Signaler

STUDY INNOVATIONSSuivre

29 Mar 2023•0 j'aime## 0 j'aime

•5 vues## vues

Soyez le premier à aimer ceci

afficher plus

Nombre de vues

0

Sur Slideshare

0

À partir des intégrations

0

Nombre d'intégrations

0

29 Mar 2023•0 j'aime## 0 j'aime

•5 vues## vues

Soyez le premier à aimer ceci

afficher plus

Nombre de vues

0

Sur Slideshare

0

À partir des intégrations

0

Nombre d'intégrations

0

Télécharger pour lire hors ligne

Signaler

Formation

Class : XII (NUCLEUS) PART TEST- 3 INSTRUCTIONS 1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2 questions of "Match the Column" type and Part-C contains 4 subjective type questions. All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. PART-A (i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. PART-B (iii) Q.1 to Q.2 are "Match the Column" type which may have one or more than one matching options and carry 8 marks for each question. 2 marks will be awarded for each correct match within a question. There is NO NEGATIVE marking. Marks will be awarded onlyif all the correct alternatives are selected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded only if all the correct bubbles are filled in your OMR sheet. 2. Indicate the correct answer for each question by filling appropriate bubble(s) in your answer sheet. 3. Use only HB pencil for darkening the bubble(s). 4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only. PART-A PART-B PART-C For example if only 'B' For example if Correct Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column will be treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimal places. e.g. 86 should be filled as 0086.00 . . . . . . . . . . choice is correct then, match for (A) is P, Q; for the correct method for (B) is P, R; for (C) is P filling the bubble is and for (D) is S then the A B C D correct method for filling the bubble is P Q R S For example if only 'B & (A) D' choices are correct then, the correct method (B) for filling the bubbles is A B C D (C) (D) The answer of the question in any other manner (such as putting , cross , or partial shading etc.) will be treated as wrong. Class - XII Part Test - 3 PART - A Select the correct alternative. Choose one or more than one. Q1. Let P & Q are two points denoting the complex number & respectively on the complex plane. Which of the following equations can represent the equationsof the circle passing through P & Q with least possible area? Z (A) Arg = Z 2 (B*) Re (Z – ) Z = 0 (C*) Z 2 + Z 2 = (D) ZZ + 2 Z + 2 Z + + = 0 [Hint: (A) is not correct as it should be + 2 Z (B) equation of circle is

