CBSE Eamination Papers (Foreign 0) Time allowed: hours Maimum marks: 00 General Instructions: As given in CBSE Sample Question Paper. Set I SECTION A Question numbers to 0 carry mark each.. Write the principal value of tan tan 9. 8. Write the value of sin sin. 5. If A is a matri, whose elements are given by aij i j, then write the value of a.. If A is a square matri and A =, then write the value of AA, where A is the transpose of matri A. 5. If A = 0 7, then write A. 6. Write the differential equation formed from the equation y = m + c, where m and c are arbitrary constants. 7. If a is a unit vector and ( a) ( a) =, then write the value of. 8. For any three vectors a, b and c, write the value of the following: a (b c) b (c a) c (a b) 9. Write the cartesian equation of a plane, bisecting the line segment joining the points A (,, 5) and B (, 5, 7) at right angles. 0. If C = 0.00 + 0.0 + 6 + 50 gives the amount of carbon pollution in air in an area on the entry of number of vehicles, then find the marginal carbon pollution in the air, when vehicles have entered in the area and write which value does the question indicate.
8 Xam idea Mathematics XII SECTION B Question numbers to carry marks each.. Prove that the relation R in the set A = {5, 6, 7, 8, 9} given by R = {(a, b) : a b, is divisible by }, is an equivalence relation. Find all elements related to the element 6.. If tan tan, then find the value of., then prove that sin y = tan. If y = cot cos tan cos. Using properties of determinants, prove the following: a ab ca b ab cb c ac bc. Differentiate the following with respect to : sin cos a b c (sin ) 5. If y = sin (log ), then prove that d y dy y 0 d d 6. Show that the function f() = is continuous but not differentiable at = 0. Differentiate tan with respect to tan, when 0. sin cos 7. Evaluate: 9 6 sin d Evaluate: log ( ) d 8. Evaluate: 0 tan sec tan d 9. The magnitude of the vector product of the vector i j k with a unit vector along the sum of vectors i j 5k and i + j + k is equal to. Find the value of. 0. + Evaluate: d 5 + 6. Find the shortest distance between the following lines: y z y 5 z 7 ; 7 6 Find the equation of the plane through the points (,, ) and (,, ) and perpendicular to the plane y + z = 0.
Eamination Papers 0 8. In a group of 50 scouts in a camp, 0 are well trained in first aid techniques while the remaining are well trained in hospitality but not in first aid. Two scouts are selected at random from the group. Find the probability distribution of number of selected scouts who are well trained in first aid. Find the mean of the distribution also. Write one more value which is epected from a well trained scout. SECTION C Question numbers to 9 carry 6 marks each.. 0 students were selected from a school on the basis of values for giving awards and were divided into three groups. The first group comprises hard workers, the second group has honest and law abiding students and the third group contains vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives, while the combined strength of first and second group is four times that of the third group. Using matri method, find the number of students in each group. Apart from the values, hard work, honesty and respect for law, vigilance and obedience, suggest one more value, which in your opinion, the school should consider for awards.. Prove that the volume of the largest cone, that can be inscribed in a sphere of radius R is 8 7 of the volume of the sphere. Show that the normal at any point to the curve = a cos + a sin, y = a sin a cos is at a constant distance from the origin. 5. Find the area enclosed by the parabola y and the line y = +. 6. Find the particular solution of this differential equation dy y d y cos, 0. Find the particular solution of this differential equation, given that when =, y =. 7. Find the image of the point having position vector i j k in the plane r ( i j k ) 0. Find the equation of a plane which is at a distance of units from origin and the normal to which is equally inclined to the coordinate aes. 8. An aeroplane can carry a maimum of 00 passengers. A profit of `500 is made on each eecutive class ticket out of which 0% will go to the welfare fund of the employees. Similarly a profit of `00 is made on each economy ticket out of which 5% will go for the improvement of facilities provided to economy class passengers. In both cases, the remaining profit goes to the airline s fund. The airline reserves at least 0 seats for eecutive class. However at least four times as many passengers prefer to travel by economy class than by the eecutive class. Determine how many tickets of each type must be sold in order to maimise the net profit of the airline. Make the above as an LPP and solve graphically. Do you think, more passengers would prefer to travel by such an airline than by others? 9. Often it is taken that a truthful person commands, more respect in the society. A man is known to speak the truth out of 5 times. He throws a die and reports that it is actually a si. Find the probability that it is actually a si. Do you also agree that the value of truthfulness leads to more respect in the society?
