Mathematics Guess Paper: 04 Class: XII Time llowed: Hours Maimum Marks: 70 General Instructions:. The question paper consists of 9 questions divided into three sections, B and C.. Section comprises of 0 questions of one mark each.. Section B comprises of questions of four marks each 4. Section C comprises of 7 questions of si marks each. 5. ll questions in Section are to be answered in one word, one sentence or as per the eact requirement of the question. 6. Use of calculators is not permitted. 7. ll questions are compulsory Section Q. Write the value of tan sin cos. Q. Write the intercept cut off by the plane + y z 5 on -ais. Q. For what value of a the vectors i j + 4 k & ai + 6 j 8k are collinear? Q 4. Write the differential equation representing the family of curves y m, where m is an arbitrary constant.
Q 5. For what value of ë is the function defined by f ( ) about continuity at? ( ) ë, if 0 continuous at 0? What 4 +, if > 0 Q 6. If ij is the cofactor of the element aij of the determinant 5 6 0 4 5 7, then write the value of a. Q 7. State the reason for the relation R in the set {,, } given by R {(, ), (, )) not to be transitive. Q 8. P and Q are two points with position vectors a b and a + b respectively. Write the position vector of a point R which divides the line segment PQ in the ratio : eternally. Q 9. The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (marginal revenue). If the total revenue (in rupees) received from the sale of units of a product is given by R() + 6 + 5, find the marginal revenue, when 5, and write which value does the question indicate. Find, if for a unit vector a, - a. + a 5. Q 0. Section B Q. Probabilities of solving a specific problem independently by and B are ½ & / and respectively. If both try to solve the problem independently, find the probability that (i) The problem is solved 66
(ii) Eactly one of them solves the problem. Q. Find the angle between the following pair of lines: + y z + + y 8 z 5 & 7 4 4 OR Find the coordinates of the point where the line through the point (, 4,) & B (5,,6) crosses the XY-plane. Q. Find the inverse of the matrices by row transformation method, if it eists. Q 4. Solve the following equation: cos tan sin cot 4 Q 5. + Differentiate tan with respect to. Q 6. Solve the following differential equation: cos dy y tan +
OR Solve the following differential equation: + dy + y cot ; 0 Q 7. Form the differential equation of the family of parabolas having verte at the origin and ais along positive y-ais. OR Find the particular solution of the differential equation dy ( ). ; y 0 when. Q 8. Maimise Z + y, subject to the constraints:, + y 5, + y 6, y 0. Q 9. Evaluate: cos Q 0. Find the equation of the plane through the line of intersection of the planes + y + z and + y + 4z 5 which is perpendicular to the plane y + z 0. Q. Evaluate: 5 + + 4 + 0 OR 68