STUDY INNOVATIONSSuivre

- Class : XII (NUCLEUS) Time : 3 hour Max. Marks : 60 INSTRUCTIONS 1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2 questions of"Match the Column" type and Part-C contains 4 subjective type questions.Allquestions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. PART-A (i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each. There is NEGATIVE marking. 1 markwillbe deducted for eachwrong answer. PART-B (iii) Q.1 to Q.2 are "Match the Column" type whichmayhave oneor more than one matching options and carry8 marks for each question. 2 marks willbe awarded for each correct matchwithin a question. ThereisNONEGATIVE marking. Markswillbeawardedonlyifallthecorrectalternativesareselected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded onlyifallthe correct bubbles are filled in your OMR sheet. 2. Indicate the correct answer for eachquestionbyfilling appropriate bubble(s) inyour answer sheet. 3. Use onlyHB pencilfor darkening the bubble(s). 4. Use ofCalculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) ofthe questionsmust bemarked byshading the circlesagainst the questionbydark HBpencil only. PART TEST-3 PART-B For example if Correct match for (A) is P, Q;for (B) is P, R; for (C) is P and for (D) is S then the correct method for filling the bubble is P Q R S (A) (B) (C) (D) PART-C Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column willbe treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimalplaces. e.g. 86 should be filled as 0086.00 . . . . . . . . . . PART-A For example if only 'B' choice is correct then, the correct method for filling thebubble is A B C D For example ifonly'B & D' choices are correct then, thecorrect method for fillingthe bubbles is A B C D The answer of the question in any other manner (such as putting , cross , or partial shading etc.) will be treated as wrong.
- Class - XII Part Test - 3 PART -A Select the correct alternative. Choose one or more than one. Q1. Let P & Q are two points denoting the complex number & respectively on the complex plane. Which ofthe following equations canrepresent the equationsofthe circle passing through P & Q with least possible area? (A) Z Z Arg = 2 (B*) Re (Z – ) Z = 0 (C*) 2 Z + 2 Z = 2 (D) Z Z + 2 Z + 2 Z + + = 0 [Hint: (A) is not correct as it should be + 2 (B) equation ofcircle is Z Z is PI Re Z Z = 0 or Re (Z – A) Z = 0 B is correct (C) is obviouslycorrect frompythagoras theorem (D) incorrect because centre ofrequired circle is 2 but centre ofgiven circle is – – coefficient of Z = – 2 Q.2 The equation to the orthogonaltrajectories ofthe systemofparabolas y= ax2 is (A*) 2 2 y 2 x = c (B) 2 y x 2 2 = c (C) 2 2 y 2 x = c (D) 2 y x 2 2 = c [Sol: ax 2 dx dy a = dx dy x 2 1 dx dy x y 2
- Orthogonaltrajectoryis dy dx x y 2 2y dy + x dx = 0 y2 + 2 x2 = C ] Q.3 Consider thefollowing regions intheplane : R1 = {(x, y) : 0 x 1 and 0 y 1} R2 = {(x, y) : x2 + y2 4/3} The area ofthe region R1 R2 can be expressed as 9 b 3 a , where a and b are integers. Then the value of (a + b) equals (A) 2 (B) 3 (C*) 4 (D) 5 [Sol. A = 1 A 1 3 1 A1 = 1 3 1 2 dx x 3 4 put x = 3 2 sin = d cos 3 2 · cos 3 2 3 6 = d ) 1 2 (cos 3 2 3 6 = 3 6 2 sin 2 1 3 2 = 6 2 3 · 2 1 3 2 3 · 2 1 3 2 = 6 · 3 2 = 9 A = 9 3 1 = 9 3 3 a = 3, b = 1 a + b = 4 Ans. ] Q.4 Imagine that you have two thumbtacks placed at two points,Aand B. Ifthe ends ofa fixed length of string are fastened to the thumbtacks and the string is drawn taut with a pencil, the path traced by the pencilwillbe anellipse. Thebest wayto maximisethe area surrounded bythe ellipse with afixedlength ofstringoccurs when I the two pointsAand Bhave the maximumdistance between them. II two pointsAand B coincide. III Aand B are placed vertically. IV The area is always same regardless ofthe location ofAand B. (A) I (B*) II (C) III (D) IV [Sol. A = ab 'a' is constant and b varies A2 = 2a2(a2 – c2) forAto be maximumc must be minimum; A& B centre as A B c 0 ellipse becomes circle ]
