep and radius of its base is 5m. If the rate of change in the volume of water in reservoir is m /min., then the rate (in m/min) at which water rises in it, when the water level is 4m, is,d Hkwfer Ec o`ùkh; 'kadokdkj Vadh] tks 10 ehvj gjh rfkk ftlds vk/kkj dh fkt;k 5 ehvj gs] esa ikuh vk jgk gsa ;fn Vadh ds ikuh ds vk;ru esa ifjorzu dh nj eh /fe- gs] rks Vadh ds Lrj esa ifjorzu dh nj (eh-/fe- esa) tcfd ikuh dk Lrj 4 ehvj gs& (1) () 8 () 1 8 (4) 1 4 Ans. () 5 Q r 10 v 1 r h 1 h tan du dt r dh dt Now vc r 5 r 4 10 dh () dt dh dt A bag contains three coins, one of which has head on both sides, another is a biased coin that shows up heads 90% of the time and the third one is an unbiased coin. A coin is taken out from the bag at random and tossed. If it shows up a head, then the probabilit that it is the unbiased coin, is,d FkSs esa rhu fldds gs] ftuesa ls,d ds nksuksa vksj fpùk gs] nwljk flddk vfhkur (biased) gs ftl ij fpùk 90% ckj izdv gksrk gs vksj rhljk vufhkur flddk gsa FkSs esa ls,d flddk ;kn`pn;k fudk dj mnkk ;ka ;fn fldds ij fpùk izdv gks] rks mlds vufhkur flddk gksus dh izkf;drk gs& (1) 8 () 5 1 () 5 4 (4) 1 CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 7

13 PAPER- (B.ARCH) of JEE(MAIN) Code-X Ans. () Required Prob. vhkh"v izkf;drk a 78. If the function f : R R, defined b f() is differentiable, then the value of a b f ( ) + f () is equal to a ;fn Qu f : R R, tks f() }kjk ifjhkkf"kr gs] voduh; gs] rks a b f ( ) + f () dk eku gs& Ans. () (1) 0 () () 4 (4) 15 a ; f() a b ; f() is differentiable function f() must be continuous f (+h) f ( h) f () 4a b + a or a b (1) Also, f'(+h) f' ( h) a() b a or a b...() Solving equation (1) & () we get a 4 & b 9 4 Now f' ( ) a & f' () a () b 6a b f' ( ) + f' () 7a b a ; Hindi. f() a b ; f(),d voduh; Qu gs f() lrr gksuk pkfg,s f (+h) f ( h) f () 4a b + a ;k a b (1) rfkk, f'(+h) f' ( h) a() b a CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 8

14 ;k a b...() leh- (1) rfkk () ls PAPER- (B.ARCH) of JEE(MAIN) Code-X vc rfkk a 4 & b 9 4 f' ( ) a f' () a () b 6a b f' ( ) + f' () 7a b Which one of the following statements is a tautolog? fueu esa ls dksulk dfku,d iqu:fdr (tautolog) gs? (1) p (p q) () (p q) q () p (p q) (4) p (q p) Ans. () p q p q q p pvq p (p q) (pvq) q T T T T T T T T F F T T F F F T T F T T T F F T T F T T Pv (p q) pv(q q) T T T T T F T T p (p q) is tautolog p (p q) iwuq:fdr gsa 80. The sum of the abscissae of the points where the curves, k + (5k + ) + 6 k + 5, (k R), touch the -ais, is equal to mu fcunqvksa ds -funsz'kkadksa dk ;ks] tgk oø k + (5k + ) + 6 k + 5, (k R), -v{k dks Li'kZ djrh gsa] gs& (1) 4 () 19 () 10 (4) 5 Ans. () Sol. k + (5k + ) + (6k + 5) curve touches -ais D 0 (5k + ) 4k(6k + 5) or 5k + 0k + 9 4k + 0k or k + 10k k 1 and 9 CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 9

15 PAPER- (B.ARCH) of JEE(MAIN) Code-X Curve can be, 1 (+1) 1 or curve can be ( + 7) 7/ Sum of abscissae Hindi. () k + (5k + ) + (6k + 5) oø -v{k dks Li'kZ djrk gsa D 0 (5k + ) 4k(6k + 5) ;k 5k + 0k + 9 4k + 0k ;k k + 10k k 1 rfkk 9 oø gks ldrk gs, 1 (+1) 1 ;k oø gks ldrk gs ( + 7) 7/ Hkqtks dk ;ks CORPORATE OFFICE : CG TOWER, A-46 & 5, IPIA, NEAR CITY MALL, JHALAWAR ROAD, KOTA (RAJ.) REG. OFFICE : J-, JAWAHAR NAGAR, MAIN ROAD, KOTA (RAJ.)-4005 PH. NO.: FAX NO. : PH.NO. : , TO KNOW MORE : SMS RESO AT WEBSITE : CONTACT@RESONANCE.AC.IN CIN : U800RJ007PLC0409 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 017 SOLUTION PORTAL Page # 0

Memory Based Questions & Solutions

PAPER- (B.E./B. TECH.) JEE (Main) 09 COMPUTER BASED TEST (CBT) Memory Based Questions & Solutions Date: 09 April, 09 (SHIFT-) TIME : (9.0 a.m. to.0 p.m) Duration: Hours Ma. Marks: 60 SUBJECT : MATHEMATICS