Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions XII - CBSE BOARD CODE (65//MT) SET - Date: 8.3.5 MATHEMATICS - SOLUTIONS. Let a iˆ 3iˆ kˆ b iˆ ˆj and a b 3 5, b a b Projection of a on b b i ˆ 3 ˆ j kˆ i ˆ ˆ j 3 5 5. iˆ ˆj kˆ ˆj. k ˆ iˆ iˆ j Now ˆj kˆ i, kˆ iˆ ˆj, iˆ ˆj kˆ ˆ ˆ iˆ iˆ ˆj ˆj k k 3
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions 3. 3 4 z 5 3 4 z 5 3 4 3 4 3 4 3 4 3 4 z 4 3 3 3 direction cosines are 3, 4, 3 3 3 4. a b a c b c 5. d d d order = order = d order degree = = 4 6. A cos B sin d d A sin B cos d A cos B sin d d A cos B sin d d d d d 7. Let a b c d a b 3 7 8 9 c d 4 5 6 4 6 3 3 a 4b a 5b 3a 6b 7 8 9 c 4d c 5d 3c 6d 4 6
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions a 4b 7, a 5b 8, 3a 6b 9 or a 4b 7... a 5b 8... a b 3... 3 and c 4 d, c 5d 4, 3c 6 d 6 c 4 d... 4 c 5d 4... 5 c d... 6 Solving equation () and () b a Solving equation (4) and (6) d c OR 3 A 5 6 5 5 A 3 3 5 3 3 3 5 6 5 Minors M, M, M 5 6, 3 M 3 5 5, M 6 5, M 5 5, 3 M 3 3, M 6 5, M 8 5 3, 3 3 33 Cofactor C, C, C, 3 C 5, C, C, 3 C, C, C 3, 3 3 33 5 Cofactor matri of A 3 Adjoint A 5 3 3 3
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions A Adj A 5 A 3 Now, 3 A A 5 5 6 5 I 3 5 A A I Hence proved 8. f 3 4 f 3 4 3 3 4 3 4 3 4 4 f 7, 3, 3 4 7, 4 Differentiabilit of 3 L.H.D. lim 3 f f 3 3 lim 3 7 3 lim 3 6 3 3 RHD lim 3 3 lim 3 f f 3 3 lim 3 3 L. H. D R. H. D 4 4
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions f is not differentiable at 3 Differentiabilit at 4 L.H.D. lim 4 4 f f 4 4 lim 4 R.H.D. lim 4 f f 4 4 lim 4 lim 4 7 4 4 4 L.H.D R.H.D function is not differentiable at 4 9. e Taking logarithms on both sides log e log ifferentiating both sides w.r.t. d d e log e d e log e d d e log. e d d d e e log 5 5
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions log tan log tan differentiating both sides w.r.t. OR d d d d d d d d d d d d d d d d d d d d. If d d d d d d d...() d d d 4 6 6
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions Diff. w.r.t. d d d d d d d 4 d Dividing b d d d d d Hence proved. d 4 cos. cos cos d cos cos d cos cos d cos cos cos cos cos cos d d cos cos d sec d cos sec sec d tan log sec tan c log sec tan tan c 7 7
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions. According to the given condition 3 3 5 4 6 3 3 3 4 45 4 5 6 8 3 3 5 5 3 Dail epense of famil A = Rs. 5 Dail epense of famil B = Rs. 5 Dail epense of famil C = Rs. 3 More children in famil hamper the growth of the societ. 3. tan tan z tan tan z cot tan z cot cot z z z z z z 8 8
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions 4. a b c b c a c a b C C C C3 a b c b c a b c c a a b c a b a b c c a b a R R R & R R R 3 a b c c a a b c b b c c a a a b c b c a b c a b a b c ab b ca bc c a ca a b c a b c ab bc ca a b c a b b c c a a b c a b b c c a So, a b c or a b b c c a Here a b b c c a a b c, a b c 9 9
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions 5. z 3 3,, z Point 3,,...() 4 z &...() 3 4,, z 3 If line () and line () are intersecting then 3 4... 3... 4 3... 5 from equation (4), from equation (5) In equation (3) L.H.S. = 3 4 R.H.S. = 4 4 L.H.S. = R.H.S. & eist Thus line are intersecting Point of Intersection 3,, 4,, 6. r i ˆ ˆj kˆ 3i ˆ ˆj 5kˆ Let r iˆ z kˆ iˆ ˆj z k ˆ iˆ ˆj k ˆ 3iˆ ˆj 5k ˆ i ˆ ˆ j z k ˆ 3 i ˆ ˆ j 5 k ˆ 3,, z 5 z 3 5 3,, z 5 The point represents Q on line is 3,, 5 P 3,, 6 direction ratios of PQ 3, 3, 5 4 Since the above line is parallel to the plane 4 3 z drs of the normal to this plane are, 4, 3 The normal of the plane is perpendicular to the given PQ
