Marking Scheme. Section A 3. 2 [1] l m n 1 n 1 cos [1] Direction ratios of the given line are 2, 1, 2.

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1 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 [/] 5A 5 5, ki k k [/] A 5A ki

2 k 0 k 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 d 4 4 d t

3 So that, d dt t dt t C ( 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 [/] 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 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) P(A B) 0. P(B) OR P(A B) P(A B) P(A) P(B) P(A B) P(A B) P(A B) 0.95

5 5. Let y R, then for any, f() y if y y Section C () ()() () 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() 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 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 7 tan cos 8 4 sin tan cos tan 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 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 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

10 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 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

11 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

12 . 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

13 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

14 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) A eists. [/] Cofactor matri of A [] 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

15 5 t X (A ) B t t (A ) (A ) 4 X X , y, z 4 A IA OR A R R R A R ( ) R R R 5R A R R R A R R A R ( )R

16 6 6 0 R R R A R R R R R 9R A R R 5R A 0 0 Hence, A 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 ()

17 7 4V 6 0 V / V d C 8V when,, 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

18 8 y 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 sin 0 0

19 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 a b c 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( ) units [] 4 OR

20 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 ) 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 y 0 0y 80 ()

21 0 0y 0 () 0 0y 50 (), y 0 Corner Points O (0, 0) A (0, 5) B (6, ) C (8, ) D (9, 0) P y (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.

22 P(E ) P(E ) P(E ) P(A E ) P(A E ) P(A E ) P(E )P(A E ) P(E A) P(E ) P(A E ) P(E )P(A E ) P(E )P(A E ) P(E A) P(E A) P(E A)

MATHEMATICS. Time allowed : 3 hours Maximum Marks: 100

MATHEMATICS. Time allowed : 3 hours Maximum Marks: 100 MATHEMATICS Time allowed : 3 hours Maimum Marks: 00 General Instructions:. All questions are compulsory.. This question paper contains 9 questions. 3. Questions 4 in Section A are very short-answer type