π = tanc 1 + tan x ...(i)

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Leon Rodgers
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1 Solutions to RSPL/ π. Let, I log ( tn ) d Using f () d f ( ) d π π I log( tnc d m log( cot ) d...(ii) On dding (i) nd (ii), we get +,. Given f() + ), For continuit t lim " lim f () " ( ) \ Continuous t. I log( tn cot ) d log d I π lim f () f( ) " + lim ( + ) " + +, true π π $ <. A cos α sin α cos α sin α G G sin α cos α sin α cos α., cos α sin α sinαcos α sinαcos α > H sinαcos α+ cosαsin α sin α+ cos α cos α sinαcos α G cos α sin α G sinαcos α cos α sin α cos α If _ i \ cos q 5. Given re sq units \ k ± q π ( k) + () ± 8 + k ± k ± + 8 k, k 6, Mthemtics ()

2 cos sin 6. Consider cos + sin d sec c π m ( ) d d + sec c d m π 7. Let r e rdius of sphere + tn tn r 9 cm nd Dr. cm To find DS, where S is surfce re of sphere 8. Given f() + + π tnc m ds d DS Dr (pr ) Dr 8pr Dr dr dr DS 8p 9. 8p.7.6p cm f () + Given intervl is (, ), i.e. < < dd, < < < < + < + < + + > f () > + [From (i)]...(ii) For incresing f () > \ For incresing t lest cb,, c re coplnr One vector cn e written s liner comintion of other two. \ λ + µ c, for some sclrs l, m.. P(A B) p, P(A B) r P(ectl one occur) P(A B) + P(A B) d d. Given eqution + cosd n d d P(A) P(A B) + P(B) P(A B) P(A B) P(A B) p r Order is nd degree is not defined s eqution cnnot e written s polnomil of differentils.. Consider t d + t tdt + ( t + ) dt t + dt dtg t + (t tn t) + C 6 + C Let t d t dt Mthemtics ()

3 . Consider tn d n+ tn d n+ tn d n tn d n nn ( + ) ( n+ ) n tn d n+ tn d n+ tn d n tn d n nn ( + ) (tn tn ) + (tn tn ) + (tn tn ) [tn (n + ) tn n] tn tn + tn tn + tn tn tn (n + ) tn n tn (n + ) tn ( n + ) tn < F tn d + ( n + ) n Hence, tn d n tn n + n n +.. Given A G 7 5 Consider A + I A n n + n G G+ G G G+ G G , 8, 56 7, + 5 8, 8 \ We hve A + 8I 8A Post multipl A, we get A A + 8IA 8AA A(AA ) + 8A 8I AI + 8A 8I A + 8A 8I A 8 (8I A) Consider G G A 8 ) G G G 5 G A G 5 G G B A C I OR A B IC B C B G B nd Adj B G Mthemtics ()

4 \ B Adj B G G...(ii) B C 7 G 5 C Adj C G \ C Adj C G G...(iii) C From (i), (ii) nd (iii), we hve 5. Consider cos q nd sin q A G G G G + 8 \ d d θ cos q ( sin q), d d θ sin q cos q sin q cos q \ d d d ' d d q dq sin qcos q tn q cos qsin q d d d ( tn q) sec dθ q d d sec q e cos o qsin q d d H π sec π cosec ^ h π θ OR sec q cosec q. Consider log d + n log d n [log log( + )] + d + < F + log d n < F+ logd n d + + ( + ) + d d + d d + Differentiting oth sides w.r.t., we get d d d + d d + d ( + ) ( + ) d d ( + ) Mthemtics ()

5 d d d d ( + ) d n + d d n [From (i)] d 6. Hence, d d f p d n d d log d ( + log ) t ( + t) t e dt t e ) dt + t ( + t) e t C + t + elog + C + log + OR log Let log t e t d e t dt + C 9Using e {() f + fl( )} d e f () + CC Consider + [ log( + ) log ] d + + > logf ph d + loge + o d t log t dt log t > t t dth t Let + t d dt t log t+ t dt t log t+ t + C 9 log e + C o e + o+ e + + o 9 7. Here,,, f(), h nh n ( ) d lim h[f() + f( + h) + f( + h)] f( + n h) h "...(ii) f() () 8 f( + h) ( + h) ( + h) ( + h + h) h h + h + 8 f( + h) ( + h) ( + h) ( + h + 8h) h h + h + 8 f( + n h) { + (n )h} { + (n )h} { + (n ) h + (n )h} (n )h (n ) h + (n )h + 8 Mthemtics (5)

