# CBSE-XII-2015 EXAMINATION. Section A. 1. Find the sum of the order and the degree of the following differential equation : = 0

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1 CBSE-XII- EXMINTION MTHEMTICS Pper & Solution Time : Hrs. M. Mrks : Generl Instruction : (i) ll questions re compulsory. There re questions in ll. (ii) This question pper hs three sections : Section, Section B, Section C. Section. Find the sum of the order nd the degree of the following differentil eqution : dy y d y dy d y y order of eqution degree of eqution hence sum of order nd degree. Find the solution of the following differentil eqution : ( y ) y ( ) dy By ( y ) y ( ) dy y dy y Integrting, we get y C y y cosθ sin θ. If, then for ny nturl number n, find the vlue of Det ( n ). sin θ cosθ cosθ sin θ Q sin θ cosθ Now. cos θ sin θ cos θ sin θ sin θ cos θ sin θ cos θ cos θ sin θ sin θcosθ cosθ sin sin θ cos θ θ Similrly we cn sy tht n cosnθ sin nθ sin nθ cosnθ sin θcosθ cos θsin θ dy Code : //, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

2 Now det ( n cosnθ ) sin nθ cos nθ sin nθ CBSE-XII- EXMINTION sin nθ cosnθ. Find vector of mgnitude which is perpendiculr to both of the vectors î ĵ kˆ nd b î ĵ kˆ. Let c be ny vector perpendiculr to nd b then c {( î ĵ kˆ ) ( î ĵ kˆ )} i j k c i j k c ( b) Now vector of mgnitude in the direction of c is given by mgnitude. ĉ (i j k) 9 i j k. Find the ngle between the lines y z nd y z. The eqution of stright lines re y z y z nd The given lines re prllel to the vectors i j k nd bi j k (i j k) nd b (i j k) Now ngle between these two vectors is given by cos θ. b b (i j k).(i j k) cos θ 9 9 cos θ 9 Hence both lines re perpendiculr.. In tringle OC, if B is the mid-point of side C nd O, OB b, then wht is OC? If O is origin then position vector of point, B nd C re, b nd w.r.t. origin then O, OB bnd OC c c nd if c divide B in the rtio :, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

3 CBSE-XII- EXMINTION then O OB OC b OC Section B cos. Evlute : sin cos I sin sin I sin I I. ( sin I. π Let tn t sec I π. sec ) sin tn t π I. (tn t ) π π I I π π I π sin cos. Evlute : sin sin cos I (sin cos ) I sin cos () (sin cos ) Let sin cos t (cos sin ) t, t, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

4 CBSE-XII- EXMINTION t π, t I t I t l n t we know tht I l n ln I ln I ln ln c 9. Let î ĵ kˆ, b î ĵ kˆ nd c î ĵ kˆ. Find vector d which is perpendiculr to both nd b nd c. d. Given tht d is perpendiculr to both nd b then d is given by d λ( b ) i j d λ k () d λ{i( ) j( ) k( )} d λ{i j k} () gin given tht c. d λ( î ĵ kˆ ).( î ĵ kˆ ) λ( ) 9λ λ Hence from eqution () the resultnt vector dis d (i j k). Find the shortest distnce between the following lines : r ( î ĵ kˆ ) λ( î ĵ kˆ ) r ( î ĵ kˆ ) µ( î ĵ kˆ ) Find the eqution of the plne pssing through the line of intersection of the plnes y z nd y z y z 9 nd is prllel to the line. From the given equtions, we observe tht first line psses through î ĵ kˆ nd prllel to α î ĵ kˆ nd second line psses through b î ĵ kˆ, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

5 CBSE-XII- EXMINTION nd prllel to β î ĵ kˆ Becuse both lines re prllel then shortest distnce between both lines re given by (b ) α SD α () where b î ĵ kˆ î ĵ kˆ î ĵ kˆ () (i j k) (i j k) SD 9 i k 9 9 Eqution of plne through the intersection of the given plnes y z nd y z 9 is y z λ( y z 9)...(i) ( λ) ( λ)y ( λ)z ( 9λ) y s per this plne is prllel to the stright line z y z Then norml of plne is perpendiculr to stright line b b c c ( λ) ( λ) ( λ) λ λ λ λ λ λ Now from eqution () the eqution of plne is y z ( y z 9) Multiply by we get y z y z 9 9y z. mn tkes step forwrd with probbility. nd bckwrd with probbility., Find the probbility tht t the end of steps, he is one step wy from he strting point. Suppose girl throws die. If she gets or, she tosses coin three times nd notes the number of 'tils'. If she gets,, or, she tosses coin once nd notes whether 'hed' or 'til' is obtined. If she obtined ectly one 'til', wht is the probbility tht she threw,, or with the die? Prob. of step forwrd p. Prob. of step bckwrd q. So by binomil distribution (p q) C.p C.p.q C.p.q C.p.q C.p.q C.q Prob. for mn so tht he is one-step wy from strting point will be C.p.q C.p.q.(.).(.).(.).(.).(.).(., CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

