Marking Scheme Section A. B. AB 6 A B 6. sin( ) cos( ) or sin( ). 4. l m n n cos 45 or 6 4 OR Direction ratios of the given line are,,. [/] Hence, direction cosines of the line are:,, or,, [/] Section B 5. An element (e, f)z Z bethe identity element, if 6. (a, b) * (e, f) (a, b) (e, f) * (a, b) (a, b) Z Z [/] i.e., if, (af + be, bf) (a, b) (eb + fa, fb) i.e., if, af + be a eb + fa and bf b fb () i.e., if, f, e 0 () Hence, (0, ) is the identity element. [/] A A 8 5 5 [/] 5A 5 5, ki k 0 5 0 0 k [/] A 5A ki
7. 7 0 k 0 k 7 0 7 0 k ( ) (sin ) I cos sec d sec d [/] sec d [/] 8. tan tan c e ( ) d ( ) ( ) e d [/] ( ) e e [f () f ()]d e f () c) [/] ( ) ( ) e ( ) C OR 4 4 ( ) I 5 Put d 4 4 d 4 5 4 5 d 4 4 d t
So that, d dt 4 5 4 4 4 t dt t C 4 4 5 5 9. ( 0) (y r) r [/] + y ry () Differentiating both sides w.r.t., we get yy ry r yy y () Substituting r from () in (), we get ( y )y y( yy ) [/] 0. ˆi ˆj kˆ a b 4 i ˆ 4j ˆ 4k ˆ ab 46 6 6 [/] Area of the parallelogram a b sq units. [/] OR The angle between the vectors a and b is given by ab cos [/] a b
4 i.e., cos (i ˆ ˆj k) ˆ (i ˆ ˆj k) ˆ () () ( ) () ( ) () i.e., i.e., cos. cos cos [/]. P(A B) P(B) P(A B) [/] P(B) P(A) P(B)[ A and B are independentevents] ( P(A)) P(B) P(A) P(B) Since, P(A B) P(A ) P(B) Therefore A and B are independent events. [/]. P (Red transferred and red drawn or black transferred and red drawn) P(A B) 0. 7 5 6 8 8 5 88 P(A B) 0. P(B) OR P(A B) 0.5 0. 0.5 P(A B) P(A) P(B) P(A B) P(A B) 0.6 0.5 0.5 P(A B) 0.95
5. Let y R, then for any, f() y if y y Section C 9 6 5 () ()() () 5 () y ( ) 6 + y 6 y 6 y 6 y 6 [0, ) foranyvalueofy Now, for y 6 R, [0, ) Hence, f() is not onto f() is not invertible. Since, 0, therefore y 6 0 i.e., y 6 i.e., y + 6 y 5 Redefining, f :[0, ) [ 5, ) makesf() 9 6 5 an onto function Let, [0, ) such that f( ) f( ) ( + ) ( + ) [( + ) + ] [( )] 0 (as ( + ) + > 0) f() is one-one Thus, f() is bijective, hence f is invertible and f :[ 5, ) [0, ) y 6 f (y)
6 Refleive: OR R is refleive, as + a.a + a > 0 (a, a) R Symmetric: If (a, b) R then, + ab> 0 + ba> 0 (b, a) R a R Hence, R is symmetric. Transitive: Let a 8, b, c Since, + ab + ( 8) ( ) 9> 0 (a, b) R also, + bc + ( ) 0 (b, c) R But, + ac ( 8) 0 Hence, R is not transitive. [] 4. sin tan cos tan 4 Lets evaluate, sin tan Put tan 4 4 tan 4 Now, sin tan 4 8 tan 7 4 To evaluate cos(tan ), put tan
7 tan cos 8 4 sin tan cos tan 4 7 5 5. a b c c b LHS a c b c a C a b b a c ac a b c c b a ca b c a a a ab b a c C C bc cc a b c b c c b a b c b c a a a b c b a c Taking (a b c ) common from C b (a b c ) c c b b c a a b a c R R R R R R b c c b (a b c ) 0 c a b a 0 a c b (a b c ) ( bc a ac ba bc) a (a b c ) (a b c) R.H.S
