Oly Oe Optio Correct Type JEE ADVANCED 0 PAPER MATHEMATICS This sectio cotais TEN questios. Each has FOUR optios (A), (B), (C) ad (D) out of which ONLY ONE is correct.. The value of (A) 5 (C) 4 cot cot (B) 5 k is k (D) 4 cotcot k cotcot ( k( k )) k cot (ta ( k ) ta k) cot(ta 4 ta ) cot ta 5 5.. Let PR iˆ ˆj kˆ ad SQ iˆ ˆj 4kˆ determie diagoals of a parallelogram PQRS ad PT iˆ ˆj kˆ be aother vector. The, the volume of the parallelepiped determied by the vectors PT, PQ ad PS is (A) 5 (B) 0 (C) 0 (D) 0 Let a ad b be the sides of the parallelogram whose diagoals be PR ad SQ, as show i the followig figure. PR a b iˆ ˆj kˆ SQ a b iˆ ˆj 4kˆ These imply that a iˆ ˆj kˆ ; b iˆ ˆj kˆ. the volume of the parallelepiped formed by abad, PT is 0 Copyright Wiley Idia Page
. Let complex umbers ad lie o circles (x x 0) + (y y 0 ) = r ad (x x 0 ) + (y y 0 ) = 4r, 0 respectively. If z 0 = x 0 + iy 0 satisfies the equatio z r, the (A) (B) (C) 7 (D) As α satisfies z z0 r, we have Also, sice satisfies z z0 r, we have z 0 0 0 0 r ( z )( z ) r 0 0 z z z r () z 0 r z0 z0 4r ( z )( z ) 4r 0 0 0 0 0 z z z 4 r () Subtractig Eq. () from Eq. (), we have That is, Also, we have That is, z0 r ( ) ( ) ( 4 ) z0 r ( )( ) ( 4 ) () Dividig Eq. () by Eq. (4), we get 0 r 0 r z ( z ) (4) 4 8 7. 7 Copyright Wiley Idia Page
4. For a > b > c > 0, the distace betwee (, ) ad the poit of itersectio of the lies ax byc 0 ad bx ay c 0 is less tha.the (A) a + b c > 0 (B) a b + c < 0 (C) a b + c > 0 (D) a + b c < 0 We kow that ax by c 0 () bx ay c 0 () Solvig, we get c x a b From Eqs. () ad (), we get y = x. That is, the poit of itersectio lies o y = x. This implies that c y a b It is give that That is, c c a b a b c a b abc a b a b c a b a b c 0 x y z 5. Perpediculars are draw from poits o the lie to the plae x + y + z =. The feet of perpediculars lie o the lie x y z x y z (A) (B) 5 8 5 x y z x y z (C) (D) 4 7 7 5 Let us cosider that x y z ad it is give that x y z. We ca write ay poit o this lie as (λ, λ, λ). which satisfies the plae ( ) ( ) ( ) 4 Thus, the poit of itersectio of plae is give by Copyright Wiley Idia Page
5 9,, C. The poit o the lie is (,, 0) ad directio ratio of AB (see the followig figure) is k. Ay geeral poit o lie AB is (k, k, k), which satisfies the equatio. (k ) + (k ) + k =. k (α, β, γ) (0,, ) Thus, the equatio of lie passig through BC is x y z 7 / 5 / x y z 7 5 6. Four persos idepedetly solve a certai problem correctly with probabilities,,,. The the 4 4 8 probability that the problem is solved correctly by at least oe of them is (A) 5 (B) 56 56 (C) (D) 5 56 56 Let us cosider that PA ( ) ; PB ( ) ; 4 PC ( ) ; 4 PD ( ). 8 P( A B C D) P( A B C D) P( A B C D) P( A) P( B) P( C) P( D) 7... 4 4 8 56 5 56 Copyright Wiley Idia Page 4
7. The area eclosed by the curves y si x cos x ad y cos x si x over the iterval 0, is (A) 4( ) (B) ( ) (C) ( ) (D) ( ) From the followig figure that depicts the area eclosed by the give curves, we have (si x cos x) dx (cos x si x) dx (si x cos x) dx / /4 / 0 0 /4 cos x si x si x cos x cos x si x / / /4 /4 / / 0 0 0 0 /4 /4 (0 ) ( 0) 0 4 ( ) 8. A curve passes through the poit,. Let the slope of the curve at each poit (x, y) be 6 y y sec, x 0. The the equatio of the curve is x x (A) si y log x (B) cosec y log x x x (C) sec y log x (D) cos y log x x x dy y sec y dx x x Substitutig y vx, we get ad as this passes through,, we have 6 dv v x v secv dx dx cosv dv x si v l x c, Copyright Wiley Idia Page 5
