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CBSE Class 12 Sample / Model Paper 2022 : Mathematics : pre -bord

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KENDRIYA VIDYALAYA SANGATHAN, RAIPUR REGION SAMPLE PAPER (SET-01) TERM-2(2021-22) CLASS-XII SUBJECT-MATHEMATICS BLUE PRINT SL. NO NAME OF THE CHAPTER SA1(2) SA2(3) LA(4) TOTAL 1. 2. 3. 4. 5. 6. Integrals Applications of Integrals Differential Equations Vectors Three-Dimensional Geometry Probability 1(2) 1(2) 1(2) 1(2) 2(4) 1(3) 1(3) 1(3) 1(3) - 1(4) 1(4) 1(4) 1 CASE BASED (2+2) 3(9) 1(4) 2(5) 2(5) 3(9) 3(8) TOTAL 6(12) 4(12) 4(16) 14(40) ` KENDRIYA VIDYALAYA SANGATHAN, RAIPUR REGION SAMPLE PAPER (SET - 01) TERM-2 (2021-22) CLASS-XII SUBJECT-MATHEMATICS TIME 2 hours MARKS-40 ___________________________________________________________________________ GENERAL INSTRUCTIONS: 1.This question paper contains 3 sections-A, B and C. Each part is compulsory. 2.Section-A has 6 short answer type (SA1) questions of 2 marks each. 3. Section-B has 4 short answer type (SA2) questions of 3 marks each. 4. Section-C has 4 long answer type (LA) questions of 4 marks each. 5. There is an internal choice in some of the questions. 6. Q 14 is a case-based problem having 2 subparts of 2 marks each. S.No. 1. Question SECTION-A 2 Marks 2 Evaluate: 03 4+9 2 OR 1+sin Evaluate: (1+cos ) 2. Write the order and the degree of the differential equation: 3. If | | = 10, | | = 2 . = 12, then find | |. 4. If the lines 1 3 = 2 2 = 3 2 and 1 3 = 1 1 = 6 5 4 4 3 + sin( 3 ) = 0. 2 2 are perpendicular to each 2 other, find the value of 5. A random variable X has the following probability distribution: X P(X) 0 0 Determine 6. 1 k 2 2k (i) k 3 2k 4 3k 5 k2 6 2 k2 2 7 7k2 + k (ii) P(X>6) A die is tossed thrice. Find the probability of getting an odd number at least 2 once. SECTION-B 1 7. Evaluate: ( 4 1) 3 8. Find the particular solution of the differential equation: 3 (1 + 2 ) = tan 1 , given that = 1 when = 0. OR Solve the differential equation: [ sin2 ( ) ] + = 0, given = 4 when = 1. 9. Let = 4 + 5 , b = 4 + 5 , and c = 3 + . Find a vector 3 and d . c = 21. d which is perpendicular to both and b, 10. Find the shortest distance between the lines whose vector equations are 3 = + + (2 + ) and = 2 + + (3 5 + 2 ) OR Find the vector equation of the plane which contains line of intersection of the planes . ( + 2 + 3 ) 4 = 0 and . (2 + ) + 5 = 0 and which is perpendicular to the plane . (5 + 3 6 ) + 8 = 0 . SECTION-C tan 11. Evaluate: 0 12. Using integration, find the area of the smaller region bounded by the ellipse 4 2 + 9 2 = 36 and the line 2 + 3 = 6. sec 4 4 OR Using integration, find the area of the region bounded by the triangle whose vertices are (1,0), (2,2) and (3,1). 13. 