6 / MTHEMTICS CBSE-XII- EXMINTION, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: (.). (.). E E E is the event tht on throwing dice or comes So ) (E P E is the event tht on throwing dice,,, comes So ) (E P If is event tht on throwing coin only one til. Then for Prob. of ectly one til if he got,,, or is be P(E /) ) E ).P( / P(E ) E ).P( / P(E ) E ).P( / P(E P(/E ), P(/E ) P(/E ).... Using elementry row opertions (trnsformtions), find the inverse of the following mtri : If, B, C, then clculte C, BC nd ( B) C. lso verify tht ( B) C C BC. Given tht we write I So operte R R we get

7 / MTHEMTICS CBSE-XII- EXMINTION, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: operte R R R, we get 9 operte R R R nd R R R we get operte R R R nd R R R 9 Hence 9, B, C Now C 9 () BC () Now (B) C

8 Now dd () nd () 9 C BC ( B) C Now from () & () ( B) C C BC Hence proved CBSE-XII- EXMINTION () (). Discuss the continuity n differentibility of the function f() in the intervl (, ). f() Defined this function we get < f() > Now test continuity t nd t f() LHL t f( ) RHL t f( ) lim ( h) h lim h Q f() f( ) f( ) So f() continuous t Now t f() LHL t lim f( h) RHL t h h lim h lim f( h) lim ( h) h Q f() f( ) f( ) hence f() continuous t So we cn sy tht f() continuous in (, ) < Differentibility Q f() > f() & f() t f ( h) f () L f () lim h h ( h) lim h h f ( h) f () Rf () lim h h lim h h Lf () Rf (), CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

9 CBSE-XII- EXMINTION Hence f() not differentible t So we cn sy tht f() not differentible in (, ) d y. If (cos t t sin t) nd y (sin t t cos t), then find. Given tht (cos t t sin t) Diff. w.r. to t ( sin t t cos t sin t) t cost...() gin y (sin t tcos t) D.w.r. to t dy [cost cost t sin t] dy t sin t...() dy dy / then Now from () & () / dy t sin t t cost dy tn t gin diff. w.r. to d y sec t. Put from () y y d d cos t. sec t t t cost. If ( b) e y/, then show tht d y dy y Given tht ( b)e y/ e y/ () b Diff. w.r. to dy e y/ y b ( b) From (), CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: 9 /

10 CBSE-XII- EXMINTION dy b y b ( b) dy b y () b gin differentite w.r. to d y dy dy ( b).b b ( b) d y b d y ( b) dy y d y dy y Hence proved Now from equ. () sin cos. Evlute : ( sin ) Evlute : ( )( ) sin cos I ( sin ) sin cos I sin sin Let t cos sin I t I ln t c sin I ln c I ln c sin I ( )( ) I ( )( ), CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

11 CBSE-XII- EXMINTION ( )( ) I ( )( ) ( )( ) I ( )( ) I ( )( ) I I I () where I Let t I t I ln t C I ln ( ) C () Now I ( )( ) () Let ( )( ) B C () ( ) B( ) C( ) ( )( ) Compre numerte ( ) B( ) C( ) t t C C C t B C B B From (). ( )( ) I ( )( ) I ln( ) ln( ) tn () Now from equ (), () nd () I ln( ) ln ln( ) I ln( ) ln tn C. There re fmilies nd B. There re men, women nd children in fmily, nd men, women nd children in fmily B. The recommended dily mount of clories is for men, 9 for women, for children nd grms of proteins for men, grms for women nd grms for children. Represent the bove informtion using mtrices. Using mtri multipliction, clculte the totl requirement, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

12 CBSE-XII- EXMINTION of clories nd proteins for ech of the fmilies. Wht wreness cn you crete mong people bout the blnced diet from this question? The members of the two fmilies cn be represented by the mtri m w c F B nd the recommended dily llownce of clories nd proteins for ech member cn be represented by clories proteins m R w 9 c The totl requirement of clories nd proteins for ech of the two fmilies is given by the mtri multipliction FR B B Hence fmily required clories nd gm proteins nd fmily B requires clories nd gm proteins.. Evlute : tn tn π π Q tn tn Q tn tn π So tn tn tn tn tn tn tn tn tn tn tn tn tn tn tn tn tn, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

13 CBSE-XII- EXMINTION 9. Using properties of determinnts, prove tht b c Let b ( b)(b c)(c )( b c). c b c b b c c b c b c b c operte R R R R R R we get b c b c b c (b )(b (c )(c b) c) (b ) (c ) b c b c operte R R R (b ) (c ) b c c b c c (b ) (c ) (b c)( b c) c c (b ) (c ) (b c) ( b c) Hence proved Section C. n urn contins red nd blck blls. Two blls re rndomly drwn, without replcement. Let represent the number of blck blls drwn. Wht re the possible vlues of X? Is X rndom vrible? If yes, find the men nd vrince of X., CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

