Maths Paper 1 Solution - Motion IIT JEE
PAPER WITH SOLUTIONMATHEMATICS PAPER - 1
MATHS [ JEE ADVANCED - 2019 ] PAPER - 1 1.SECTION -1 (Maximum Marks : 12)This section contains FOUR (04) questions.Each question has FOUR options ONLY ONE of these four options is correct answer.For each question, choose the option corresponding to the correct answer.Answer to each question will be evaluated according to the following marking scheme.Full Marks: 3 If ONLY the correct option is chosen.Zero Marks:0 If none of the options is choosen (i.e. the question is unanswered)Negative marks :–1 In all other casesA line y mx 1 intersects the circle (x – 3)2 (y 2)2 25 at the points P and Q. If themidpoint of the line segment PQ has x - coordinateSol.is correct ?(1) – 3 m – 14(2) 6 m 8 3, then which one of the following options5(3) 4 m 6(4) 2 m 4mAB. mcm – 13 1 5m 2 m . –13 3 5 15 3m m 18 – 1 15m – 3m2 –18 0m2 – 5m 6 0m 2, m 3 2 m 42. sin4 1 sin2 I M–1Let M 2cos4 1 cos where ( ) and ( ) are real numbers, and I is the 2 2 identity matrix. If * is the minimum of set { ( ): [0,2 )} and * is the minimum of the set { ( ): [0,2 )}then the value of * * is(1)Sol. 2916(2) 37161 sin4 1 sin2 I M–1M 2cos4 1 cos M I M–1(3) 1716(4) 3116
M2 M I sin4 sin4 1 sin2 sin4 1 sin2 1 sin2 1 0 24242cos 1 cos cos cos4 0 1 1 cos 1 cos sin8 – 1 – sin2 – cos2 – cos2 sin2 sin4 sin8 – 2 – cos2 sin2 sin4 .(1)sin2 cos2 sin4 cos4 cos6 (1 cos2 ) sin4 1 cos2 cos4 1 cos2 1 cos 2 sin4 cos4 1 –1sin2 2 211 22for equation (1)sin8 – 2 – cos2 sin2 – sin4 sin2 – 2 – sin2 cos2 – sin4 (sin4 cos4 ) – 2 – sin2 cos2 – sin4 cos4 min 1 –2– –2 –114sin2 2 –sin2 416116 sin2 4 sin2 2 4 14 712– sin2 2 416 7 1 79 28 937 .9 4 164 161616 *min *min 3.4 37 8 29 1616Let S be the set of all complex numbers z satsfying z – 2 i 5 . If the complex number z0 11 is such that z 1 is the maximum of the set z 1 : z S , then the principal argument of 0 4 z0 z0z0 z0 2i is(1)Sol. 2(2)3 44 z – 2 i 15 for max of z0 1 min z0 – 1 (3) 4(4) 2
1 –1 1Now use parametric coordinate 135 mCA tan 1 1 , 1 5 P(z0) 2 5. 2 2 55 z0 2 2 , 1 2 4 z0 z0 arg z z 2i 0 0 10 arg i 10 arg(–i) 4.5 2 5 i 2 1 arg i 2The area of region {(x,y) : xy 8, 1 y x2} is(1) 16loge 2 –Sol. 4 2 2 arg 2i 2 1 1xy 8143(2) 8 loge2 –& 1 y x273(3) 8loge2 – 43(4) 16 loge2 – 6
2A 8 x2 1 dx 1 8 x 1 dx22x38 8ln x 2 1 6A 3 1 8 1 A 3 3 8(ln 8 – ln2) – 7 A 7– 7 16 ln 23A 16 ln2 – 143SECTION -2 (Maximum Marks : 12)This section contains EIGHT (08) questions.Each question has FOUR options ONE OR MORE THAN ONE of these four option(s) is (are)correct answer(s).Answer to each question will be evaluated according to the following marking scheme.Full marks: 4 If only (all) the correct option(s) is (are) chosen;Partial Marks: 3 If all the four options are correct but ONLY three options are chosenand both of which are correctPartial Marks: 1 If two or more options are correct but ONLY one option is chosenand it is a correct option.Zero Marks: 0 If two or more options is chosen (i.e. the question is unanswered)Negative Marks: –1 in all other casesFor example, in a question, if (A),(B) and (D) are the ONLY three options corresponding tocorrect answer, thenchoosing ONLY (A), (B) and (D) will get 4 markschoosing ONLY (A) and (B) will get 2 marks