6 Sustituting in (ii), we get lim h[(8) + (h + h + 8) + (h + h + 8) {(n ) h + (n )h + 8}] h " lim h[h { (n ) } + h{ (n )} + 8n] h " lim h " h ( n ) n( n ) ( ) h h n < $ + n + 8nF 6 ] nh hg( nh)( nh h) lim < + 5( nh h)( nh) + 8nhF h " ( h)( )( h) lim < + 5 ( h)( ) + 8 F [From (i)] h " Consider d ( + )d d + d d dv Let v v + d d From (i), dv v + d v + v dv v + v d v+ v dv v + v dv cv + m c m v + log v + + log v v + d d log + log C log C C + C...(ii) v C v + + Given when From (ii), C + + is prticulr solution. Mthemtics (6)

7 d 9. Consider eqution cos d Here, P(), Q() cos Integrting fctor (I.F.) e $ d e Solution is (I.F.) "(. IF.) Q (), d e Consider I e cos d e cos d e sin sin e $ d e cos cos sin + < e $ d n e $ d d n F I e sin 9 e cos 9 I I e ( sin cos ) 9 9 I e ( sin cos ) Sustituting in (i), we get e e ( sin cos ) + C ( sin cos ) + Ce is required solution.. Generl point on the line B(k, 6k +, k) DR s of AB: k, 6k +, k i.e., k 5, 6k, k + z k(s) is 6 If AB is prllel to the plne + + z 5 then (k 5) + (6k ) + (k ) 9k 5 + k + k 8 5k 5 k From (i) coordintes of point B re (, 6 +, ), i.e. B(, 8, ) A(,, ) + + z 5 \ Distnce AB ( ) + ( 8 ) + ( ) + 5+ units + z 6 B Mthemtics (7)

8 . Generl eqution of plne through the points (,, ) is ( ) + ( ) + c(z ) ( ) + + cz This plne psses through the point (,, ) (ii) nd plne (i) mkes ngle π with plne + \ $ + $ + c $ π cos + + c ( + ) + + c + + c c c c [from (ii)] c ±...(iii) Sustituting in (i) from (ii) nd (iii), we get Plne is ( ) + ± z + ± z + ± z DR s of norml to the plne re,, ±. A: even numer ; B: odd numer A : B : \ P(A), P(B) Proilit of r, successes is p, q, n r P(r) C r d n r d n ; r,,, P() C d n d n ; P() 7 C d n d n P() C d nd n ; P() 8 7 C d n d n 7 Tle for proilit distriution nd clcultion of men is s follows: X P(X) X P(X) / PX ( ) / X P( X) 7 7 Men SX P(X) Mthemtics (8)

9 . S : getting si P(S) 6, P(S) 6 5 P(A) P() + P() + P(5) +... P(S) + [P(S)] P(S) + [P(S)] P(S) d n + d n P(B) P() + P() + P(6) +... P(S) P(S) + [P(S)] P(S) + [P(S) 5 ] P(S) $ + d n $ + d n $ Decision of referee to sk cptin of tem A to strt first is not fir s whosoever strts first will hve more chnces of winning. + n n. Let P(n) : A n G n n P() : A + G G A, true Let P(k) e true, i.e. + k k A k G k k To show P(k + ) is true, i.e. + ( k + ) ( k + ) k + k A k + G G k + ( k + ) k + k k Consider A k + A k A + k k k G G [from (i)] + 6k k 8k+ k k + k G G k+ k k + k k + k \ P(k) true P(k + ) is true, P() is lso true, so sttement is true for ll positive integers the principle of mthemticl induction. Consider, f() + OR + f() [ ] + + [Performing C C + C ] Mthemtics (9)

