R Prerna Tower, Road No, Contractors Area, Bistupur, Jamshedpur 8300, Tel (0657)89, www.prernaclasses.com Jee Advance 03 Mathematics Paper I PART III MATHEMATICS SECTION : (Only One Option Correct Type) This section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 4. Let f : [ /, ] R (the set of all real numbers) be a positive, non constant and differentiable function such that f ' (x) < f (x) and f ( / ) =. Then the value of f ( x ) dx lies in the interval e (A) (e, e) (B) (e, e ) (C), e / (D) 0, e f ( x ) 4. (D) f ' (x) < f (x) < f ( x ) (Q f (x) is (+)ve function) f ( x ) < dx ln f (x) < x + c f (x) < ke f ( x ) x Q f ( / ) < ke ke > k > / e x e ( e ) f ( x ) dx < k e dx = k / / e ( e ) f ( x ) dx < e / e 0 < f ( x ) dx <. / 4. A curve passes through the point (, π / 6). Let the slope of the curve at each point (x, y) be (y / x) + sec (y / x), x > 0. Then the equation of the curve is (A) sin (y / x) = log x + ( / ) (B) cosec (y / x) = log x + (C) sec (y / x) = log x + (D) cos (y / x) = log x + ( / ) 4. (A) dy / dx = (y / x) + sec (y / x) y = vx dy / dx = v + x (dv / dx) v + x (dv / dx) = v + sec v dv / sec v = dx / x sin v = ln x + c / = 0 + c c = / sin (y / x) = log x + ( / ). JEE Advanced 03 (0 Jun 3) Question & Solutions Paper I www. prernaclasses.com
43. Perpendiculars are drawn from points on the line feet of the perpendiculars lie on the line. (A) x y z 5 8 3 (B) (C) x y z 4 3 7 (D) x + y + z to the plane x+ y + z = 3. The 3 x y z 3 5 x y z 7 5 43. (D) x + y + z 3 = r P L (hkw) P = (r, r, 3r) = (x 0 y 0 z 0 ) = Variable point on the line L. h x 0 k y 0 w z 0 ( x 0 + y 0 + z 0 3 ) 4 r 6 4 r = = = + 3 3 3 3 Here (h, k, w) is a variable point on the locus joining the feet of the perpendiculars from L. h = r (4r / 3) + = r / 3 k = r (4r / 3) + = (7r / 3) + w = 3r (4r / 3) + = (5r / 3) + h k w. Answer is (D). 7 5 44. Let PR = 3 i ˆ + j ˆ k ˆ and SQ = i ˆ 3 j ˆ 4 k ˆ determine diagonals of a parallelogram PQRS and PT = i ˆ + j ˆ + 3 k ˆ be another vector. Then the volume of the parallelopiped determined by the vectors PT, PQ and PS is (A) 5 (B) 0 (C) 0 (D) 30 44. (C) T S R θ P Q PR = 3 i ˆ + j ˆ k ˆ, SQ = i ˆ 3 j ˆ 4 k ˆ, PT = i ˆ + j ˆ + 3 k ˆ Volume of parallelopiped = (vector area of parallelogram). PT i j k r r = ( d d ). PT = 3. PT 3 4 = { i ( 0 ) j ( 0 ) + k ( 0 )}. PT = { 0 + 0 30 } = 0 Q volume is +ve. JEE Advanced 03 (0 Jun 3) Question & Solutions Paper I www. prernaclasses.com
3 n 45. The value of cot cot + k is n = k = (A) 3 / 5 (B) 5 / 3 (C) 3 / 4 (D) 4 / 3 3 45. (B) cot ( + n ( n + )) = tan n = 3 n = 5 Exp.= cot cot = 5 / 3. 3 ( n + ) n = tan ( + n ( n + ) 4 ) tan ( ) 3 = tan 5 46. For a > b > c > 0, the distance between (, ) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than. Then (A) a + b c > 0 (B) a b + c < 0 (C) a b + c > 0 (D) a + b c < 0 c c 46. (A) Point of intersection is, a + b a + b a + b + c < a + b a + b > c a + b c > 0. a + b + c 0 < < a + b 47. Let complex numbers α and lie on circles (x x α 0 ) + (y y 0 ) = r and (x x 0 ) + (y y 0 ) = 4r, respectively. If z 0 = x 0 + iy 0 satisfies the equation z 0 = r +, then α = (A) / (B) / (C) / 7 (D) / 3 47. (C) z 0 = α α α = r + α = α + α α α 7 α + 8 = 0 α = or / 7 α α = or. 7 48. The number of points in (, ), for which x x sin x cos x = 0, is (A) 6 (B) 4 (C) (D) 0 48. (C) f(x) = x x sin x cos x f '(x) = x ( cos x) f '(x) < 0 for x < 0 & f '(x) > 0 for x > 0 f(x) is decreasing before x = 0 and increasing after x = 0, also f (0) =. f (x) = 0 is satisfied by two points in (, ) JEE Advanced 03 (0 Jun 3) Question & Solutions Paper I 3 www. prernaclasses.com