Evaluate: ( + )( + ) ( + )( + 4) Q. cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes hours on grinding/cutting machine and hours on the sprayer to manufacture a pedestal lamp. It takes hour on the grinding/cutting machine and hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 0 hours and the grinding/cutting machine for at the most hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs. ssuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maimize his profit? Section C of the vol- Q. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8 ume of the sphere. 7 5 Q 4. If 4, find. Using solve the system of equations y + 5z + y 4z 5 + y z OR The cost of 4 kg onion, kg wheat and kg rice is Rs 60. The cost of kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion kg wheat and kg rice is Rs 70. Find cost of each item per kg by matri method. 69
Q 5. The area between y and 4 is divided into two equal parts by the line a, find the value of a. Q 6. Find the equation of the plane through the line of intersection of the planes + y + z and + y + 4z 5 which is perpendicular to the plane y + z 0. OR Let the vectors a a i + a j + ak b b i + b j + bk c c i + c j + c k Then show that a ( b + c) ( a b) + ( a c) 4 Q 7. Find the limit of sums: ( - ) Q 8. The management committee of a residential colony decided to award some of its members (say ) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matri method, find the number of awardees of each category. part from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards. Q 9. Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X. 70
Solutions Section nswer : π 6 tan sin cos tan sin cos cos π tan sin 6 π tan sin tan tan π tan tan π nswer : + y z 5 Divide the equation by 5 + y z 5 5 5 + y z 5 5 5 Thus, the intercepts cut off by the plane are 5,5, 5 7
nswer : Let P i j + 4 k Q ai + 6 j 8k a Q i + j 4k a Thus, a 4 nswer 4: We have, y m On differentiation dy m y m The differential equation representing the family of curves y m, is dy y 0 nswer 5: f ( ) ( ) λ, if 0 4 +, if > 0 If f is continuous at 0, then + 0 0 lim f ( ) lim f f 0 λ lim λ( ) lim 4 0 0 + 0 λ (0 0) 0 + 0 0 0 + This is not possible. 7
Therefore, there is no value of λ for which f is continuous at 0 t, f 4 + 4 + 5 ( ) lim 4 + 4 + 5 Thus, lim f f Therefore, for any values of λ, f is continuous at nswer 6: 5 Let 6 0 4 5 7 a 5 ( 8 0) 6 4 5 Thus, a. 5 0 nswer 7: Let {,, }. R {(, ), (, )}. We know that, (, ), (, ), (, ) R Hence, R is not refleive. 7
Now, s (, ) R and (, ) R, then R is symmetric. (, ) and (, ) R lso, (, ) R Thus, R is not transitive. Therefore, R is symmetric but neither refleive nor transitive. nswer 8: ( a + b) ( a b) Position vector of point R a + b a + b a+ 4b nswer 9: Total revenue, + 6 + 5 dr Marginal revenue, 6 6+ t 5, dr Thus, 6 5 6 + 66 Marginal revenue 66 R nswer 0: We have,. + 5 a 5 ( a) ( a) 5 a + ( a ) 5 + 4 74 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More
Section B nswer : P () and P (B) (i) Problem is solved independently by and B, Thus, ( ') P P ( ') P ( B) P B P. P( B) P B. 6 Probability (problem is solved) P ( B) ii) P ( B) P () + P (B) P (B) + 6 Probability (eactly one of them solves the problem). ( ') +. ( ') P P B P B P + + 6 75
nswer : + y z + 8 5 & + y z 7 4 4 y z + + y 4 z 5 and 7 4 a i + 7 j k a i + j + 4k ngle between the given pair of lines,cosθ a. a ( i + 7 j k ) i + 4 j + 4k + 7 4 4 + 8 8 4 4 a. a a a a a 7 + + 4 + 49 + 9 6 + + 4 + 4 + 6 4 cosθ 6 4 9 θ cos 4 9 76
OR (,4,) & B (5,,6) The equation of line passing throgh above points is y 4 z 5 4 6 y 4 z 5 XY plane means Z0 y 4 0 5 + 5 5 y + 4 5 5 So, Co-ordinates of the point where the line through the points & B crosses the XY-plane is,,0 5 5 nswer : I 0 0 0 0 0 0 R R R 0 0 0 5 0 0 0 0 77