14. A plane meets the , and - axes at A, B and C respectively, such that the centroid of the triangle ABC is (1, 2, 3). Find the Vector and Cartesian equations of the plane. CASE BASED 4 Read the following text and answer the following questions on the basis of the same: The reliability of a COVID PCR test is specified as follows: Of people having COVID, 90% of the test detects the disease but 10% goes undetected. Of people free of COVID, 99% of the test judged COVID negative but 1% are diagnosed as showing COVID positive. From a large population of which only 0.1% have COVID, one person is selected at random, given the COVID PCR test, and the pathologist reports him/her COVID positive. (i) What is the probability of the person is having actually COVID positive given that he is tested as COVID positive? 2 (ii) What is the probability of the person selected will be diagnosed as COVID positive? 2 KENDRIYA VIDYALAYA SANGATHAN, RAIPUR REGION SAMPLE PAPER (SET - 01) TERM-2 (2021-22) CLASS-XII SUBJECT-MATHEMATICS MARKING SCHEME S.No. 1. Answers 2 3 1 0 4+9 2 = 6 [tan 1 3 ] Marks 1 2 3 2 0 1 = 24 OR 1+sin 1 1 (1+cos ) = (2 2 2 + tan 2) 1 = tan 2 + 2. = 4, . 3. = | || | = 20 = 5 , so = 5 . 12 3 1+1 1 4 4 | | = | || || | = 10 2 5 = 16. 1 4. 1 2 + 1 2 + 1 2 = 0 = 1+1 5. K=10 10 7 1 1 17 P(X>6) = 100 6. 1 1 1 1 7 Probability of getting an even number three times = 2 2 2 = 8 Probability of getting an odd number at least once = 1 8 = 8 7. 1 3 = ( 4 1) 4 ( 4 1) Let 4 = 4 3 = 1 3 1 1 = = ( 4 1) 4 ( 4 1) 4 ( 1) 1 1 1 = 4 ( 1 ) 1 1 1 1 4 1 = 4 | 8. 1 1 tan + = 1 + 2 1 + 2 1 P= 1+ 2 Q= tan tan 1 I.F= solution is 1 1+ 2 4 |+C 1 1 1 1 2 1 2 1 2 y tan 1 ( +1) tan = 1 +1 1 + +1 1 2 OR = 2 ; Put = , = + 1 2 = cot = log| | + To get c = 1 To get the solution: cot = log| | + 1 i.e., cot = log| | 9. A vector to both and is = 21 21 j 21 k Let = (21 21 j 21 k ) ; . = 21 63 21 + 21 = 21 = 1 3 = 1 (21 21 j 21 k ) = 7( j k ); So, 3 10. 1 = + , 1 = 2 + , 2 = 2 + , 2 1 2 1 2 1 1 2 1 2 1 1 1 2 1 2 = 3 5 + 2 2 1 = 1 1 2 = 3 7 and | 1 2 | = 59 1 =| = ( 2 1 ). ( 1 2) | | 1 2| 1 10 59 OR Let the equation of the required plane be 1 . ( + 2 + 3 ) 4 + [ . (2 + ) + 5] = 0 This is perpendicular to the plane . (5 + 3 6 ) + 8 = 0 . 7 Therefore, 5(2 + 1) + 3( + 2) 6(3 ) = 0 = 19 1 Therefore,vector equation of the plane is 1 . (33 + 45 + 50 ) 41 = 0 11. tan = 0 2 sec Let I = 0 2 = 0 ( ) 2 ( ) = 0 ( ) 2 1 2 sin 2 2I = 0 2 = 0 = [ ] 2 2 2 0 1 0 1 = 2 2 1 1 2 12. 2 I= 4 To draw the correct graph 3 36 4 2 Required area= 0 3 = 3 2 3 1 2 1 3 6 2 0 3 1 2 OR To draw the graph and finding the equations of the sides 1 = 2( 1), = 4 , = ( 1) 2 2 3 31 Required area= 1 2( 1) + 2 (4 ) 1 2 ( 1) 3 =2 13. 14. 