14 Red Blck CBSE-XII- EXMINTION (i),, (ii) yes, is rndom vrible it vries from to urn. No.of bll P() blck R () P() P(R ) P R B (b) P() P(R ) P R P(B ) P R B B (c) P() P(B ) P B (iii) Men ( ) P P P (iv) Vrince P ()( ) P(). P(). P() 9 () () () () (). Solve the following liner progrmming problem grphiclly. Minimise z y subject to the constrints y y y, y The objective function is z y The constrints re y () y () y (), y (), CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

15 CBSE-XII- EXMINTION (, ) B C(, ) D(, ) E(, ) The shded unbounded region is the fesible region of the given LPP. The vertices of the unbounded fesible. Region re B(, ) C(, ) nd D(, ) nd E(, ) t B(, ) z t C(, ) z t D(, ) z t E(, ) z The minimum of,,, is. We drw the grph of the inequlity y < The corresponding eqution is y < nd this psses through L, nd M, The open hlf plne of y < is shown in the figure. This hlf plne hs no point in common with the fesible region Minimum vlue of z nd occurs when nd y. Determine whether the reltion R defined on the set R of ll rel numbers s R {(, b) :, b R nd b S, where S is the set of ll irrtionl numbers}, is refleive, symmetric nd trnsitive. Let R R nd * be the binry opertion on defined by (, b) * (c, d) ( c, b d). Prove tht * is commuttive nd ssocitive. Find the identity element for * on. lso write the inverse element of the element (, ) in. Let R be the given reltion nd defined s R {(, b) :, b, R nd b s} where S is set of irrtionl number then (i) Refleive : Let R then R S So R therefore R is refleive (ii) symmetric Let, b R nd Let, b then Rb S, CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

16 CBSE-XII- EXMINTION So Rb Q S So br/ Q Rb br/ So R is not symmetric (iii) Trnsitive : Let, b nd c Now Rb S brc S nd Rc S R/ c Q Rb & brc R/ c Therefore reltion R is not trnsitive hence reltion nd is refleive but not symmetric nd not trnsitive. Let (, b), (c, d), (e, f) be ny three elements of (i) Commuttive : Given tht (, b) * (c, d) ( c, b d) (c, d b) (c, d) * (, b) Q (, b) * (c, d) (c, d) * (, b) Therefore * is commuttive (ii) ssocitive : By the given definition (c, d) * (e, f) (c e, d f) then (, b) * {(c, d) * (e, f)} (, b) * (c e, d f) (, b) * {(c, d) * (e, f)} ( c e, b d f) () Now (, b) * (c, d) ( c, b d) then {(, b) * (c, d)} * (e, f) ( c, b d) * (e, f) ( c e, b d f) () hence by () nd () (, b) * {(c, d) * (e, f)} {(, b) * (c, d)} * (e, f) So * is ssocitive (iii) Let (, y) be the identity element in (, b) * (, y) (, b) by definition (, b y) (, b) nd b y b nd y hence (, ) be the identity element of * Now let (p, q) be the inverse element of (, ) in of opertive * then by definition of inverse we sy tht (, ) * (p, q) (, ) ( p, q) (, ) p nd q p nd q so inverse element is (, ). Tngent to the circle y t ny point on it in the first qudrnt mkes intercepts O nd OB on nd y es respectively, O being the centre of the circle. Find the minimum vlue of (O OB). dy. Show tht the differentil eqution ( y) y is homogeneous nd solve it lso., CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

17 CBSE-XII- EXMINTION Find the differentil eqution of the fmily of curves ( h) (y k) r, where h nd k re rbitrry constnts.. Find the eqution of plne pssing through the point P(,, 9) nd prllel to the plne determined by the points (,, ), B(,, ) nd C(,, ). lso find the distnce of this plne from the point. P(,, 9) (, y, z ) ( ) C (,, ) (,, ) B (,, ) Eqution of the plne which is pssing through the point P is Where, n B C (It's norml of the plne) Q B OB O b î ĵ kˆ C OC O c î kˆ î ĵ kˆ n B C î( ) ĵ( ) kˆ ( ) î ĵ kˆ r. n. n Eqution of plne is r.( î ĵ kˆ ) ( î ĵ 9 kˆ ).( î ĵ kˆ ) r.( î ĵ kˆ ) y z y z () Distnce between the point nd the plne () is () ( ) () ( ). If the re bounded by the prbol y nd the line y m is sq. units, then using integrtion, find the vlue of m., CP Tower, Rod No., IPI, Kot (Rj.), Ph: - Website : Emil: /

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