8 Xam idea Mathematics XII Set II Only those questions, not included in Set I, are given 9. If p is a unit vector and ( p) ( p) = 8, then write the value of. 0. Write the principal value of tan tan 7. 6 9. Differentiate the following with respect to : (sin ) + ( cos ) sin 0. Find a vector of magnitude 6, perpendicular to each of the vectors a b and a b, where a i j k and b i j k.. Prove that the relation R in the set A = {,,,..., } given by R = {(a, b) : a b is divisible by }, is an equivalence relation. Find all elements related to the element.. Evaluate: d. 8. Find the area of the region bounded by the parabola y and the line y =. 9. Show that the differential equation ( y) dy = ( + y) is homogeneous and solve it. d Set III Only those questions, not included in Set I and Set II are given. 9. Write cot, > in simplest form. 0. If a is a unit vector and ( a) ( a) = 9, then write the value of. 9. + Evaluate: d. 5 + 6 0. Using properties of determinants, prove the following: a ab b b a ab a b a b a b ( a b ). Find a unit vector perpendicular to each of the vectors a b and a b, where a i j k and b i j k.. Differentiate the following with respect to : sin sin tan, 0 sin sin 8. Find the area of the region bounded by the two parabolas y = a and = ay, when a > 0. 9. Show that the differential equation y e / y d y e / ( y ) dy 0 is homogeneous. Find the particular solution of this differential equation, given that when y =, = 0.
Eamination Papers 0 85 Solutions. tan tan 9. Let sin sin sin = tan tan 8 8 = tan tan = 8 8 5 5 sin Set I SECTION A 8, sin 5 9 5 = sin sin sin { } sin 6 sin sin sin 5 5 5 sin sin 5 5. a 6 =. AA' = A. A' = A. A= A = =. [Note: AB A. B and A, where A and B are square matrices.] 0 5. Here A = 7 T 7 7 0 Adj A = 0 Also A 0 0 A 7 0 A = 7 0 7 0 A T 6. Here y = m + c dy Differentiating, we get m d Again, differentiating we get d y d 0, which is the required differential equation
86 Xam idea Mathematics XII 7. Given: a. a =.. a a. a. a = a ( ) 5 a 5 8. a ( b c) b ( c a) c ( a b) = a b a c b c b a c a c b = a b a c b c a b a c b c = 0 9. One point of required plane = mid point of given line segment. = 5 5 7,, = (,, 6) Also D.r s of Normal to the plane =, 5, 7 5 =,, Therefore, required equation of plane is ( ) ( y ) ( z 6) 0 y z 6 or y z 0. We have to find i.e. [ C ( )] Now C() = 0.00 + 0.0 + 6 + 50 C ( ) 0. 009 0. 0 6 C ( ) 0. 009 9 0. 0 6 = 0.08 + 0. + 6 = 6.0 This question indicates how increment of vehicles increase the carbon pollution in air, which is harmful for creature.. Here R is a relation defined as SECTION B R {( a, b): a b is divisible by } Refleivity Here (a, a) R as a a = 0 = 0 divisible by i.e., R is refleive. Symmetry Let (a, b) R ( a, b) R a b is divisible by a b m b a m b a is divisible by (b, a) R Hence R is symmetric Transitivity Let (a, b), (b, c) R Now, (a, b), (b, c) R a b, b c are divisible by a b m and b c n