R R 5 0 0 0 0 0 5 5 0 0 R R R 0 0 0 0 5 5 0 0 R R + R R R + R 4 0 5 5 0 0 0 5 5 0 5 5 R R + R 4 0 5 5 0 0 0 5 5 0 0 5 78 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More
R R R 5 R 5 4 0 5 4 5 0 0 5 5 0 0 5 5 0 0 0 0 5 5 0 0 5 5 5 R R R 5 5 5 R R R 0 0 0 5 5 0 0 0 0 5 5 0 5 5 0 0 0 0 5 5 0 0 5 5 5 5 5 5 R R 0 0 0 5 5 0 0 0 5 5 0 0 5 5 5 0 5 5 0 5 5 5 5 5 79
nswer 4: cos tan sin cot ( 4 ) cos tan sin cot 4 4 cos tan sin tan 4 cos tan sin tan π 4 cos tan cos tan π 4 cos tan cos tan On comparing On comparing π 4 tan tan π 4 tan tan 4 π tan + tan 4 π tan + tan 4 + tan 4 π + 4 tan π 4 + 4 tan + 4 π 4 tan π 4 + 4 + 4 π tan 4 π tan 4 + 4 + 44 4 4 0 4 0 4 4 80
nswer 5: Let y + tan Put tan ------------------------------- equation y y + tan tan tan sec tan tan y sec tan tan y cos si n tan sin si n cos y tan cos sin, sin sin cos y tan tan y Put value of from equation tan y Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More 8
On differentiation nswer 6: dy ( + ) dy cos + y tan dy + sec y sec.tan dy + Py Q P Q Where sec, sec tan P sec tan I. F e e e Now, multiplying the equation by I.F dy tan tan + sec y e e ( sec.tan ) tan dy tan tan e e sec y e sec.tan tan tan y. e e ( sec.tan ) Let tan t sec + dt y e t tan t. t. e dt... t Let I t. e dt t d t I t e dt t. e dt dt dt t I t. e e dt t t I t. e e + C 8
tan t t.. ( tan ) Now, equation becomes as y e t e e + C. tan. + tan tan tan y e e e C y Ce+ tan OR + dy + y cot dy cot + y + + dy + Py Q cot Where P, Q + + Now, + I.Fe P e... Equation + Let t dt... Equation Put value of equation in equation dt logt + I.Fe t e t Multiplying both sides by I.F + dy + y cot 8
d y. ( + ) cot On integrating both sides ( ) y + cot y + log sin C + nswer 7: Verte (0, 0) The equation of the parabola 4 ay... On differentiation 4 ay ' ay ' a y ' Put value of a in equation 4 y y ' ' y y y ' y y ' y 0 84 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More
This is the required differential equation. OR dy ( )., y 0, dy dy y Now, ( )( + ) B C + + + + ( ) +..( + ) +.. B c + + + B + + c By putting 0,, We get 85
, B, C + + B + C ( ) y + + + y + log log+ + log y log + log ( ) y log + C Now, we know that y 0, 0 log C + C log Now the equation becomes as; y log +log 86
nswer 8:, + y 5 + y 6 y 0 Points of shaded region are: (6, 0) B (4, ) C (, ) It can be seen that the feasible region is unbounded. Thus, Z is not the maimum value. Z has no maimum value. 87
nswer 9: Let I cos d ( cos ) I cos I cos I cos I cos I cos cos I cos I d cos cos cos.... I cos I equation. d + 88
cos cos Put value of in equation cos cos cos I I cos cos cos 4 4 cos 4 4 cos + C 4 4 I cos+ cos I cos I nswer 0: Equations of planes are: + y + z and + y + 4z 5 The equation of the plane passing through the intersection of the given planes is: ( + y + z ) + λ ( + y + 4z 5) 0 ( λ ) ( λ ) y ( λ ) z ( λ ) + + + + 4 + 5 + 0 The direction ratios of this plane are: a (λ + ) 89
b (λ + ) c (4λ + ) The given plane y + z 0 is perpendicular to the equation of desired plane Direction ratios of plane, y + z 0 are: a b c Both planes are perpendicular. Thus, a a bb c c + + 0 ( λ ) ( λ ) ( λ ) + + + 4 + 0 λ + 0 λ Equation of desired plane becomes as: + + + y + 4 + z 5 + 0 z + 0 z + 0 This is the required equation of the plane. nswer : 5 + + 4 + 0 d Let + ( 4 0 + ) + B + 5 + 4 + B + 5 + 4+ B+ 90 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More