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions a a b b c c 3 4 3 3 5 4 3 4 5 8 8 4 OR A P B (3,, ) (, 4, 3) Point P is 3 4 3,,,, The normal of the plane is parallal to the line AB its direction ratio will be a 3 b 4 6 c 3 4 equation of plane in cartesian form is z 6 4 4 6 6 4 z 4 6 4 z 6 3 z 3 equation of plane in vector form is r iˆ 3 ˆj kˆ 3
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions 7. S,, 3,... ns Let A : Event that the number on the card is divisible b 6. B : Event that the number on the card is divisible b 8. C : Event that the number on the card is divisible b 4. A B C 6,,8, 4, 3, 36, 4, 48, 54, 6, 66, 7, 78, 84, 9, 96 n A 6 8, 6, 4, 3, 4, 48, 56, 64, 7, 8, 88, 96 nb 4, 48, 7, 96 nc 4 A BC n A B C 6, 8,, 6, 8, 3, 3, 36, 4, 4, 54, 56, 6, 64, 66, 78, 8, 84, 88, 9 P 6 or 8 but not 4 5 8. Let I sin d sin d d sin d sin d d d sin d sin d sin d sin d
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions sin d 9. sin sin sin c sin sin c sin sin c 4 4 e d We know that : as the limit of a sum a h f ( ) d lim h f ( a) f ( a h) f ( a 3 h) f ( a ( n ) h)...() b a We know that h, nh b a n n Here f ( ) e a a f ( a) a e a e. e ( a h) a h f ( a h) ( a h) e a h ah e. e. e ( a h) a 4h f ( a h) ( a h) e a 4h 4 ah e. e. e h ( a ( n ) h) f ( a ( n ) ) ( a ( n ) h) e a ( n ) h f ( a ( n ) h) a ( n ) h a ( n ) h e. e. e put all the values of f ( a), f ( a h), f ( a h) f ( a ( n ) h) in equation () a a h a a h f ( ) d lim h a e. e a h ah e. e. e 4 4 a. 4h. ( ) ( ) ( ) q. n h a h ah e e e a n h a n h e e e h f ( ) d lim h na h 3 ( n ) ah 3... n a h 4h ( n ) h e. e e e e 3 3
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions a h n ( ) n ( n ) (n ) n( n ) e a f ( ) d lim h na h ah e. e h h 6 e a q f ( ) d lim h na n ( nh h) ( nh h) ah n.( nh h) e. e h nh ( e ) h e a a nh nh( nh h) ( nh h) a( nh)( nh h) e. e h( e ) e d lim nha h h 6 e We know that nh b a, and h, n a lim ( b a) ( b a) (( b a)) e d b a a h 6 Here a, b. a a b a b a e e e ( ba) ( )( ). ( ) ( ) ( ).( ) ( )( ) e. e 4 e lim e h 6 e 4 8 e ( e ) e e 6 3 6 5 e 8 e e 3 5 OR I tan sec cosec d sin cos cos sin d I sin d...() sin d sin I d...() 4 4
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions Adding equation () and () f f I sin d sin d cos d sin sin I I 4. Slope d d m ( ). ( ) ( ) ( ) and dm ( ). ( ) ( ). ( ). 4 d ( ) ( ) ( ) ( ) 4 ( ) ( ) [3 ] 4 ( ) 3 (3 ) ( ) 5 5
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions Now consider dm d (3 ) 3 ( ) (3 ) ; 3; 3 3 3 d m ( ) (6 6) ( 6 ) 3( ) 6 d ( ) 6( ) ( ) ( 3) 4 ( ) d m d at 6 () ( ) 6 d m 6 ( 3) (3 ) d 4 ( 3) at 3 6 4 3 4 4 64 6 d m 6 ( 3) (3 ) 3 d 6 4 ( 3) at 3 Slope is maimum when = Now Point,. d d This is a homogeneous equation Put v d dv v d d equation () becomes dv v v d v...() dv v v d v 6 6
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions dv d v v v dv v v v d v dv d v v v dv v d d dv v Integrating on both sides dv v d v log v log c log log log c log log log c log e c e ce c This is the general solution of the given equation. OR d sin tan d d tan d sin sin d sin cosec d cos sin cos d d cosec cos 7 7