oosing1.LetONLY (A) and (D) will get 2 marksONLY (B) and (D) will get 2 marksONLY (A) will get 1 markONLY (B) will get 1 markONLY (D) will get 1 markno option (i.e., the question is unanswered) will get 0 marks; andany other combination of options will get –1 markdenotes a curve y y(x) which is in the first quadrant and let the point (1,0) lie on it. Letthe tangent toat a point P intersect the y - axis at YP. If PYYP has length 1 for each point P on, then Which of the following options is/are correct ?(1) xy' –21 x(3) xy ' 1 xSol.2 0 1 1 x2 loge(4) y x 03,4Equation of Tangent at PY – y dy X x dxFor YP (X 0)YP y – xdydxdistance YpP 12dy x y y xdx 2 1 1 x2(2) y loge x 1 1 x2 1 x2
2 dy x2 1 1 dx 2 dy dx 1 1x2dy1 x2 dxx dy option 1 and 31 x2dxxx sin y cos cos d sin y 1 sin2 d sin y cos ec sin d y ln cos ec cot cos C 1 1 x2 ln 1 x2 Cy x x P as (1,0) c 0 1 1 x2y ln x 2. 1 x2 option (2), (4) Define the collections {E1,E2, E3.} of ellipse and {R1, R2, R3 .} of rectangles as follows :x2 y2 1 ;94R1 : rectangle of largest area, with sides parallel to the axes, inscribed in E1;E1 :En : ellipsex2 y2 1 of largest area inscribed in Rn – 1, n 1;a2n bn2Rn : rectangle of largest area, with sides parallel to the axes, inscribed in En , n 1Then which of the following options is/are correct?(1)The eccentricities of E18 and E19 are NOT equal(2) The distance of a focus from the centre in E9 is532
N(3) area of R 24 , for each positive integer Nnn 1(4) The length of latus rectum of E9 is2.16(3),(4)x2 y2 194E1 l 6 cos b 4sin Area 12 sin2 Amax 12sin2 12 2 4E2 : a 32(i) e2 1 ; b 22; a 3 ; r 12; b 2; r b2eccentricities of all ellipse will be equala2 1 5(ii) for E9 ; e and a 3 3 2 distance of focus from centre812
ae 355 16 316(iii) sum of area of rectangles 12 6 3 .12A 1 1 24222(iv) L.R. 3.2ba1 2 2 16 12.161641 1 3 6162 0 1 a 1 1 1 1 2 3 8 6 2 Let M and adj M where a and b are real numbers. Which of the 3 b 1 5 3 1 following options is/are correct ?(1) det (adjM2) 81(3) (adj M)–1 adj M–1 –MSol.(2) a b 3 1 (4) if M 2 , then – 3 3 2,3,4 0 1 a 1 1 1 123 and adj M 8 6 2 M 3 b 1 5 3 1 2 3b ab 1 1 1 1 1 8 62 8 6 2 adj M b 6 5 3 1 3 1 2 – 3b – 1; ab – 1 1b – 6 – 5 ; a 2b 1 0 1 2 1 2 3 Now M 3 1 1 M 8 – 10 –2 a b 3 option (2) adj(M2) M2 2 M 4 16(3) (adjM)–1 adj(M–1) option(3)
adj(M–1) adj(M–1) 2adj(M–1) 2( M–1 M) 1 2 2 M – M 1 M 2 (4) 3 0 1 2 1 2 3 3 1 1 1 2 3 2 1 2 3 23 1 1 – 1 1 – 34.option (4)Let and be the roots of x2 –x – 1 0, with For all positive integer n, define n n,n 1 an b1 1 and bn an – 1 an 1 , n 2Then which of the following options is/are correct ?(1) a1 a2 a3 . an an 2 – 1 for all n 1(2) bn n n for all n 1 (3)bn 10n 889 1089n 1 (4)an 10nn 1Sol.1,2,4x2 – x – 1 0an n n (2) b1 11 51 5, 22bn an –1 an 1
n 1 n 1 n 1 n 1 bn n 1 1 2 n 1 1 2 n 1 2 n 1 2 5 5 5 5 n 1 n 1 2 2 5 n 5 n n n (i) a1 a2 a3 . an 2 . n 2 . n 1 n 1 1 n1 2 – – 1 0 2 – 1 1 1 2 1 n 2 1 n 2 n 2 2 n 2 n 2 n 2– an 2 – 1
b(3) nn 10 n 10 n n 10n 2 10 102 . 10 1 1 1010 10 10 10 2 100 10 10 212 100 10 189(4)an 10n1 10 10 1 10 10 8989 5.Let f : R R be given by x5 5x4 10x3 10x2 3x 1,x 0 2x x 1,0 x 1; 2 3821 x 3 x 4x 7x ,33 f(x) 10x 3 x 2 loge x 2 x 3 , Then which of the following options is /are correct ?Sol.(1) f is increasing on , 0 (2) f is onto(3) f' has a local maximum at x 12,3,4(4) f' is NOT differentiable at x 1
