SOLUTIONS & ANSWERS FOR KERALA ENGINEERING ENTRANCE EXAMINATION-2018 PAPER II VERSION B1

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1 SOLUTIONS & ANSWERS FOR KERALA ENGINEERING ENTRANCE EXAMINATION-8 PAPER II VERSION B [MATHEMATICS]. Ans: ( i) It is (cs5 isin5 ) ( i). Ans: i z. Ans: i i i The epressin ( i) ( ). Ans: cs i sin cs i sin ( i) The number is i i ; i 5. Ans: z cs i sin cs i sin cs i sin i 6. Ans: i ω a cube rt unity z 9 z 9 ω 9 ω 9 ω ω (z 9 z 9 ) ( ω) b ia i(a ib) (b ia) i (a ib) i ( i) i. Ans: 9 α, β are rts 5 α β 5, αβ α β β α 8. Ans:, ( α β) αβ 5 6 αβ 9 α β. β α equatin required is 9 9 Equatin is (y ) i.e. Y a X where X Y y a 9. Ans: cus is, y i.e., 8 > < < () < () < () ( ) ( ) < < ( ) ( ) lim lim 8 8 ( ) ( ) By squeeze therem, ( ) lim 8. Ans: () () lim. Ans: ( ) () ( ) () ' ( ) lim Equatin is ( ) (y ) 6 e 6 6 ( ) ( y ) 6

2 . Ans: (, ) Equatin is Y ax where Y y, X a cus is, y i.e. (, ). Ans: y y D. Ans: 5p 5q r (p q r) p q r 6p q r 6p 6q r q q p p I α is the rt, α α 5. Ans: 6. Ans: 58 (y) () (y) () n () n n () Sum last ceicients sum st ceicients sum all ceicients Ans: NA z z z z i. Ans: sin It is lim cs. Ans:. Ans: (). Ans: cs sin sin sin is the th dd n required sum sin cs y sin λ cs sin cs sin cs sin cs sin y cs. Ans: ± 5 5,. Ans: 6 It is C ( ). C ( ) ( ) ( ) 8. Ans: 6 6( ) 9(y ) Fci are ±, y 5. Ans: i.e. ± 5 5, Thus a a a 6. Ans:.5 P(A B C) P A ( ) P( B C) P( A B C) Mean 6 8 M D.5. Ans:

3 ~p 8, npq q, p ; n 6 6 P(X ) 6 C 8. Ans: 9 6 sinα csα b c, sinα csα a a b c sin α cs α a a b ac a i.e. a b ac The number is 5 C Ans: P(MS)., P(M).6 P(MS) P(S/M) P(M) 5. Ans: area Ans: ( ) 9 y 8 y ( ) 8 6 e 8 6 e. Ans: 8 i New sum ( ) 5( ) 9. Ans: S T A I C N Ι ΙΙ 9 Required prbability Ans: 9 B G Required number 6 C C 6 C C 6 C C 6 c Ans:. Ans: 5 There are distinct letters required number P (6 ) 8 5[(M(5) ] remainder is 5 8. Ans: b ac b cannt be r There are nly tw cases where rts are real Ttal number cases! 6 required prbability 6 9. Ans: 9 lim lim 9. Ans: a b ac

4 . Ans: y 6. Ans:. Ans: h () (g()). g () h () () g () 6 y It is cs sin sin cs [ ( 5 )] cs 5 cs sin cs 5 Minimum where y & y meet i.e. when, y required minimum value. Ans:, y Equatin is (y ) (y ) i.e. ( ) (y ) Asympttes are, y i.e., y. Ans: lg(6) () lg6. Ans: NA 6 9. Ans: 5. Ans: ( ) n 5 m n m m sin m lim n n sin n lim lim ( ) 8. Ans: e 5 e m m is the prbability density! P(X ) e m m 9. Ans: 5. Ans: P(X ) e m m! m 6 P(X 6) e m 6! m m P (X 6) e e 6! 5 6 The selected numbers can be (, ) r (, ) r (, ) r... r (9, ) Ttal number cases 9 Nw, a b a b (a b) ab a b ab a b (ab ) (k ), since the prduct tw cnsecutive integers is even. a b a b k, dd integer Number avurable cases 9 Prbability I a, b are the semi aes the ellipse & a > b the circle is r radius a ab required prbability a b a

5 5. Ans: e 8 9 m 6 ( ) 5 5. Ans: 6 m 8 m m m a b i j 6k ( i j k) b c 8i j 6k ( i j k) ( a b) ( b c) i 8(i j k) required area j k ( ) 5 r 5 5. Ans: b ab, b b a ab a b (a ) (b ) a r b a gives imaginary rts r a 58. Ans: sinθcsθ sinθcsθ sin θ cs θ (sinθ csθ) (sin θ sinθ csθ cs θ) (sinθ csθ) ( ). 59. Ans: 9 5. Ans: 9 a b c ( b.c c.a a.b ) 9 6 b. c b.c } Required prbability Ans: a b a. b csθ csθ θ 55. Ans: 9 6. Ans: 9! 6 9i i ! 5! λ 9 µ λ 6 µ λ µ 9 λ µ 6 λ 8 λ µ 8 λ µ Ans:, 6. Ans: and rder degree 6. Ans: d ( )