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions d cosec d sec This is the linear differential equation in the form of d p Q d where P cosec Q sec log tan Pd cosec d I. F e e e tan I. F Solution is tan I. F Q I. F d c sec d c tan tan sec d c tan tan Put tan t in R. H. S sec d dt tan t dt c tan t dt c tan t c tan t c tan tan c...() where then 4 8 8
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions in equation () tan c 4 c solution is tan tan tan tan. Vector equation of plane is given as r.( iˆ ˆj kˆ ) 6 and r.(iˆ 3 ˆj 4 kˆ ) 5 The cartesian form of the above equations are 3 6 and 3 4z 5 Equation of plane passing through line of intersection of planes is s s ( z 6) ( 3 4z 5)...() Since the above plane passes through point (,, ) It should satisf the above equation. ( 6) ( 3 4 5) putting the value of ( 3) (4) 3 4 3 in equation () we get 4 3 ( z 6) ( 3 4z 5) 4 4 4 4z 84 6 9 z 5 3 6z 69 therefore the equation of plane in vector form is r.(iˆ 3 ˆj 6 kˆ ) 69 9 9
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions 3. a a is an even number ( a, a) R a A R is refleive Again ( a, b) R a b is even ( b a) is even b a is even ( b, a) R R is smmetric. Net, let ( a, b) R and ( b, c) R a b is even and b c is even a b is even and b c is even ( a b) ( b c) is even a c is even a c is even ( a, c) R R is transitive. Hence, R is an equivalence relation. 3, 3 5 and 5 4 are all even, therefore, all the elements of {, 3, 5} are related to each other Again 4 is even, therefore, all the elements of {, 4} are related to each other. However,, 4 3, 3, 3 4, 5 3 and 5 4 are all odd, therefore, no element of the set {, 3, 5} is related to an element of {, 4}.
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions 4. 3 ( ) if 3 ( ) if 3 if Plotting,, 3 and 3 Area of shaded region d d 3d 4d (3 ) d (3 ) d ( ) d ( ) d 3d d 3d d d d d d 3 3 4 ( 3) (6 ) 4 ( )
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions 3 3 6 3 7 6 4 sq. unit 5. Let he purchases items of A and items of B. Acc. to the condition 5 5 5,...() 6...(),...(3), (4) z 5 5 Dividing equation () b 5 we get 5
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions (, ) z 5 5 (, ) z 5() 5() (, 6) z 5() 5(6) 9 (, 5) z 5() 5(5) 5 (, ) z 5() 5() Maimum Thus he should purchase items of and 5 items of. OR Let and be the number of packets of food P and Q respectivel. Obviousl,. Mathematical formulation of the given problem is as follows: Minimise Z 6 3 (vitamin A) subject to the constraints 3 4 (constraint on calcium), i.e. 4 8... () 4 46 (constraint on iron), i.e. 5 5... () 6 4 3 (constraint on cholesterol), i.e. 3 5... (3),... (4) Let us graph the inequalities () to (4). The feasible region (shaded) determined b the constraints () to (4) is shown in figure and note that it is bounded. The coordinates of the corner points L, M and N are (, 7), (5, ) and (4, 5) respectivel. Let us evaluate Z at these points : Corner Point Z 6 3 (, 7) 8 (5, ) 5 Minimum (4, 5) 85 From the table, we find that Z is minimum at the point (5, ). Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if 5 packets of food P and packets of food Q are used in the special diet. The minimum amount of vitamin A will be 5 units. 3 3
Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions 6. Let us define the following events: E = taking course of mediatation and oga. E = taking a course of drugs A = the patient gets a heart attack. Here P E ; P E 4 3 8 P A / E 4 5 3 P A/ E B Baes theorem P E / A P E P A / E / / P E P A E P E P A E 8 8 3 8 58 4 9 4 4