Eamination Papers 0 87 a b b c ( m n) ( a c) = k k m n ( a c) k ( a c) is divisible by ( a, c ) R. Hence R is transitive. Therefore, R is an equivalence relation. The elements related to 6 are 6, 8.. Refer to Q, Page 9. Given y cot cos tan cos y tan cos tan cos y tan cos y y cos sin cos sin y cos cos sin y tan Note: tan cot, R sin cos, [, ] and tan cos, 0. Refer to Q 6, Page 0.. Refer to Q 8, Page 88. 5. Refer to Q 56, Page 98. 6. Here f() = For continuity at = 0 cos sin sin y cos cos cos lim f( ) lim f( 0 h) lim f( ) lim f( 0 h) 0 h0 = lim f( h) h0 0 h0 = lim f( h) h0 = lim = lim ( h) h h h h 0 = lim( h h) h h0 = lim h h 0 h 0 = lim h h 0 = lim( h) h0 = 0...(i) = 0...(ii)
88 Xam idea Mathematics XII Also f(0) = 0 0 = 0...(iii) (i), (ii) and (iii) lim f( ) = lim f( ) f( 0) 0 0 Hence f( ) is continuous at 0 For differentiability at 0 f L.H.D. = lim ( 0 h ) f ( 0 ) f( h) f( 0) = lim h0 h h0 h = lim ( h ) h 0 0 h h 0 = lim h0 h h0 h h = lim = lim h 0 h h0 L.H.D. =...(iv) f Again R.H.D. = lim ( 0 h ) f ( 0 ) h0 h f = lim ( h ) f ( 0 ) h h 0 0 = lim h0 h h 0 h h h h = lim = lim h 0 h h 0 h = lim. h0 R.H.D. =...(v) From (iv) and (v) L.H.D. R.H.D. Hence, function f() = is not differentiable at 0 i.e., f() is continuous but not differentiable at 0. Let u tan v tan We have to find du dv Now, u tan Let tan tan tan sec u tan = tan tan tan = tan cos sin = tan cos. cos cos sin cos
Eamination Papers 0 89 u = sin cos = tan = tan sin sin cos sin = tan = tan tan = cos tan Differentiating, both sides w.r.t. we get du...(i) d ( ) Also, v tan dv d du dv ( )...(ii) 7. Let sin cos I = sin d 9 6 Let sin cos z (cos sin ) d dz Also (sin cos ) z (sin cos ) z (sin cos sin. cos ) z sin z sin z dz I ( z ) = dz 5 6z = dz 6 =.. log 5 6 5 z 5 5 z C z = 5 log z C = 5 (sin cos ) log 0 5 z 0 5 (sin cos ) Let I = log( ) d = log( ).. d = log( ). d = log( ) ( ) d C
90 Xam idea Mathematics XII. log( ) d = d = log( ) log( ) C 9 6 8. Refer to Q. 8, Page 99. 9. Let a i j k ; b i j 5 k ; c i j k From question a b c b c a ( b c) b c b c ( ) i 6j k b c ( ) 6 ( ) a ( b c) = = 6 = i j k 6 = ( 6) i ( ) j ( 6 ) k = 8i ( ) j ( ) k Putting it in (i), we get 8i ( ) j ( ) k = Squaring both sides we get 6 6 8 6 8 = 8 8 0. Let I = d 5 6 ( 8) ( ) ( ) 96 = 5 5 = d 5 6 5 6 6 5 5 5
Eamination Papers 0 9 = d = I = I 5 5 Now I = d ( )( ) 5 5 d = 5 6 5 5 d = ( ) ( ) 5 5 Let = ( )( ) = A B If = 5 = B B = 5 If 0 = A A = 0 5 5 ( )( ) = 5 0 I = 0 5 d 5 5 d ( 6) 5 5 d ( )( )...(i) = 0 log 5 log C I = 0 log 5 log C (using (i) ) 5 5 = A ( ) B( ). Refer to Q, Page 7. Let the equation of required plane be a( ) b ( y ) c( z ) 0 (i) passes through (,,) also a ( ) + b ( ) + c ( + ) = 0 a + b + 5c = 0 Again, plane (i) is perpendicular to plane y z 0 a b c 0 From (ii) and (iii) a b c 8 0 5 6 a b c 8 7 a 8, b 7, c Putting in (i) we get 8 ( ) 7 ( y ) ( z ) 0 8( ) 7( y ) ( z ) 0 8 7y z 6 7 0 8 7y z 9. Let X be no. of selected scouts who are well trained in first aid. Here random variable X may have value 0,,.