8 6. Let u ln u sin sin ln du u d sin ln cos du d sin sin cos ln Let v sin( ) () Put t log t ln dt t d ln dt d ( ln ) () dv d dt cos( ) (where t ) d dv d cos ( ) ( ln ) () (using ()) Since, y u + v Therefore, dy du dv d d d dy d sin sin cos ln ( ln ) cos ( ) OR y y y 4 log( t t ) log ( t ) log( t ) dy dt (t) 4t t t tan t
9 d dt t 4t dy dy / dt t 4t d d / dt (4t) d dt t d y d dt d 4 ( t ) 4( t ) 7. y cos(m cos ) dy m sin(m cos ) d Squaring both sides, we get dy m sin(m cos ) d dy ( ) m sin (m cos ) d Differentiating again, dy ( ) m ( y ) d dy d y dy dy ( ) ( ) m y d d d d 8. The given curve is d y dy ( ) m y 0 d d y 4 5 Let the required normal be at (, y ) dy Slope of the tangent d
0 m slope of the normal m Slope of the line 9 Since normal is perpendicular to the line. Therefore, m m 9 dy d (, ) 9 ± Hence, the points are (, 6) and (, 4) Equations of normals are: and y 6 ( ) i.e., + 9y 55 9 y 4 ( ) i.e., + 9y 5 9 9. I 4 ( ) d 4 ( ) d ( ) Put t So that d dt (t ) dt I t(t ) Now, t A B C t(t ) t t (t ) t A(t ) Bt(t ) Ct () On comparing the coefficients of like terms in (), we get A, B 0, C
I t (t ) dt (t ) log t C log C 0. I d Now, d d I + I (say).() I Let f() d f( ) f () ( ) f() is odd function. Hence, I 0 () Also, Let I d g() g( ) ( ) g( ) g() g() is even function I d 0 0
. ln ln ln 0 I ln () From (), () and (), we get I ln cos yd ( e )sin y dy 0 d e sin y dy cos y e sin y d dy e cos y ln(e ) ln cos y lnc ln(e + ) ln cos y C e C cos y () e Ccos y e kcos y Substituting 0, y 4 in (), we get + kcos 4 k e cos y is the particular solution. d dy y tan y tan y y y tan y d dy y tan y OR I.F e cot y dy y ln ylnsin y e
I.F Solution of the D.E. is: ln(ysin y) e I.F (QI.F)dy ysin y ysin y ysin dy ysin y y( cos y) ( cos y)dy ysin y ycos y sin y C sin y ycos y C ysin y. Let r ai ˆ bj ˆ ckˆ be the required vector. Since, r q therefore, a b c 0 () Also, p, q and r are coplanar. [p, q r] 0 0 a c 0 () a b c Solving equation () and () i.e., a b c 0 0 a b c a b c r i ˆ j ˆ k ˆ r Unit vector r ˆi ˆj kˆ ˆr r Required vector 5 rˆ 5(i ˆ ˆj k) ˆ
4. Vector equation of the line passing through (,, ) and (, 4, ) is r a (ba) where a ˆi j ˆ kˆ and b i ˆ 4j ˆ kˆ r ˆi j ˆ k ˆ ( 4i ˆ ˆj) () [] Equation of z-ais is Since ( 4i ˆ ˆj) kˆ 0 r kˆ () line () is to z-ais. Section D 4. A 0 A ( ) () ( 4) 6 8 7 0 A eists. [/] Cofactor matri of A 4 7 7 5 [] 7 A Adj A 7 5 A 7 4 Now for given system of equations. 0 y 4 z 5 t (A )X B t X (A ) B
5 t X (A ) B t t (A ) (A ) 4 X 7 4 7 7 5 5 4 X 7 7 68 4, y, z 4 A IA OR 0 0 5 0 0 A R R R 0 0 6 0 5 0 0 A R ( ) R 0 0 6 0 R R 5R 5 0 0 A R R R 0 0 6 0 0 9 5 5 A R R 0 5 0 6 0 0 5 0 A R ( )R 0 9 5 5
6 6 0 R R R 0 5 0 A R R R 0 9 5 5 0 9 0 R R 9R 0 5 0 A R R 5R 0 0 0 0 7 9 0 0 0 5 7 A 0 0 Hence, 7 9 0 A 5 7 5. Let the length and breadth of the base. Also let the height of the godown y. Let C be the cost of constructing the godownand V be the given volume. Since cost is proportional to the area, therefore C k[ 4y], where k 0 is constant of proportionality () y V(constant) () V y () Substituting value of y from equation (), in equation (), we get C V 4V k 4 k dc 4V k 6 d For maimum or minimum value of S dc 0 d ()