0+ +0 1 1 0+0+ Here, = 1, = 2, = 3, 3 3 3 = 3, = 6, =9 Therefore, the equation of plane is 3 + 6 + 9 = 1 . . , 6 3 + 2 18 = 0 which in vector form is . (6 3 + 2 ) = 18 E1: Person actually having COVID, E2: Person actually not having COVID and A: person tested as positive ( ) ( / ) (i) P(E1/A) = ( ) ( / 1 )+ ( 1) ( / ) = 0.0826 (ii) 1 1 Let the coordinates of A, B and C be ( , 0,0), (0, , 0) (0,0, )respectively. Therefore, the equation of plane is + + = 1 +0+0 1 1 2 2 P(A) = ( 1 ) ( ) + ( 2 ) ( ) = 0.01089 1 2 1 1 1 1 1 1 1 1 2 1 2 KENDRIYA VIDYALAYA SANGATHAN, REGIONAL OFFICE RAIPUR SAMPLE PAPER (SET 02) TERM 2 (2021-22) CLASS XII SUB: MATHEMATICS BLUE-PRINT S.No. Chapter SA-1 (2 Marks) SA-2 (3 Marks) LA (4 Marks) Total 1 INTEGRAL 1(2) 1(3) 1(4) 3(9) 2 APPLICATIONS OF INTEGRALS 1(4) 1(4) 3 DIFFERENTIAL 1(2) EQUATION 1(3) - 2(5) 4 VECTOR ALGEBRA 1(2) 1(3) - 2(5) 5 3-D GEOMETRY 1(2) 1(3) 1(4) 3(9) 6 PROBABILITY 2(4) - 1(4) 3(8) 6(12) 4(12) 4(16) 14(40) TOTAL KENDRIYA VIDYALAYA SANGATHAN, REGIONAL OFFICE RAIPUR SAMPLE PAPER (SET- 02) TERM -2 (2021-22) CLASS XII SUB: MATHEMATICS (041) TIME- 2 hours M.M:40 ________________________________________________________________________ GeneralInstructions: 1. This question paper contains three sections A, B and C. Each part is compulsory. 2. Section A has 6 short answer type (SA1) questions of 2 marks each. 3. Section B has 4 short answer type (SA2) questions of 3marks each. 4. Section C has 4 long answer type questions (LA) of 4markseach. 5. There is an internal choice in some of the questions. 6. Q14 is a case-based problem having 2 subparts of 2 marks each. S.No. Question Marks SECTION A 3 1. 2 2 Find 3 dx OR Find Find the sum of degree and order of differential equation 2 4 2 2 2 2 = 1+ [( ) ] 3 4 5 6 Find the unit vector perpendicular to each of the vectors : = 4 +3 + and = 2i j + 2k 5 Show that the plane 5 2 = 1 contains the line 3 = = 2 A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2balls are drawn at random from the bag one-by-one without replacement. Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack. 2 2 2 2 SECTION B 7 Find 3 3 8 Find the general solution of the following differential equation: ( + 2 2 ) = 0 3 OR Find the particular solution of the following differential equation: = 1+ 2 + 2 + 2 2 given that = 1 when = 0 9 10 11 12 13 If 0 , . = . , = , then show that = . Find the shortest distance between the following lines: = ( + )+ (2 + + ) =( + + 2 ) + (4 + 2 + 2 ) OR Find the vector and the Cartesian equations of the plane containing the point + 2 and parallel to the lines = ( +2 + 2 ) + (2 3 + 2 )=0 and =(3 + 2 )+ ( 3 + )=0 SECTION C Evaluate : 0 tan sec +tan Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y2 = x and the x-axis. OR Using integration, find the area of the region {( , ):0 3 , 2+ 2 4} Find the foot of the perpendicular from the point (1,2,0) upon the plane x 3y+2z=9. Hence,find the distance of the point(1,2,0) from the given plane. 