Sr. No. Marking Scheme (Mathematics XII 07-8) Answer Section A., (, ) A A: (, ) A A: (,),(,) Mark(s). -5. a iˆ, b ˆj. (or an other correct answer). 6 6 ( ), () ( ) ( ). Hence, is not associative. Section B 5. sin sin sin sin sin 6 6 6. [ + ½] 5 9 0 5 5 8 P [ ½ ] 5 7. [ ½ ] Let cos. Then,, 0,, cos The given epression on LHS cos sin cos cos cos( ) cos cos( ) 0 RHS [ ½ ] 8. d [ / ] Let. Then. d d d ( ) 0.00 0.0005. d [ / ] decreases b 0.0005. 9. sin (sin cos ) [ / ] e d e d cos cos [ / ] sin cos e ( ) d e (sec sec tan ) d cos cos e sec c e ( f ( ) f ( )) d e f ( ) c
0. a b a b 0 a b 0 () a b[ ] 0 a b[ ] () Substituting this value, for a in the equation (), we get, b[ ] b 0 [ ]. Hence verified. a iˆ ˆj kˆ, b iˆ ˆj kˆ, a b 5, b 6. The required Projection (vector) of a on b a b b b 5 ( ˆ ˆ ˆ ). 6 i j k. P( A B) P( B / A) P( A B) 0. 0.6 0. PA ( ) P( A B) 0. P( A / B) PB ( ) 0.80 0 Section C. [ / ] [ / ] [ / ] [ / ] [ ] ( ). [ ] ( ). Since, f is differentiable at, f is continuous at. Hence, lim f ( ) lim( ) lim ( ) lim( ) f a b a b f() = As f is continuous at, we have a + b =... () f ( h) f () a( h) b Lf () lim lim h0 h h0 h a ah ah b lim lim( ah a) (using ()) h0 h h0 = a f ( h) f () ( h) Rf () lim lim. h0 h h0 h As f is differentiable at, we have a =, i. e., a = and b =. [ / ] [ / ] [ / ] [ / ] [ / ] [ / ]
sin lim f( ) lim 0 0 sin( a ) sin lim 0 sin( a ) ( a ) a ( a ) sinb e lim f( ) lim 0 0 b sinb e sin b lim 0 sin b b f(0) =. For the function to be continuous at 0, we must have lim f ( ) lim f ( ) f (0) i.e., we must have a 0; a 0 0 b ma be an real number other than 0. 5. log( ) log( ) [log( ) log ] [ ] () ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) (using ()) ( ) ( ). Hence, proved. 6. When = 0, we have ( ) ( + + ) ( ) = 0, i.e., = or. d ( ) 6 d d ( )(,0) d d ( )(,0) 7. d The required equations of the tangents are 0 = - ( ) or, = - + and 0 = 7( ) or, = 7. ( ) Domain f = (, ) f( ). ( ) ( ) ( )( ) f ( ) 0 0 [ as (, )]. ( ve)( ve) In (-, 0), f ( ) ve. Therefore, f is strictl decreasing in (-, 0]. ( ve)( ve) In (0, ), f () ve. Therefore, f is strictl increasing in [0, ). [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ ]
7. We have C( ) 5 600,0 0. For the time being we ma assume that the function C() is continuous in [0, 0]. C( ) 90 600 ( 0)( 0) C( ) 0 if 0 or 0. But, 0, 0 (0,0). Therefore, the maimum or the minimum value will occur at the points. C(0) 500, C(0) 000. Hence, the person must place the order for 0 trees and the least amount to be spent = Rs 000. Value: The person cares for a health environment despite being economicall constrained. [] [] [] [] 8. sec sin sin cos d d d cos ec cos ( sin ) ( sin ) ( sin ) t dt [sin t cos d dt] ( t) ( t) t A B C t A( t)( t) B( t) C( t) ( t) ( t) t ( t) t (an identit) Put t = -, - = B, i.e., B = -½. Put, t =, = C, i.e., C = ¼. Put t = 0, 0 = A + B + C, which gives A =. [ + / ] Therefore the required integral dt dt dt t ( t) ( t) log t log t c t log sin log sin c sin 9. The given differential equation is e d ( e ) d, (0) or, e d d or, ( ), which is linear in. e d d e ( ) I. F. = e d e log Multipling both sides b the I. F. and integrating, we get, e d e c e c (the general solution). When = 0, =. 0 e c c. Hence, the required particular solution is e e. e [ + /] [ / ] [ / ]
d The given differential equation is ( ) d ( ) d or, f ( ), d hence, homogeneous. Put = v d dv v. d d The equation becomes dv v v or, v dv d d v v v v d v d or, dv or, [ ] dv v v v v v v ( ) v d Integrating, we get dv dv v v ( v ) ( ) v or, log( v v ) tan log c or, log( ) tan c (the general solution). 0. a ( a b c) a 0 ab ac 0 ab c a b ( a b c) b 0 b c abab b c c a a b c a ( b c) a ( a b) 0 [As the scalar triple product of three vectors is zero if an two of them are equal.]. General point on the first line is (,, ). General point on the second line is (,, ). Direction ratios of the required line are,,. Direction ratios of the same line ma be,,. Therefore, () ( ) k sa k, ( ) k, ( ) k k, k, k 9 k 9,, which satisf (). Therefore, the direction ratios of the required line are 6,0, or,,0,7. Hence, the required equation of the line is z. 0 7 [ ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ] [ / ]