On equating coefficients 5 5 4 B + 5 4 B + B 7 Now, 5 + + 4 + 0 5 ( + 4 ) 7 + 4 + 0 ( + 4) 5 7 + + + + 4 0 4 0 B... Equation 5 ( 4) ( + 4) 4 0 + + Let 4 0 + + t + dt Now, ( + 4) 5 4 0 5 + 4 + 0 5 dt t 5 ( 4 + 0 ) + + + 9
B 7 7 7 + 4 + 0 + 4 + 4 + 6 ( ) ( + ) + ( 6 ) 7log + + 4 + 0 + Put value of & B in equation B 5 4+ 0 + 7log + + 4+ 0 + C+ OR ( + )( + ) ( + )( + 4) ( 4 + 0) ( + )( + 4) ( 4 + 0) + B C + D Let ( + )( + 4) ( + ) ( + 4) + 4 + 0 + B + 4 + C + D + 4 + 0 + 4 + + 4 + + + + B B C C D D 4 + 0 + + + + 4 + + 4 + C B D C B D 9
Equating the coefficients of,,, and constant term, we obtain + C 0 B + D 4 4 + C 0 4B + D 0 On solving these equations, we obtain 0, B, C 0, and D 6 Now, ( 4 + 0) ( + )( + 4) 6 + + + ( ) ( 4) 6 + ( + ) ( + ) 6 + tan + tan tan tan + + C C nswer : Let the cottage industry manufacture pedestal lamps Wooden shades y s per the question, 9
The profit on a lamp Rs 5 Profit on the shades Rs Total profit, Z 5 + y The equations are: + y + y 0 0 y 0 Now, 94
The corner points are: (6, 0) B (4, 4) C (0, 0) The values of Z are: The maimum value of Z is at (4, 4). Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maimize his profits. Section C nswer : Let r radius of the cone h height of the cone Let V the volume of the cone V π r h 95
Height of the cone, h R + B h + R R r h + R R r V π r R R+ r V π r + R π r R r dv r π rr + π r R r + π r. dr R r dv π r π rr + π r R r dr R r ( ) dv π r R r π r π rr + dr R r dv π rr π r π r π rr + dr R r dv π rr π r π rr + dr R r dv dr 0 π rr π r π rr + 0 R r π rr π r π rr R r R r R R r 8 r R 9 d V dr < 0 96
Thus, by second derivative test, the volume of the cone is the maimum at r 8 R 9 h + R R r 8 h + R R R 9 8 h + R R 9 R 4R h R+ V V V π r h 8 4R π R 9 8 Vol of sphere 7 nswer 4: 5 4 4 + 4 + 6 + 4 + 5 6+ 5 0 97
0 9 5 Now, 0 adj 9 5 0 9 5 0 9 5 Now, y + 5z + y 4z 5 + y z 5 4 y 5 z 98
5 y 4 5 z 0 y 9 5 z 5 5 + 6 y 45 69 + z 5 + 9 y z, y, z OR Let the cost of onions per kg be Rs. Cost of wheat per kg be Rs. y Cost of rice per kg be Rs z Given equations are: 4 + y + z 60 + 4y + 6z 90 6 + y + z 70 4 60 4 6 y 90 6 z 70 X B X B 99
4 4 6 6 4 6 6 + 4 4 90 40 50 0 0 0 5 0 0 0 0 0 Thus, 0 5 0 adj 0 0 0 0 0 0 adj 0 5 0 0 0 0 50 0 0 0 Now, X B 0 5 0 60 X 0 0 0 90 50 0 0 0 70 450 + 700 y 800 400 50 z 00 + 900 + 700 00 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More
0 5 0 60 X 0 0 0 90 50 0 0 0 70 450 + 700 y 800 400 50 z 00 + 900 + 700 50 y 400 50 z 400 5 y 8 z 8 5, y 8, z 8 Thus, The cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg. nswer 5: The line, a, divides the area bounded by the parabola and 4 into two equal parts. rea OD rea BCD Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More 0
We can observe that the given area is symmetrical about -ais. Thus, rea OED rea EFCD rea of region OED is a 0 a 0 y a a 0 rea of region EFCD is 4 a 4 a 8 a Now, rea OED rea EFCD 0 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests
( a) 8 a ( a) ( a) ( a) 8 a 8 a 4 ( 4) nswer 6: Equations of planes are: + y + z and + y + 4z 5 The equation of the plane passing through the intersection of the given planes is: ( + y + z ) + λ ( + y + 4z 5) 0 ( λ ) ( λ ) y ( λ ) z ( λ ) + + + + 4 + 5 + 0 The direction ratios of this plane are: a (λ + ) b (λ + ) c (4λ + ) The given plane y + z 0 is perpendicular to the equation of desired plane Direction ratios of plane, y + z 0 are: a b c 0