3 3 4 4 4 14 CASE STUDY A shopkeeper sells three types of flower seeds A1 , A2 and A3 . they are sold as mixture where proportions are 4:4:2 respectively. Their germination rates are 45%, 60% and 35% respectively. Calculate the probability (i) of a randomly chosen seed to germinate (ii) that it is of the type A2 given that a randomly chosen seed does not germinate. 2 2 KENDRIYA VIDYALAYA SANGATHAN, REGIONAL OFFICE RAIPUR SAMPLE PAPER (SET 02) TERM -2(2021-22) CLASS XII SUB-MATHEMATICS MARKING SCHEME S.No. ANSWE RS 1 dx = 3 3 ,put tanx =t, = 3 1 + 2 dx, put 1 + 2 = , = ( 1) 2 1 1 = 5 3 + 5 3 OR Put = , = = ( 1) + = (1 log ( )) +c Order = 2, degree =1 ,sum = 2 = |4 3 1| = 7 - 6 -10 , 2 1 2 1 | | = 185 , = 2 3 = MARKS 3 3 185 4 5 d.r.s. of line 3, 1 ,-1,d.r.s of normal to the plane1 ,-5 ,-2 3(1) +1(-5) +(-1)(-2) = 0 :Let X be the random variable defined as the number of red balls. ThenX = 0,1 3 2 1 P(X=0) = = 4 1 3 3 2 3 1 1 2 1 1 2 2 The required probability = P((The first is a red jack card and Thesecond is a jack card) or (The first is a red non-jack card and The second isa jack card)) 2 3 24 4 1 = + = 52 7 51 52 51 1 1 1 2 1 1 1 1 1 P(X=1) = + = 4 4 4 3 2 ProbabilityDistributionTable: X P (X) 6 1 26 = 3 = 2 = 2 = 3 + 1 1 1 1 2 = + | + | + 1 = + | + | + 2 2 8 1 +2 2 = - = 2 1 1 2 . = = 2 + OR = (1 + 2 ) (1 + 2 ) (1 + 2 ) 1 1 OR = (1 + 2 ) 1 (1 + 2) = (1 + 2 ) +c 3 tan = + + 3 = tan 1 1 = 4 3 tan 1 = + + 3 4 1 9 10 1 1 Solution: We have .( )=0 ( )= 0 or ( ) = or ( ) Also, ( )=0 ( )= 0 or ( ) = or ( ) = or ( ) ( ) ( ) Hence, = . |( ) | The lines are parallel. The shortest distance = | | |(3 ) (2 + + )| = 4+ 1+ 1 (3 ) (2 + + )=|0 0 3|= 3 +6 2 1 1 Hence, the requiredshortestdistance=3 5units 6 1 1 1 1 1 2 1 1 2 1/ 2 1 1/2 1 1/2 OR Since, the plane is parallel to the given lines, the cross product of the vectors 2 3 + 2 and 3 + will be a normal to the plane (2 3 + 2 ) ( 3 + )= |2 3 2|=3 3 1 3 1 The vector equation of the plane is .(3 3 )=( + 2 ).(3 3 ) or, .( )=2 and the Cartesian equation of the plane is x z 2 =0 11 I= 0 1 12 dx + = 0 dx 1+ ( ) = 0 dx 1+ 2I= 0 dx 1+ = 0 ( 1 12 2 x)dx = [ + ] 0 = ( 2) 2 12 1 y2 Solving + = 2 and = x to get point of intersection as (1,1) and (4,2) Correct fig 1 2 Reqd area 0 + 1 (2 ) 2 3 2 1 = [ ] + [2 3 2 1 0 = + = 3 2 7 6 2 2 ] 1 OR Solving = 3 and 2+ 2 = 4 to get point of intersection as (1, 3) and ( -1 ,- 3) Correct fig 1 2 Reqd area = 0 3 + 1 4 2 = 3 2 2 3 1 1 2 1 = 1 1 [ 2 ]10 + [ 4 2 + 4 2 sq units 2 sin 1 ] 2 1 1 1 1 1 13 : The equation of the line perpendicular to the plane and passing through the point (1, 2, 0) is 1 2 = = 1 3 1 2 The coordinates of the foot of the perpendicular are ( + 1, 3 + 2,2 ) for some These coordinates will satisfy the equation of the plane. Hence, we have + 1 3( 3 + 2) + 2(2 ) = 9 =1 The foot of the perpendicular is (2, -1, 2). 