ad hece c si y l x x 9. Let f :, such that (the set of all real umbers) be a positive, o-costat ad differetiable fuctio f ( x) f ( x) ad f. The the value of (A) (e, e) (B) (e, e ) e e (C), e (D) 0, 0. Let That is, which implies that f :, such that x ye dy y dx f ( x) dx lies i the iterval x dy x e ye dx d x ( ye ) 0, dx is a decreasig fuctio. As x, we have x e ye y() e x x e y y() e / x x e dx ydx y() e 0 / / / 0 / e ydx (the set of all real umbers) be a positive, o-costat ad differetiable fuctio f ( x) f ( x) ad f. The the value of (A) (e, e) (B) (e, e ) e e (C), e (D) 0, f ( x) dx lies i the iterval / Let us cosider that f ( x) x xsi x cos x Copyright Wiley Idia Page 6
f ( x) x xcos x si x si x x( cos x) f ( x) is icreasig whe x 0; f ( x) is decreasig whe x 0. f (0) f ( ) f ( ) as show i the followig graph, the umber of poits i (, ), for which x xsi x cos x 0, is two. Oe or More tha Oe Optios Correct Type This sectio cotais 5 multiple choice questios. Each questio has four choices (A), (B), (C) ad (D) out of which ONE OR MORE are correct.. A rectagular sheet of fixed perimeter with sides havig their legths i the ratio 8 : 5 is coverted ito a ope rectagular box by foldig after removig squares of equal area from all four corers. If the total area of removed squares is 00, the resultig box has maximum volume. The the legths of the sides of the rectagular sheet are (A) 4 (B) (C) 45 (D) 60 V (8 x)(5 x) x 4 46 0 x x x Differetiatig with respect to x, we get dv x 9x 0 0 at x 5 dx 60 050 0 6 5 0 (6 5)( ) 0 For λ =, the legths of sides (as show i the followig figure) are obtaied as 45, 4.. Let S kk ( ) 4 ( ) k. The S ca take value(s) k (A) 056 (B) 088 (C) 0 (D) Copyright Wiley Idia Page 7
S 4 k ( ) kk ( ) k S 4 5 6... (4 ) (4 ) (4 ) (4 ) S ( ) (4 ) (7 5 ) (8 6 ) ( 9 ) 0 )... (4 ) (4 ) (4 ) (4 ) S ( ) (4 ) (7 5) (8 6)... (4 4 ) (44) S.4 (4) [... 4 ] From the value give i optio (A), we get 4 (4) 056 4 64 4 64 0 8 From the value give i optio (B), we get which is ot possible. 4(4 + ) = 088, From the value give i optio (C), we get 4(4 + ) = 0, which is also ot possible. From the value give i optio (D), we get for = 9. 4(4 + ) =. A lie l passig through the origi is perpedicular to the lies l : ( t) iˆ ( t) ˆj (4 t) kˆ, t l : ( s) iˆ ( s) ˆj ( s) kˆ, s The, the coordiate(s) of the poit(s) o l at a distace of 7 from the poit of itersectio of l ad l is (are) 7 7 5 (A),, (B) (,,0) 7 7 8 (C) (,, ) (D),, 9 9 9 The equatio of the lie, l, is The equatio of the lie, l is The equatio of the lie, l, is x y z a b c y z4 t Copyright Wiley Idia Page 8
x y z s The directio ratio of the lie, l, is give by iˆ ˆj kˆ iˆ ˆj kˆ The equatio of the lie, l, is x y z The poit of itersectio of l ad l are as follows: t () t () Substitutig the value of t, we get ( ) That is, 4 6 The poit of itersectio is (,,) 7 7. ( s ) ( s ) ( s ) 7 4s 4s 6 4s 4s s 7 9s 8s 0 0 0 s, 9 Thus, the itersectio poits are obtaied as 7 7 8 (,, 0) ad,, 9 9 9. 4. Let f ( x) xsi x, x 0. The for all atural umbers, f ( x) vaishes at (A) A uique poit i the iterval, (B) A uique poit i the iterval, (C) A uique poit i the iterval (, ) (D) Two poits i the iterval (, ) f ( x) xsi x f ( x) si x xcos x 0 tax x It is clear from the followig graph that f ( x) has oe root i root i (, )., ad f ( x) also has oe Copyright Wiley Idia Page 9