9 Xam idea Mathematics XII Now P( X 0 ) = P( X ) P(X = ) = 0 C 50 C 0 C 50 0 C 50 C 0 9 8 50 9 5 0 C 0 0 C 50 9 0 9 50 9 Now distribution table is as 87 5 = 0 5 X 0 P() 8 5 0 5 87 5 Now Mean = i p = 0 8 0 87 i 5 5 5 = 0 7 = 9 5 5 5 A well trained scout should be disciplined. Let no. of students in I st, nd and rd group to, y, z respectively. From question y z 0 y y z 0 The above system of linear equations may be written in matri form as AX B where 0 A 0, X y, B z 0 A = 0 Now A = ( ) ( 0) ( 8 0) ( ) 8 5 0 0 0 0 A = ( ) ( 8 0) 8; A = ( ) A = ( ) ( ) 5; A = ( ) 5 A = ( ) ( ) 0; A = ( ) 0 0 A = ( ) ( 0 ) ; A 0 = ( )
Eamination Papers 0 9 8 Adj A = 5 5 0 T 5 8 5 0 Now AX = B X = A B. 5 0 y 8 5 5 z 0 0 5 y z y z 5 0 65 80 65 0 5, y, z 5 5 5 0 Apart from these values, the school should consider disciplined behaviour for awards.. Refer to Q, Page 5. Given a cos a sin y a sin a cos d = d a sin a ( cos sin ) = a sin a cos a sin = a cos dy and d = a cos a ( sin cos ) = a cos a sin a cos = a sin dy dy d = d a sin tan d a cos d Slope of tangent at tan Slope of normal at = cot tan Hence equation of Normal at is y ( a sin a cos ) cot ( a cos a sin ) y a sin a cos cot cot ( a cos a sin ) 0 cos cos y a sin a cos a a cos 0 sin sin cos y sin a 0...(i) Distance from origin (0, 0) to (i) = 0.cos 0.sin a = a cos sin 5. Refer to Q, Page 9. 6. Given dy y = cos d y, 0
9 Xam idea Mathematics XII dy y d y cos sec y Dividing both sides by y sec dy y.. d d y tan d Integrating both sides w.r.t. we get. d y d d d tan tan y C y tan C For particular solution when =, y =, we have tan C y tan + = C C = Hence Particular Solution is tan y 7. Refer to Q 6, Page 50. Since, the required plane is at inclined to the coordinate aes. dy d y C unit distance from the origin and its normal is equally d and Normal vector of required plane = li mj nk where l cos, ; m cos, ; n cos, n (normal unit vector of plane) i j k = = i j k = i j k
Eamination Papers 0 95 Hence equation of required plane r. n d r. i j k = r ( i j k ) = ( i yj zk ).( i j k ) 9 y z 9 8. Let there be tickets of eecutive class and y tickets of economy class. Let Z be net profit of the airline. Here, we have to maimise z. 80 75 Now Z 500 00y 00 00 Z = 00 00y...(i) According to question 0...(ii) Also y 00...(iii) 00 5 00 0...(iv) Shaded region is feasible region having corner points A (0, 0), B (0,0) C (0, 60), D (0,80) Now value of Z is calculated at corner point as Corner points Z = 00 + 00y (0, 0) 8,000 (0, 0) 6,000 (0, 60) 6,000 (0, 80) 60,000 Hence, 0 tickets of eecutive class and 60 tickets of economy class should be sold to maimise the net profit of the airlines. Yes, more passengers would prefer to travel by such an airline, because some amount of profit is invested for welfare fund. 9. Let E, E and E be three events such that E = si occurs E = si does not occurs E = man reports that si occurs in the throwing of the dice. Now P (E ) = 6, P (E ) = 5 6 Maimum