. Let E = Bag I is chosen, E = Bag II is chosen, E = Bag III is chosen, A = The two balls drawn from the chosen bag are white and red. P( E ) P( E) P( E), P( A E ), P( A E), P( A E). 6 6 9 9 B Baes s Theorem, the required probabilit = P( E) P( A E) 9 9 6 P( E A). 99 P( Ei) P( A Ei) 6 6 9 9 i. Let X denotes the random variable. Then X = 0,,. 6 6 c 60 c c c P( X 0), P( X ), P( X ). 0 0 0 c 95 c 95 c 95 [ ] [ + /] i p i i p i i p i 0 60/95 0 0 [ / ] /95 /95 /95 /95 6/95 /95 total 8/95 /95 Mean = i p i i 8 95 5 [ / ] Variance = ( ). 95 5 75 i pi i pi i i [ +/ ] Section D. f g: defined b f g( ) f ( g( )) f ( 5) ( 5) 7 Let, ( D f g ) such that f g( ) f g( ) 7 7. Hence, f g is oneone. Let ( Codomain f g ). Then for an f g () = if + 7 =, i.e., if, = 7, i.e., 7 =, 7 Hence, for ever ( Codomain ), f g ( Df g ) which ( D f g ). 7 such that f g( ). Hence, f g is onto. Since, f g is both one-one and onto, it is invertible. [ ] [ / ]
( f g) : defined b ( ) (9) 9 f g 7. Let ab, such that a = 0, b 0. ( f g) ( ) 7 Then ab a b 0 b b, ba b, ab ba Let ab, such that a 0, b = 0. Then ab a, ba b a 0 a a, ab ba Let ab, such that a = 0, b = 0. Then ab a 0, ba b 0, ab b a. Now we need to check whether is commutative. One more case is needed to be eamined. Let ab, such that a 0, b 0. Then ab a b, ba b a and a b ma not be equal tob a, e.g., (-) =, (-) =, hence, (-) (-). Thus is not commutative. The element e will be the identit element for if a e = e a = a for all a. ae a provided e = 0 and ea a provided e = 0 (As 00 0 and 0a 0 a a for a 0). Hence, 0 is the identit element for. 5. A ( 6) ( )( ) ( 7) 6 0. Hence, A eists. Let cij represent the cofactor of (i, j) th element of A. Then, c, c 6, c 9, c 9, c 6, c 5, c 5, c, c. 9 5 adja = 6 6 9 5 9 5 A 6 6 6 9 5 The given sstem of equations is equivalent to the matri equation AX B, where X, B. z 0 X ( A ) B ( A ) B 6 9 6 9 6 5 6 Hence, =, =, z = 6 6 5 0 6 [ / ] [ ] [ ] [ ] [ ]
0 0 0 0 0 0 0 0 A 0 0 A ( R R) 0 0 0 0 0 0 0 0 0 0 0 A ( R R R ) 0 0 0 0 0 0 0 0 A ( R R R ) 0 0 0 0 0 0 A ( R R R, R R R ) 0 0 A X 0 A 0 0 0 6., 0., 0 Solving, simultaneousl, 0 ( )( ) 0 ( lies in quadrant I). [] [/] [] The required area = the shaded area = ( ) 0 d [ sin ] 0 [ ] 0 ( ) sq units. 6 [ ] [ +/ ]
7. The given definite integral = I = d d cos cos f ( ), f ( ) f ( ). cos cos Hence, f is odd. Therefore, d 0 cos g ( ), ( ) ( ). cos g cos g Hence, g is even. Thus d d. cos cos Hence, I = d cos sin 0 0 0 d sec sec d d dt t d dt tan tan tan t 0 0 0 0 ( ) t dt 6 [tan t] 0 [ ] 6 8 Let f() = +. Then the given definite integral = f ( ) d lim h f ( rh), where nh. n h0 n r f rh rh rh r h rh ( ) ( ) ( ) 0 [tan sec ] n( n )(n ) n( n ) f rh h r h r n h h n 6 n n n ( ) 0 0 r r r n( n )(n ) n( n ) f ( ) d lim h[h h 0 n] n 6 h0 h0 nh( nh h)( nh h) nh( nh h) lim[ 0 nh] n 6 ( h)(8 h) lim[ 7 ( h) 0 ] h0 [ ] [ ] [ + /] [ + /]
8. The general point on the given line z 9 is (,, 9 ). The direction ratios of the line parallel to the plane + z - =0 intersecting the given line and passing through the point (-,, -) are,, 9 and ( ) ( )( ) ( 9 ) 0. 7 The point of intersection is (,, ). 7 59 The required distance = ( ) ( ) ( ) unit. 9. Let = the number of units of Product to be produced dail = the number of units of Product to be produced dail To maimize P = (9.) + (8 0.9) = 7.8 + 7. subject to the constraints: 90, or 080, 80, or 8 90, 00, 0, 0. 8 [ ] [ ] At the point P (0, 0) 0 (00, 0) (0, 0) 70 (00, 0) 560 (80, 0) 5 The maimum profit = Rs..