14 Hence, the required distance = 14 Let A1: seed A1 is chosen, A2: seed A2 is chosen & A3: seed A3 is chosen 3 E: seed germinates and : seed germinates P(A1) = 4 4 2 45 , P(A2) = , P(A3) = , P(E/A1) = 10 10 10 100 35 P(E/A3) = 100 (i) (ii) , P(E/A2) = 60 100 P(E) = ( 1 ) ( ) + ( 2 ) ( ) + ( 3 ) ( ) = 0.49 1 ( 2 / ) = 2 ) 2 ( 1 ) ( )+ ( 2 ) ( )+ ( 3 ) ( ) 1 2 3 ( 2 ) ( 1 1 , 2 3 16 = 51 2 KENDRIYA VIDYALAYA SANGATHAN, RAIPUR REGION SAMPLE PAPER (SET-03) TERM-2(2021-22) CLASS-XII SUBJECT-MATHEMATICS BLUE - PRINT SL. NO NAME OF THE CHAPTER SA1(2) SA2(3) LA(4) TOTAL 1. 2. 3. 4. 5. 6. Integrals Applications of Integrals Differential Equations Vectors Three-Dimensional Geometry Probability 1(2) 1(2) 1(2) 1(2) 2(4) 1(3) 1(3) 1(3) 1(3) - 1(4) 1(4) 1(4) 1 CASE BASED (2+2) 3(9) 1(4) 2(5) 2(5) 3(9) 3(8) TOTAL 6(12) 4(12) 4(16) 14(40) KENDRIYA VIDYALAYA SANGATHAN, RAIPUR REGION SAMPLE PAPER (SET - 03) TERM-2 (2021-22) CLASS-XII SUBJECT-MATHEMATICS TIME 2 HOURS MARKS-40 ___________________________________________________________________________ GENERAL INSTRUCTIONS: 1.This question paper contains 3 sections-A, B and C. Each part is compulsory. 2.Section-A has 6 short answer type (SA1) questions of 2 marks each. 3. Section-B has 4 short answer type (SA2) questions of 3 marks each. 4. Section-C has 4 long answer type (LA) questions of 4 marks each. 5. There is an internal choice in some of the questions. 6. Q 14 is a case-based problem having 2 subparts of 2 marks each. S.No. Question Marks SECTION-A 1. sin Evaluate: dx, 1+sin 2 2 0<x< 2 OR Evaluate 2 2 2. Find the direction cosines of the vector + 2 + 3 . 3. Solve: + 2y = 3 2 4. Find the cartesian equation of the line which passes through the point (-2,4,-5) and parallel to the line 2 2 1 2 = 1 3 = 3 4 5. A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find the probability that none is red . 6. If A and B are two events such that P(A)= 4 , P(B) =2 and P(A )= 8 . Find 1 1 1 2 2 P(notA and notB). SECTION-B 7. 2 Evaluate : dx (1+ 2 )(3+ 2 ) 3 8. 3 Solve the differential equation: x cos( ) = y cos( ) + x OR Solve the differential equation: ( 2 +1) +2xy = 2 + 4 9. If and are two unit vectors and is the angle between them prove that 1 sin = | |. 10. Find the shortest distance between the two skew lines 3 =8 9 + 10 + (3 16 + 7 ) =15 + 29 + 5 + (3 + 8 5 ) OR Find the length and foot of the perpendicular from the point (1,3/2,2) to the plane 2x-2y+4z=0. SECTION-C tan 4 Evaluate : dx 2 11. 12. 2 0 sec +tan Find the area bounded by the curve y= 2 and the lines y=4. OR Find the area of the region included between the parabola y= 13. 