10 \ f() f() [() + () + ] [ + + ] 5. Given A { W : } nd R {(, ) : is multiple of } For refleive: Let for A (, ) R is multiple of Hence, refleive. For smmetric: Let for, A For trnsitive: Let for,, c A is multiple of, true. (, ) R is multiple of ( ) is multiple of is multiple of \ (, ) R ( ) R. Hence, smmetric. (, ) R nd (, c) R is multiple of nd c is multiple of ( ) is multiple of nd ( c) is multiple of ( ) + ( c) c is multiple of c is multiple of (, ) R \ (, ) R, (, c) R (, c) R. Hence, trnsitive. As reltion R is refleive, smmetric nd trnsitive. \ R is n equivlence reltion. Set of elements relted to (, ) R \ [] {, 6, } is multiple of l + l, 6, A 6. Let length nd redth of the rectngle e m nd m respectivel. An equilterl tringle is long the redth. Perimeter of window m \ + Are of window, (A) + < F + [from (i)] A 6 + Mthemtics ()

11 On differentiting oth sides, w.r.t., we get da 6 + d For mimum or minimum re, da d d n 6 6 m 6 da d n d da d H < \ for 6 6, re is mimum. Sustituting in (i), we get e o m Hence, for redth 6 Now nd 8 6 m nd length m, re is mimum. 6 OR Hpotenuse (h) AB AP + PB AP cosec θ AP cosec θ BP sec θ BP sec θ From (i), h cosec θ + sec θ...(ii) dh ( cosec θ cot θ) + (sec θ tn θ) dθ dh For minimum h, cosec θ cot θ sec θ tn θ dθ tn θ tn θ c m / dh [cosec θ( cosec θ) + cot θ( cosec θ cot θ)] + [sec θ sec θ + tn θ sec θ tn θ] dθ dh > for tn θ d n dθ h is minimum for tn θ d n Mthemtics ()

12 sin θ Putting in (ii), we get h / / / + / / + /, cos θ + ( / + / ) / 7. Given curves:, $ ( +, < nd, $ ( +, < / / / + / + / / / / / / q + Plotting the grph of the curves we notice we hve to find the shded re. Y + X O X + Y Point of intersection of + nd + is + + nd for nd Are ( + d ) + ( ) d ( + ) d ( ) d + G + G + G G c + m + (6 ) 5 + sq units c + m+ c m ( ) + c m Mthemtics ()

13 8. Let skilled helpers nd unskilled helpers re emploed. Then LPP is Minimise Z 5 + suject to the constrints, On plotting the inequtions, we notice shded portion is fesile solution. Possile for minimum Z re 68 A d, n, B(6, ), C d, n 5 Points Z 5 + Vlue X Y 8 6 O Y 68 Cc, 5 m B(6, ) A c,m X A d, n B(6, ) Minimum 68 C d, n Z is minimum t B(6, ). Since region is unounded; we drw grph of ineqution 5 + < < 86. Since grph of ineqution < 86 does not hve n point common with fesile region. So, B(6, ) represents minimum, i.e. 6,. \ 6 skilled workers nd unskilled workers must e emploed for minimum cost of ` Given points re A(,, ); B(5,, 6); C(5,, ); D(7,, ) If points re coplnr, then 6 AB AC AD@ 6 AB AC AD@ AB ( 5 ) it+ ( ) tj + ( 6 ) kt i t + ( ) t j + k t AC ( 5 ) it+ ( ) tj + ( ) kt it kt AD ( 7 ) it+ ( ) tj + ( ) kt it+ tj kt (9) ( ) (7) + () Mthemtics ()

14 OR it tj kt LHS # c 5 i t ( 6 5) t j( 9 ) + k t ( + ) it + 9 tj + 7 kt #( # c) it tj kt i t ( 7 8) t j( ) + k t ( 8 + ) 5 it tj + 9 kt 9 7 RHS ^ $ ch _ it tj + kt i$ _ it+ tj kt i 6 ^ $ h _ it tj + kt i _ it tj + 5kt i ^ $ ch ^ $ h c _ it tj + 5kt i 8_ i t + t j k t i 9it+ 6tj 5kt 6it 8tj + 5kt 5i t t j + 9k t Hence, # ^# ch ^ $ ch ^ $ hc Mthemtics ()

EXPECTED ANSWERS/VALUE POINTS SECTION - A

EXPECTED ANSWERS/VALUE POINTS SECTION - A 6 QUESTION PPE ODE 65// EXPETED NSWES/VLUE POINTS SETION - -.... 6. / 5. 5 6. 5 7. 5. ( ) { } ( ) kˆ ĵ î kˆ ĵ î r 9. or ( ) kˆ ĵ î r. kˆ ĵ î m SETION - B.,, m,,, m O Mrks m 9 5 os θ 9, θ eing ngle etween