5. For matrices M ad N, which of the followig statemet(s) is (are) NOT correct? (A) N T MN is symmetric or skew symmetric, accordig as M is symmetric or skew symmetric (B) MN NM is skew symmetric for all symmetric matrices M ad N (C) MN is symmetric for all symmetric matrices M ad N (D) (adj M) (adj N) = adj(mn) for all ivertible matrices M ad N (N T MN) T = N T M T (N T ) T = N T M T N (A) If M is skew symmetric, the (N T MN) T = N T MN; therefore, it is cocluded that it is skew symmetric. If M is symmetric, the (M T MN) T = N T MN; therefore, it is cocluded that it is symmetric. Hece, optio (A) is correct. (B) T T T ( MN NM ) ( MN) ( NM ) N M T T T T M N T T T T ( M M N M ) ( MN NM ) it is cocluded that it is skew symmetric ad hece optio (B) is correct. (C) (MN) T = N T M T. Symmetricity ad skew symmetricity deped o the ature of M ad N, therefore, optio (C) is icorrect. (D) adj(mm) = adj(n) adj M, therefore, optio (D) is icorrect. Iteger Aswer Type This sectio cotais FIVE questios. The aswer to each questio is a SINGLE DIGIT INTEGER ragig from 0 to 9, both iclusive. x y 6. A vertical lie passig through the poit (h, 0) itersects the ellipse at the poits P ad Q. 4 Let the tagets to the ellipse at P ad Q meet at the poit R. If Δ(h) = area of the triagle PQR, 8 max ( h) ad mi ( h), the 8. /h /h 5 Copyright Wiley Idia Page 0
x y S 4 Let P ad Q be (h, β) ad (h, β), respectively (see the followig figure). R is 4,0 h. 4 h h h 4 h 4 h (4 h ) h d That is, 0, from which it is clear that Δ is decreasig. That is, dh 7. The coefficiets of three cosecutive terms of / 5 45 8 9 () 8 8 9 5 5 x are i the ratio 5 : 0 : 4. The =. ( ) Let us cosider that the cosecutive terms be t r +, t r + ad t r. tr 0 t 5 r ( 5) ( r) r r 0 () Also we have t t r r 4 0 5r 6 0 () Solvig Eqs. () ad (), we get = 6. 8. Cosider the set of eight vectors V ai bj ck ˆ a b c ˆ ˆ :,, {,}. Three o-coplaar vectors ca be chose from V i P ways. The p is. Copyright Wiley Idia Page
The eight vectors are as show i the followig figure. The total umber of vectors is give by C 56 The total umber of coplaar vectors is give by 8 That is, Hece, p = 5. (6 ) = 4 5 56 4 9. Of the three idepedet evets E, E ad E, the probability that oly E occurs is α, oly E occurs is β ad oly E occurs is γ. Let the probability p that oe of evets E, E or E occurs satisfy the equatios ( ) p ad ( ) p. All the give probabilities are assumed to lie i the iterval (0, ). The Probability of occurrece of E =. Probability of occurrece of E P( E ) P( E ) P( E ) () P( E ) P( E ) P( E ) () P( E ) P( E ) P( E ) () P( E ) P( E ) P( E ) (4) Dividig Eq. () by Eq. (4), we get Also PE ( ) PE ( ) PE ( ) PE ( ) PE ( ) PE ( ) PE ( ) PE ( ) PE ( ) (5) Copyright Wiley Idia Page
4 5 4 5 4 Dividig Eq. () by Eq. (4), we get PE ( ) ( ) PE ( ) PE ( ) PE ( ) 5 4 PE ( ) 5 4 PE ( ) PE ( ) 5 4 PE ( ) 6 6 6( ) PE ( ) PE ( ) 6( ) PE ( ) 6 PE ( ) 6( ) Probability of occurrece of E = 6. Probability of occurrece of E 0. A pack cotais cards umbered from to. Two cosecutive umbered cards are removed from the pack ad the sum of the umbers o the remaiig cards is 4. If the smaller of the umbers o the removed cards is k, the k 0 =. (6) The smallest value of for which For 50, we have k = 5 ad thus ( ) 4 ( ) 448 49 ( ) k ( k) 75 4 5 k 0 5. 75 Copyright Wiley Idia Page