x5 5x 4 10x3 10x2 3x 1x 0 20 x 1 x x 1 2 38ƒ(x) x 4x2 7x 1 x 33 3 10x 3 (x 2) n(x 2) x 3 ƒ is onto Range R ( n (x–2) contains all real values) 5x 4 20x3 30x2 20x 3x 0 2x 10 x 1ƒ '(x) 2x2 8x 71 x 3 1 n(x 2) 1x 3 ƒ is not diff. 20x3 60x2 60x 20 2ƒ " x 4x 8 1 x 2 20(1 x)3 2ƒ " x 4x 8 1 x 2ƒ"ƒ'6.x 3x 00 x 11 x 3x 30-1-1x 00 x 11 x 3 0 Not always oneThere are three bags B1, B2 and B3. The bag B1 contains 5 red and 5 green balls, B2 contains 3 redand 5 green balls, and B3 contains 5 red and 3 green balls. Bags B1, B2 and B3 have probabilities334,andrespectively of being chosen. A bag is selected at random and a ball is chosen at10 1010random from the bag. Then which of the following options is/are correct ?(1) Probability that the chosen ball is green, given that the selected bag is B3, equals(2) Probability that the selected bag is B3 and the chosen ball is green equals31038
(3) Probability that the selected bag is B3, given that chosen ball is green, equals(4) Probability that the chosen ball is green equalsSol.7.51339801, 4P(B1 ) 334P(B2 ) P(B3 ) 1010101.P(G1 B3) 33 882.P(B3 G) 4133.P(B3 G) 124 39134.P(G) 3 53 54 3 12 15 1239 10 10 10 8 10 88080In a non-right angled triangle PQR, let p,q,r denote the lengths of the sides opposite to theangles at P,Q,R respectively. The median from R meets the side PQ at S, the perpendicularfrom P meets the side QR at E, and RS and PE intersect at O. If p 3 , q 1, and the radiusof the circumcircle of the PQR equals 1, then which of the following options is/are correct?(1) Length of RS 72(3) Radius of incircle PQR Sol.1,2,3(2) Length of OE 32 32 (4) Area of SOE 16312
sin LawPRQP 2RsinP sin 13 2sinP sin sinP sin 3 P 602 P 1201 302 150000 P 120 , 30 , R 30(1)(2)RS 122 3 2 2(1)2 1 Eq. of RS : (y – 0) 7Ans 121 041x 3 y –x 333 3 34 3 1 Hence coordinate of O : 2 , 6 OE (3)r S1611. 3.2233 1 1 2(2 3)23(2 3)2(4)1 2320 13211634114
8.10 1 31412116114 3 2 2 4 24 12 3 1 1 1 1 1 1 4 6 4 4 12 3 23.4 24 48Let L1 and L2 denote the lines ˆ R andˆ 2k),r ˆi ( ˆi 2j ˆr µ(2iˆ ˆj 2k),µ Rrespectively, If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, thenwhich of the following options describe(s) L3?Sol. 2ˆ t(2iˆ 2jˆ t Rˆ k),(1) r (2iˆ ˆj 2k)9 2ˆ t(2iˆ 2jˆ t Rˆ k),(2) r (4iˆ ˆj k)9 1ˆ t(2iˆ 2jˆ t Rˆ k),(3) r (2iˆ k)3 ˆ t Rˆ k)(4) r t(2iˆ 2j1,2L1 x 1y 0z 0 122L2 xyz 2 1 2x y z a b cL3 L1 & L2L3 (L1 L2)L3 ˆˆ 3k) L3 (6iˆ 6jLet any point on L1 is ( 1, 2 , 2 )Let any point on L2 is B (2 , , 2 )DR(s) of AB will be2 –1, – –2 , 2 –2 But D.R. of AB are6, 6, –3 or 2, 2, –1 2 1 2 2 2 k(let)22 1
2 –1 2k– –2 2k2 – 2 –kSolve (1) & (3) 3k 13Put .(1).(2).(3)3k 1in equation (2)312k 2 3 Put & in eq. (3) 12k 2 3k 1 2 2 k 0 3 3 k 29 2 3 1 9 3 2 113 93 2 812 2 22 9 3 3 39 A (– 1, 2 , 2 )2 2 1 1, , 999 8 2 2 A , , 9 9 9 B (2 , – , 2 ) 4 2 4 B ,, 9 9 9 Equation of L3 can be 2ˆ t(2iˆ 2jˆ t Rˆ k),L3 r (4iˆ ˆj k)9 2ˆ t(2iˆ 2jˆ t Rˆ k),or L3 r (2iˆ ˆj 2k)9