96 Xam idea Mathematics XII P E, P E E 5 E 5 5 We have to find P E E P E P E P E ( ). E = E P( E ). P E P( E ). P E E E = 6 6 5 5 5 6 5 = 0 0 5 = 9 Everybody trust a truthful person, so he receives respect from everyone. 9. Given ( p).( p) 8 0. tan tan 7 SET II.. p p. p. p 8 8 9 7 = tan tan 6 6 9. Let u sin and v (cos ) sin = tan tan = 6 6, 6 Given differential equation becomes y = u + v dy du dv...(i) d d d Now u (sin ) Taking log on both sides, we have log u log sin Differentiating w.r.t., we get du. u d. cos log sin sin du u( cot log sin ) d du (sin ) cot log sin d...(ii) Again v (cos )
Eamination Papers 0 97 Taking log on both sides we get log v sin. log cos Differentiating both sides w.r.t., we get dv. sin. ( sin ) log(cos ). cos v d cos dv sin v cos. log cos d cos sin sin (cos ) cos. log(cos ) cos dv d sin (cos ) {log(cos ) tan } From (i), (ii) and (iii) dy sin (sin ) { cot log sin } (cos ) {log(cos ) d 0. a b i j k a b j k Now vector perpendicular to ( a b) and ( a b) is i j k 0 Required vector = 6 = ( 6 ) i ( 0) j ( 0) k = i j k ( i j k ) ( ) ( ). We have the relation R = ( a, b): a b is divisible by 6 ( i j k ) 6 ( i j lk ) 6 ( i j k ) 6 We discuss the following properties of relation R on set A. Refleivity For any a A we have a a 0 which is divisible by ( a, a) R a R So, R is refleive Symmetry Let (a, b) R a b is divisible by a b = k [where k n]...(iii) tan }
98 Xam idea Mathematics XII a b k b a k b a is divisible by b, a R So, R is symmetric Transitivity Let a, b, c A such that (a, b) R and (b, c) R. a b is divisible by and b c is divisible by a b = m and b c = n m, n N a b = m and b c = n ( a b) ( b c) ( m n) a b b c ( m n) a c ( m n) a c ( m n) a c is divisible by ( a, c) R. So, R is transitive Therefore, R is an equivalence relation.. Refer to Q. 0, Page 87. 8. Given curves are y...(i) and y...(ii) Obviously, curve (i) is right handed parabola having verte at (0, 0) and ais along +ve direction of -ais while curve (ii) is a straight line. For intersection point of curve (i) and (ii) ( ) 8 6 0 6 0 8 6 0 ( 8) ( 8) 0 ( 8)( ) 0, 8 y, Intersection points are (, ), (8, ) Therefore, required Area = Area of shaded region = ( y ) dy y dy = ( y ) y = 6. 6 6 8 = 0 7 6 9. Refer to Q. 8, Page 79. = 8 sq. unit X' Y O Y' (, ) (8, ) X
Eamination Papers 0 99 9. cot Let sec sec Now, cot = cot = cot SET III sec = tan cot cot = = sec 0. Given: ( a).( a) 9 6 a 6. a 9 a 9 9 9 00 5 9. Let I = d 5 6 0 9 = d [ 5 6 ] 5 6 0 0 9 = d d 5 6 0 9 = d 6 0 9 = d ( ) ( ) 0 9 = d ( )( ) = d d d = log log C 0. Refer to Q 5, Page 0.. Given a i j k b i j k 0 9 0 9 A B ( ) ( ) 0 9 A ( ) B ( ) Putting we get B Putting we get A
500 Xam idea Mathematics XII a b = ( i j k ) ( i j k ) = 5i 6j k a b ( 6i j k ) ( i j k ) = 7i 6j k Now, perpendicular vector of ( a b) and ( a b) = i j k 5 6 7 6 ( ) i ( 0 ) j ( 0 ) k = i j k = ( i j k ) ( i Required unit vector = j k ) ( ) ( ). Let y tan = i j k i j k sin sin sin sin cos sin cos sin = tan cos sin cos sin cos = tan = tan cot sin = tan tan y = 8. Refer to Q 8, Page 9. 9. Refer to Q. 6 CBSE (Delhi) SeI-I. dy d 0 = 0 0 0,,