Both planes are perpendicular. Thus, a a bb c c + + 0 ( λ ) ( λ ) ( λ ) + + + 4 + 0 λ + 0 λ Equation of desired plane becomes as: + + + y + 4 + z 5 + 0 z + 0 z + 0 This is the required equation of the plane. b b i+ b j+ b k c c i + c j + c k OR ( b + c) ( b ) + c i + b + c j + b + c k i j k a b + c a a a b + c b + c b + c 04 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More
a a i+ a j+ a k + + + + + + ( a ) b ac ab ac i ab ac ab ac j ab ac ab ac k... i j k a b a a a b b b + + ( a b a b ) i ( a b ab ) j ( ab a b ) k... i j k a c a a a c c c + + ( a c a c ) i ( a c ac ) j ( ac a c ) k... Now, add equation & a b + a c a b a c+ a b a c i ( ab + ac+ a b + a c ) j + ( ab + ac a b a c ) k Thus, LHS RHS a b + c a b + a c nswer 7: 4 ( ) 4 4 4 Let I I I B We know that b { } f ( ) lim h f ( a) f ( a h) f ( a h). f a n h a h 0 + + + + + + +... b a where h n 05
h n 4 lim h f () f. f ( n ) h 0 + + + + + n n ( n ) n lim + +. +... + +.. + n n n n n n lim n + + n n n {... ( ) } + ( +. +. ) n... n n ( ) 9 n n n 6 n n lim + n n + n n 6 n ( n )( n ) ( n ) 6 lim + n + n n n 6n 9n + lim n + n + n n n 6n 9n + lim + + n n n 9 lim n + + + n n n + + [ ] 06 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests
Now, B 4 lim h. ( n ) n + + + + n n n ( n ) n lim n + + + + + +... times n.... n( n ) lim n n + n n ( n ) lim n + n n n lim + n n lim n + n + 5 Thus, I B+ 5 I 7 I 07
nswer 8: Let the awardees for honesty Let the awardees for helping others y Let the awardees for keeping the colony neat & clean z s per the question,equations become as + y + z + y + z y + z 0 y z 0 X B 6+ + 4 [ ] [ ] [ ] 9 + 7 0 9 0 dj 0 7 ( adj) 9 0 0 7 08 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests nd Much More
Now, X B X B 9 0 X 0 7 0 08 99 + 0 X 0 0 + + 84 + 99 + 0 9 X 5 y 4 z 5, y 4, z 5 The awardees for honesty The awardees for helping others 4 The awardees for keeping the colony neat & clean 5 The other category can be responsibility. nswer 9: P(X ) P (, ) 6 P(X ) P (, ) + P (, ) 6 8 P(X 4) P (, ) + P (, ) + P (, ) 6 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests
P(X 5) P (, 4) + P (, ) + P (, ) + P (4, ) 4 6 9 P(X 6) P (, 5) + P (, 4) + P (, ) + P (4, ) + P (5, ) 5 6 P(X 7) P (, 6) + P (, 5) + P (, 4) + P (4, ) + P (5, ) + P (6, ) 6 6 6 P(X 8) P (, 6) + P (, 5) + P (4, 4) + P (5, ) + P (6, ) 5 6 P(X 9) P (, 6) + P (4, 5) + P (5, 4) + P (6, ) 4 6 9 P(X 0) P (4, 6) + P (5, 5) + P (6, 4) 6 P(X ) P (5, 6) + P (6, 5) 6 8 P(X ) P (6, 6) 6 Hence, required probability distribution is: X ip X i E (X) 5 5 + + 4 + 5 + 6 + 7 + 8 + 9 + 0 + 6 8 9 6 6 6 9 + 8 6 5 5 7 0 5 + + + + + + + + + + 8 6 9 6 6 9 6 8 0 Get CBSE Syllabus, NCERT Solutions, Sample papers, Practice papers, Guess Papers, Solved Question Papers, Online Tests 7 nd Much More
5 5 + + 4 + 5 + 6 + 7 + 8 + 9 + 0 + 6 8 9 6 6 6 9 + 8 6 5 5 7 0 5 + + + + + + + + + + 8 6 9 6 6 9 6 8 7 E (X ) X P ( X ) i i 5 5 4 + 9 + 6 + 5 + 6 + 49 + 6 + 8 + 00 + 6 8 9 6 6 6 9 + 44 8 6 4 5 49 80 5 + + + + 5 + + + 9 + + + 4 9 9 6 9 8 9 54.8 6 Var X E X E X 54.8 49 5.8 Standard deviation Var X 5.8.4 nd Much More