1.Sol.Section - 3This section contains SIX (06) qeustions. The answer to each question is a NUMERICAL VALUE.For each question, enter the correct numerical value of the answer using the mouse and theon-screen virtual numeric keypad in the place designated to enter the answer. If the numericalvalue has more than two decimal places, truncate/roundoff the value to TWO decimal places.Answer to each question will be evaluated according to the following marking scheme;Full Marks: 3 If ONLY the correct numerical value is enteredZero Marks: 0 in all other cases.Three lines are given by r ˆi, R r (iˆ ˆj), R ˆ v Rr v(iˆ ˆj k),Let the lines cut the plane x y z 1 at the points A, B and C respectively. If the area ofthe triangle ABC is then value of (6 )2 equals .0.75 ˆr µ ˆi ˆjr ˆi ˆj kr ˆix y z 1Ist linex ,y 0,z 0 1 A(1,0,0)For 2nd Linex µ, y µ, z 0 2µ 1 1 1 B , ,0 2 2 1 1 1 Similarly C 3 , 3 , 3 1 AB AC Area of 21 1 ˆ 1 ˆ 2 ˆ 1 ˆ1 ˆ i j i j k 2 22 333 ijk1 1 10 2 2 2 2 1 13 3 3 1 ˆ 1 ˆ 1 ˆ 1 j i k 6 2 6 6 ˆ1 ˆi ˆj k 2 6 5 6 1 3;2 36 (6 )2 3 .754 312
2.2.Let S be the sample space of all 3 3 matrices with entries from the set {0,1}, Let the eventsE1 and E2 be given byE1 {A S : det A 0} andE2 {A S : sum of entries of A is 7}If a matrix is chosen at random from S, then the conditional probability P(E1 E2) equals1/2Sample space 29P E1 E2 P(E1/E2) P E2 E2 : sum of entries 7 '7' one and '2' zero1 1 18 99!1 1 1total E2 367!2!21 0 01 1 11 1 1for A to be zero both zeros should by in same row or column0 1 0 (3 3)2 181836 P(E1/E2) 1 1 11 1 00 1 13.Sol.12 1(1) –1(–1)Let 1 be a cube root of unity . Then the minimum of the set { a b c 2 2 : a, b, cdistinct non-zero integers} equals .3a b c 22 a b c a b a 12 1 1 1 4 32222 c b2 c2 ab bc ca a b 2 (b c)2 (c a)2
4.Let AP(a; d) denote the set of all the terms of an infinite arithmetic progression with first term and common difference d 0, IfAP(1;3) AP(2;5) AP(3;7) AP(a;d) then a d equals .Sol.157First APa 1, common diff. 3Second APa 2, common diff. 5Third APa 3, common diff. 7Now on AP whose first term and common diff. is common of all three 1 (n–1)3 2 (m–1)5 3 (k–1)73n 1 m5(i)3n 2 k7andm and k are integerSo at n 18m 11 and k 8first term of AP 1 (18–1)3 52Common diff. LCM (3,5,7) 105 a d 157 /45.If I Sol.42I 2dxthen 27I2 equals . sin x / 4 (1 e )(2 cos 2x) 4 (1 e 4sin xdx)(2 cos 2x)Apply King x –x2I 4 4esin x2 (1 esin x )(2 cos 2x) ; 2I dx 2 cos 2x 44 24dx24sec2 xdx I , tan x t 0 1 2 sin2 x 0 1 tan2 x 2 tan2 x12dt2 2 0 1 3t3 27 4 42723 1tan–1 3t 0 2 2 I 3 3 3 3
6.Let the point B be the reflection of the point A(2,3) with respect to the line 8x – 6y – 23 0. Let6. A and B be circles of radii 2 and 1 with centres A and B resepectively.Let T be a commontangent to the circles A and B such that both the circles are on the same side of T. If C is thepoint of intersection of T and the line passing through A and B, then the length of the linesegment AC is .10For B 2(16 18 23)x 2 y 3 64 368 6x 2 y 3 2( 25) 86100x 2 y 3 1 x 6 and y 6862B (6, 6)Now for 'C' external division in ratio r1 : r22.6 1.22.6 1.3b 2 12 1a 10,b 9a AC 82 62 64 36 100 10
MATHS [ JEE ADVANCED - 2019 ] PAPER - 1 SECTION -1 (Maximum Marks : 12) This section contains FOUR (04) questions. Each question has FOUR options ONLY ONE of these four options is correct answer. For each questio
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