- Q.5 Consider the expansion , (a1 + a2 + a3 + ....... + ap)n where nN and n p. The correct statements are (A*) num ber of di f f erent t er m s i n t he expansi on i s , n+p 1Cn (B)co-efficientofanyterminwhichnoneofthevariables a 1,a2,......,a p occurmorethanonce i s' n ' (C*)co-efficien tofanyterminwhichnoneoft hevariables a 1,a2,.... ..,a p occurmor ethanonce i s n ! (D*)Numberoftermsinwhichnoneofthevariables a 1,a2,......,a p occurmorethanonceis n p . [ S o l . (a1 +a 2 +a 3 +.....+a p)n; p n wehavepbeggarsandn identicalcoinstobedistributed. n n Ø Ø ...... Ø Ø 0 0 ...... 0 0 Total ways = n + p – 1Cn ( A )i sc o r r e c t a g a i n degreeofeverytermwillbenandp n,hencenumberoftermswillbeasmanyas thenumberofwaysinwhichntermscanbeselectedoutofp number of terms = pCn ( D )i sc o r r e c t a g a i n coe ffic ientofthet ermsay a 1,a2,a3. ...a n= terms n ! 1 !..... 1 ! 1 ! n (usingthetheoryofmultinomialse.g.coefficientofa 2b3c4 in(a+b+c) 9is ! 4 ! 3 ! 2 ! 9 ] Q.6 The probabilitythat a positive two digit number selected at randomhas its tens digit at least three more thenitsunit digit is (A*) 14/45 (B) 7/45 (C) 36/45 (D) 1/6 [Sol. n (S) = 9 · 10 = 90 6 , 5 , 4 , 3 , 2 , 1 , 0 9 2 , 1 , 0 5 1 , 0 4 0 3 possible not is 2 or 1 ) units ( y ) Tens ( x n (A) = 1 + 2 + 3 + ..... + 7 = 2 8 · 7 = 28 p = 90 28 = 45 14 Ans. ] PART - B MATCH THE COLUMN INSTRUCTIONS: Column-Iand column-IIcontains fourentrieseach. Entries ofcolumn-Iare to be matchedwith some entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II. Q1. Column–I Column–II (A) The differentialequationofallparabolas having (P) 1 their axisofsymmetrycoinciding withthe axis of x has its degree (B) The differentialequation ofallparabolaseach of (Q) 2 which has a latus rectum'4a' &whose axes are parallelto x-axis is ofdegree (C) Degree ofthe differentialequationy= a a x e 1 , (R) 3 a being the parameter is
- (D) The polynomialf (x) satisfies the condition f (x+ 1) = (S) does not exist x2 + 4x. The area enclosed by y = f (x – 1) and the curve x2 + y = 0, is k 2 16 where k is [Ans: (A) P;(B) P;(C) S; (D) R] [Sol: (A) equation (x a)2 + y2 = (x b)2 [ S = (a , 0) ; D : x = b ] y2 = (b2 a2) + 2x (a b) differentiate twice to get y d y dx 2 2 + dy dx 2 = 0 ; y 2 2 dx y d + 2 dx dy = 0 (B) Equation to the familyof parabolas is (y– k)2 = 4a(x h) 2(y – k) dx dy = 4a (y – k) dx dy = 2a (y – k) 2 2 dx y d + 2 dx dy = 0 2a 2 2 dx y d + 3 dx dy = 0. Hence the degree is 1. (C) DE = dx dy x – y ln dx dy – x = 0 (D) f (x + 1) = x2 + 4x put x = x – 2 f (x – 1) = (x – 2)2 + 4(x – 2) f (x – 1) = x2 – 4 A = 2 2 0 2 2 dx )] 4 x ( ) x [( = 2 2 0 2 dx ) x 2 4 ( = 3 2 16 Ans. ] Q2. Column–I Column–II (A) The name ofthe conic represented bythe (P) Ellipse equation px + qy = 1, where p, q R, p, q > 0 is (B) Two parabola y2 = 4a (x – 1), and x2 = 4a(y – 2) (Q) Circle always touch each other. 1 and 2 being variable parameters. Then, their points ofcontact lie on a (C) Ifparabola oflatus-rectuml, touches a fixed equal (R) parabola parabola, the axes ofthe two curves being parallel, then the locusofthe vertexofthemoving curve is (D) Froma point P tangents are drawnto the parabola (S) hyperbola y2 = 4ax. Ifthe chord ofcontact ofthese tangents touches the rectangular hyperbola x2 – y2 = a2, then the locus ofP is [Ans: (A) R;(B) S;(C) R; (D) P] [Sol: (A) We have, px + qy = 1