14. i). ii). 3 4 3 2 4 and the line 3x-2y+12=0. Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes . ( + 2 ) = 5 and . ( 3 + 3 + ) = 6. Let X denote the number of college where you will apply after yours result and P(X=x) denotes your probability of getting admission in x number of college.It is given that , = 0 1 2 , = 2 P(X=x) = { (5 ), = 3 4 0, > 4 Where k is a positive constant. Based on the above information answer the following: Find the value of k. What is the probability that you will get admission in at least two college. 4 2 2 KENDRIYA VIDYALAYA SANGATHAN, RAIPUR REGION SAMPLE PAPER (SET - 03) TERM-2 (2021-22) CLASS-XII SUBJECT-MATHEMATICS MARKING SCHEME S.No. 1. Answers 1 + sin2x = (sin + cos ) 2 = sin x + cos x sin 1+sin 2 dx = =-log t + C = - log (sin x +cos x ) + C OR For writing 2. ( 2 + 2 ) 2 2 3. I = tanx cotx + c Direction ratios of the vector 1,2,3 1 2 3 For finding direction cosines , , 14 5. 1 12 14 I. F. = 2 5 2 4. 1 1 and separating it 14 For finding correct solution y = 5 + C Direction ratios of the line -2,-3,4 Cartesian equation of the line +2 4 +5 = = 2 3 4 8 7 6 1 1 1 1 1 5 + - = 4 2 8 8 P(not A and not B ) =P(A ) =1-P(AUB)= I= 2 3 (1+ 2 )(3+ 2 ) dx 2 Getting result by partial function For taking y=vx and finding = v+x dv/dx For finding xdv/dx= 1/cosv For finding correct solution sin( ) = log| | OR ( 2 +1) +2xy = 2 + 4 2 2 +4 + = 2 +1 2 +1 1 1 8 Let 2 = t dt=2xdx I= (1+ )(3+ ) 8. 1 12 1 P(A)= 4 , P(B) =2 and P(A )= 8 P(AUB) = P(A)+ P(B)- P(A )= 7. 12 8 6. 1 12 1 P(none is red)= 15 X14 X13 = 65 Marks 1 is a linear differential equation 1 1 1 2 Getting integrating factor = 2 +1 Multiplying I.F. and finding solution 9. 10. 2 | |2 =| |2 +| | -2 . =1+1-2cos =4 2 2 Showing result Shortest distance= 1 1 2 ) 1 ).( 1 ( 2 | 1 2| | 1 2 | Finding 2 1 and 1 2 Getting result OR 1 3/2 2 = 2 = 4 =k 2 X= 2k+1 y= -2k+3/2 z=4k+2 Putting these values in equation of plane and finding the value of K For finding Foot=(x,y,z) For finding length 11. Let I = 0 2I= 12. tan dx = 0 sec +tan 0 sec tan dx +tan ( 2) ( ) tan( ) sec( )+tan ( ) ( ) tan dx = 0 sec +tan dx Getting result I= 2 For correct figure 4 Required area= 2 0 4 =2 0 = 32/3 squnit OR For finding limit x=-2 to x=4 For correct figure 4 12+3 A= 2 2 4 3 2 2 4 1 1 1 1 1 1 1 2 1 1 1 12 1 2 For integrating and writing correct area 27 sq unit. 13. 14. Let a, b and c be the drs of the req. line a b + 2c = 0 and 3a + 3b + c = 0 finding a ,b and c in the terms of arbitrary constant a=-7k, b= 5k , c=6k finding vector equation of line = + 2 + 3 + (-7 + 5 + 6 ) i) k+4k+2k+k = 1 K=1/8 ii) P(getting admission in atleast two colleges)= 4k+2k+k =7k= 7/8 2 1 1 1 1 1 1

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