6 6. Ans: 6. Ans: 5 ( ) d ( ) d 65. Ans: 66. Ans: ( tan ( )) tan tan ( ) d ( ) d ( ) d Q ( ( ) ) d ( ) d 6 ( ) d 5 ( ) ( ) e d ( ) e d e C e ( ' )d e () C ( ) ( ) (6 ) 5 Remainder 6. Ans: 5 Ceicient 5 8 C 68. Ans: 5 5 cs θ sin θ cs θ sin θ 69. Ans: Maimum value C a b 69 z Area. Ans: 9 8. Ans:, z () cs () sin () 56 By sectin rmula,. Ans: A 9 8, a b c b c c a a b a b c. Ans: b ay b c c a a b Midpint (a, b) satisy nly b ay y. Ans: a b 5. Ans: (8, 8) Equatin line line b ay ab any line parallel t b ay ab passing thrugh (a, b), k ab y a b a a Area a a 8 a a a 8

7 a () 8 a 6 a ± 6 a a centrid, (8, 8) 8. Ans: i j kˆ b î 6ĵ 6kˆ a î yĵ zkˆ a & b cllinear a. b 6. Ans:, Ans: ± 5 a a. Ans: 9 8. Ans: 5 ( a ) ( a) (a 9 a) ±5 a a 5a 9 ±5 a ±5 a ±5 9 r a r verte is (a, a ),, Fr pair straight line abc gh a bg ch g g a b a b ( a. b) 8. Ans: 8. Ans: 5 i a b a b P r 6 : 5 P r 8 : 56! ( 56 r 6 )! 5! ( 5 r ) 5 r r! 8 y y 5 d 5 5 tanθ 5 5 tan 9 θ 5 h ab a b 9. Ans:, y, z y z y z 5 y z back substitutin 8. Ans: sin(5 ) Ans: 86. Ans: 9 C. C C C C 8 ( ) ( ) 6 ceicient ( 6 C ) ( 6 C ) Ans: y y 5 (, ) (, 5) r 9

8 88. Ans: (,, ) ( h) (y k) r ( ) (y ) y y y y 5 Let P(, y, ) be a pint n y plane P(, y, ) A (,, ) B (,, ) C (,, ) PA ( ) y 9 PB ( y) PA PB PC PC y ( ) 9 (PA PC ) 9 (PA PB ) 6y y 89. Ans: () and () ( y) () (y) gives () 9. Ans: 9 n n ( a k) 6( ) k a a a... a n 6 ( n ) a (... n ) 6( n ) n ( ) a n 6 a ( n ) 6( n ) a 6 a a n C r 6 () n C r 8 () n C r 6 () ( ) () ( ) ( ) 9. Ans: n r () n 5r (5) Slving () and (5) n 9 () ( ) (, ) () is a cnstant () 9. Ans: ( ) ( ) lim 9. Ans: 5 lim () lim () lim 5 h 9. Ans: ( h) ( ) h limit eist () () 5 () 6 6 6( 5 6), () < at Maimum at 95. Ans: ( ) 96. Ans: ( ), Ι [ ( ) ] cs d C () sin C C d sin Als Ι Ι sin d ( ) d sin ( sec sec tan )d [ tan sec ] sin sin d cs

9 9. Ans: 98. Ans: 99. Ans: tan tan sec sec [( ) ( )] [ ] ( ) Ι ( ) sin Ι d () sin cs cs Ι d () sin cs () () Ι d Ι Ι sin t dt a lim sin. lim cs Area sin d d sq. units. Ans: y y y The slutin is y y y. Ans: NA i i [ i ] i i i ω. Ans: ( i ) ( i) ω 6 ( ) i ω 6 ( ) i ω ( ) i i Real part ( ) e lim Applying L- Hspitals rule e lim Applying L- Hspital s rule again e lim sin cs. Ans: C sin ( sin cs )( sin ) d sin ( sin cs ) [ ( sin ) ] [ ( sin ) ] ( sin cs ) d sin cs d sin cs d [ ( sin cs sin cs ) ] cs [ sin cs ] d [ ( sin cs ) ] ( sin ) ( ) t ( ) dt t ( t ) ( t ), where t sin cs t dt t dt ( t) ( t ) t t t sin cs C C t ( sin cs ) sin cs C sin. Ans: r. ( i j k) 5 i j k n ( i j k) C

10 5. Ans: 6. Ans: 6. Ans: Equatin is r. n d ( i j k) 5 r. A B tan A B A B cs sin sin A sinb cs A cs B A B A B cs cs A B tan Acst Bsint d A sin t B cs t dt d 6A cs t 6B sin t dt 6 n n A.M 8. Ans: 9. Ans: 9 C n C... Cn n n n Variance irst natural number is 99 number element in S 9 9. Ans: A. Ans:, A A 6 6 A ( 9) ( 6) 8 diagnal element A are 6 sum ( ) 6 R R R R C C C, C C C 9 ( 9) ( ) ( ) 9,,. Ans: ; 5 [ ] 5 [ ] [ ] [ 6 8] 6 8 ;, 6, S {H, H, H, H, H 5, H 6, T, T, T, T, T 5, T 6} P(cin shws head and die shw ). Ans:. Ans: AA Ι

11 5. Ans: 6 (6 6) ( 6 6) 6 ( ) a b c d e 5a b c d e ( ) d sin d ( cs ) sin cs ( ] 9. Ans: 5 6y z d 6. Ans: a b b c c a c a b a a b c b a b c C c a b c a (a b c) b c. Ans: () ( ) C C C C ( ) ( ) C C C ( ) ( ) ( ) (5) C y z Equatin is. Ans: 5(5) ( ) ( ) (y ) ( ) (z ) (6 8) 5 5 6y z 6 5 6y z 5 6y z t n 5n t 6 t 5 S [ 6 5] 5( 5) 8. Ans: () cs cs sin cs C C C C sin cs ( ) [cs (sin cs) cs( cs)] ( ) [cs sincs cs cs cs ] sincs