- ( px + qy )2 = 1 px + qy + xy pq 2 = 1 (px + qy – 1)2 = 4pq xy p2x2 – 2pq xy + q2y2 – 2px – 2qy + 1 = 0 Oncomparing thisequationwiththe equation a = p2, b = q2, c = 1, g = –p, f = –q and h = –pq = abc + 2fgh –af2 – bg2 – ch2 = p2q2 – 2p2q2 – p2q2 – p2q2 – p2q2 = –4p2q2 0 and, h2 – ab = p2q2 – p2q2 = 0 Thus we have D 0 and h2 = ab So the givenrepresents a parabola (B) Equate dx dy to get xy= 4a2 which is a hyperbola (C) Let the vertexofthe moving parabola is (h, k) the the equationof the moving parabola (y– k)2 = –4a(x – h) .........(1) Equation offixed parabola is y2 = 4ax .........(2) where 4a = l since these two parabolas touch each other Solving eq. (1) and (2) 2y2 – 2ky +k2 – 4ah = 0 D = 0 k2 = 8ah k2 = 2lh y2 = 2lx (D) Let P (h, k). Chord ofcontact to parabola is yk = 2a (x + h) 2ax – yk + 2ah = 0 Equation oftangent to hyperbola at anypoint (asec, atan ) is xsec – ytan – a = 0 On comparing equations (1) and (2) sec 2a = tan k = a ah – 2 sec = h a , tan = 2h k So, 2 2 4h k h a 2 2 = 1 k2 + 4h2 = 4a2 So the locus of(h, k) is 2 2 4a y a x 2 2 = 1 Therefore ans is Ellipse]
- PART - C Q1. Let Z = 18 + 26i where Z0 = x0 + iy0 (x0, y0 R) is the cube root of Z having least positive argument. Find the value of x0y0(x0 + y0). [Ans. 12] [Sol. Let Z = 18 + 26i let r cos = 18 and r sin = 26 r2 = 324 + 676 = 1000 r = 10 10 tan = 18 26 = 9 13 ; hence 2 , 0 ; 3 6 , 0 Z1/3 = [ 10 10 (cos + i sin )]1/3 = 10 3 sin 3 cos i now tan = ) 3 ( tan 3 1 ) 3 ( tan ) 3 tan( 3 2 3 = 2 3 t 3 1 t t 3 where t = tan 3 13(1 – 3t2) = 9(3t – t3) 13 – 39t2 = 27t – 9t3 9t3 – 39t2 – 27t + 13 = 0 3t2(3t – 1) – 12t(3t – 1) – 13(3t – 1) = 0 (3t – 1)(3t2 – 12t – 13) = 0 tan 3 = 3 1 3 6 , 0 sin 3 = 10 1 and cos 3 = 10 3 Z1/3 = 3 + i x0 = 3 and y0 = 1 x0y0 (x0 + y0) = 12 Ans. ] Q2. A die isweighted such that the probabilityofrolling an n isproportionalto n2 (n= 1, 2, 3, 4, 5, 6). The die is rolled twice, yielding the numbers a and b. The probability that a < b in lowest form is q p where p + q = [Sol. Given P(1) = K ; P(2) = 22K ; P(3) = 32K ; P(4) = 42 K ; P(5) = 52K ; P(6) = 62K Total = 91 K = 1 K = 91 1 ; P(1) = 91 1 ; P(2) = 91 4 and so on Let three eventsA, B, C are defined as A : a < b B : a = b C : a > b By symmetry, P(A) = P(C). Also P(A) + P(B) + P(C) = 1 ....(1) since P(B) = 91 25 )] i ( P [ 6 1 i 2 2P(A) + P(B) = 1 ( from(i) ) P(A) = 2 1 [1 – P(B)] = 91 33 Ans. 33 + 91 = 124 ] Q.3 If the expression z5 – 32 can be factorised into linear and quadratic factors over realcoefficients as (z5 – 32) = (z – 2)(z2 – pz + 4)(z2 – qz + 4) then find the value of (p2 + 2p). [10] [Ans. 4]
- [Sol. z5 – 32 = (z – z0)(z – z1)(z – z2)(z – z3)(z – z4) [T/S, Q.24, Ex-1, Complex] Where zi's , i = 0, 1, 2, 3, 4 are given by zi = 2 5 m 2 sin 5 m 2 cos i (using Demoivre'sTheorem) with m = 0, 1, 2, 3, 4, we get z0 = 2, amd z1 = 5 2 sin 5 2 cos 2 i ; z2 = 5 4 sin 5 4 cos 2 i ; z3 = 5 6 sin 5 6 cos 2 i z4 = 5 8 sin 5 8 cos 2 i hence, z5 – 32 = (z – 2) 4 z 5 2 cos 4 z2 4 z 5 4 cos 4 z2 z z ) ( z ) z )( z ( using 2 p = 5 2 cos 4 = 4 cos 72° = 4 sin 18° p2 + 2p = [16sin218° + 8 sin18°] = 8[1 – cos 36° + sin 18°] = 8 4 1 5 4 1 5 1 = 4 Ans.] Q.4 A movable parabola touches the x and the y-axes at (1, 0) and (0, 1). The locus of the focus of the parabola is ax2–ax + ay2 – ay + 1 = 0 where a is [Ans: 2] [Sol: Since the x-axis and y-axis are two perpendicular tangents to the parabola and both meet at the origin, the directrixpassesthroughthe origin. Let y= mx be the directrix and (h, k) be the focus. FA = AM 2 2 k 1 h = 2 m 1 m .............(1) and FB = BN
- 2 2 1 k h = 2 m 1 m .............(2) Fromequations (1) and (2), we get (h – 1)2 + h2 + k2 + (k – 1)2 = 1 2x2 – 2x + 2y2 – 2y + 1 = 0 is the required locus. ]