7 4V 6 0 V / V d C 8V when,, 6 8 0 d / / V (8V) Cisminimumwhen and y 6. The given curves are y () y () Solving equation () and (), we get y y y, y (as y 0) Substituting value of y in () we get () + 9 i.e., () and () intersects at (9, ) Required Area (y )dy y dy 0 0
8 y 9 9 9 The given curves are y y 0 9 sq units. OR y 8 () y () Solving () and () 8 y y y, 4 y (as y > 0) Substituting y in () we get 4 or Required Area 8 d d ( ) d d 0 0 8 8 sin 0 0
9 4 4 0 8 0 8 4 4 sq. units 7. The two given lines are and y 4 z 4 () y z 4 () Let a, b, c be the D.R s of the normal to the plane containing the line (). Therefore, equation of plane is a( ) b(y 4) c(z 4) 0 () a b c 0 (4) (Required plane contains line ()) a 4b c 0 (5) (line () is parallel to the required plane) a b c 8 4 4 a b c 6 7 6 Putting, a 6, b 7, c 6 in (), we get 6 ( ) 7 (y 4) 6 (z 4) 0 6 7y 6z 98 0, which is the required equation of the plane Since line () is parallel to required plane SD between two lines Perpendicular distance of the point (, ) from the plane. i.e., SD SD 6( ) 7() 6( ) 98 6 7 6 9 units [] 4 OR
0 The given line y z () is coplanar with the line determined by the planes y z 8 0 () and y 4z 0 (), if we are able to show there eists a plane passing through intersection of planes () and () containing the line (). Equation of the plane passing through the intersection of planes () and () is ( y z 8) k( y 4z ) 0 (4) [] We find, value of k for which the plane given by (4) passes through the point (,, ) lying on line (). Substituting the coordinates of the point (,, ) in (4), we get ( 8) + k( 4 ) 0 4 0k 0 k 7 0 Putting, 7 k in (4) we get 0 7 ( y z 8) ( y 4z ) 0 0 4 y z 0 (5) [] Now we find value of aa bb cc, where a, b, c and D.Ratios of the line () and a, b, c and D. Ratios of the normal to the plane (5) aa bb cc (4) () ( ) i.e., aa bb cc 0 which implies line () lies in plane (5) Hence the two lines are coplanar and the equation of the plane containing them is 4 y z 0 8. Let the manufacturer make and y quantity of toy A and toy B respectively. Subject to Ma P 50 60 y 0 0y 80 ()
0 0y 0 () 0 0y 50 (), y 0 Corner Points O (0, 0) A (0, 5) B (6, ) C (8, ) D (9, 0) P 50 + 60y 0 00 480 50 (Ma) 450 [] Hence, Ma Profit is ` 50, at 8 and y 9. Let A be the event thatcar delivered to firm needs service and tuning. Also let E, E and E be the events ofcar being rented from agencies X, Y and Z respectively.
50 0 0 P(E ) P(E ) P(E ) 00 00 00 9 0 P(A E ) P(A E ) P(A E ) 00 00 00 P(E )P(A E ) P(E A) P(E ) P(A E ) P(E )P(A E ) P(E )P(A E ) 0 0 00 00 0 P(E A) 50 9 0 0 0 0 00 00 00 00 00 00 P(E A) P(E A) 0 0 0 8