49. The area enclosed by the curves y = sin x + cos x and y = cos x sin x over the interval [0, π / ] is (A) 4( ) (B) ( ) (C) ( + ) (D) ( + ) π / 4 49. (B) Area = ((sin x + cos x ) (cos x sin x )) dx + ((sin x + cos x ) (sin x cos x )) dx 0 π / 4 π / cos xdx = sin + 0 xdx = ( ) π / 4 50. Four persons independently solve a certain problem correctly with probabilities /, 3 / 4, / 4, / 8. Then the probability that the problem is solved correctly by at least one of them is (A) 35 / 56 (B) / 56 (C) 3 / 56 (D) 53 / 56 π / π / 4 50. (A) Probability of solving the problem correctly by at least one of them 3 = 4 4 8 35 56 JEE Advanced 03 (0 Jun 3) Question & Solutions Paper I 4 www. prernaclasses.com
SECTION (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE or MORE are correct. 5. For 3 3 matrices M and N, which of the following statement(s) is(are) NOT correct? (A) N T MN is symmetric or skew symmetric, according as M is symmetric or skew symmetric (B) MN NM is skew symmetric for all symmetric matrices M and N (C) MN is symmetric for all symmetric matrices M and N (D) (adj M) (adj N) = adj (MN) for all invertible matrices M and N 5. (CD) (A) (N T MN) T = N T M T (N T ) T = N T MN (since M T = M) (A) (MN NM) T = (MN) T (NM) T = N T M T M T N T = (MN NM) skew symmetric. (C) (MN) T = N T M T = NM (since N T = N, M T = M) MN is not always symmetric. (D) As adj (MN) = adj M. adj N statements (C) and (D) are incorrect. 5. A line l passing through the origin is perpendicular to the lines l + t i ˆ + + t j ˆ + + t ) k ˆ : ( 3 ) ( ) ( 4, < t < l + s i ˆ + + s ) j ˆ + + s ) k ˆ : ( 3 ) ( 3 (, < s < Then, the coordinate(s) of the point(s) on l at a distance of 7 from the point of intersection of l and l is (are) (A) (7 / 3, 7 / 3, 5 / 3) (B) (,, 0) (C) (,, ) (D) (7 / 9, 7 / 9, 8 / 9) 5. (BD) L : ( 3 i ˆ ˆ j + 4 k ˆ ) + t ( i ˆ + ˆ j + k ˆ ) L : ( 3 i ˆ + 3 ˆ j + k ˆ ) + s ( i ˆ + ˆ j + k ˆ ) Let a, b, c be DR of req. line L a + b + c = 0 a + b + c = 0 a : b : c = : 3 : Eq. of L : ( 0 i ˆ + 0 ˆ j + 0 k ˆ ) + λ ( i ˆ 3 ˆ j + k ˆ ) Point of int. of L and L Any point of L: (λ, 3λ, +λ) and L : ( λ + 3, λ,, λ + 4) For point of int λ = λ + 3, 3λ = λ, λ = λ + 4 λ =, λ = point of int (, 3, ) Any point of L : (k + 3, k + 3, k + ) (k + 3 ) + (k + 3 + 3) + (k + ) = ( ) 7 k =, k = 0 / 9 points are (,, 0), ( 7 / 9, 7 / 9, 8 / 9). JEE Advanced 03 (0 Jun 3) Question & Solutions Paper I 5 www. prernaclasses.com
53. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 5 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 00, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are (A) 4 (B) 3 (C) 45 (D) 60 53. (AC) V = (5x y) (8x y) y [y side of square, clearly y = 5 for max. volume] V = 0 x y 46xy + 4y 3 dv / dy = 0x 46 xy + y = 0 when y = 5, 0x 9 5x + 300 = 0 x 46x + 30 = 0 x = 3, 5 / 6. If x = 3 sides : 45, 4. If x = 5 / 6 : (Not possible since y > 8x). 4 n k ( k + ) 54. Let S n = ( ) k. Then S n can take value(s) k = (A) 056 (B) 088 (C) 0 (D) 33 54. (AD) S n = + 3 + 4 5 6 + 7 + 8... + (4n ) + (4n) = 3 + 4 + 7 + 8 5 6... + (4n ) + (4n 3) + 4n (4n ) = 4 [ + 3 + 6 + 7 + 0 + +... (4n ) + (4n )] = 8( + 3 + 5 +... + (n )) + 4(3 + 7 + +... + (4n )) = 6n + 4n = 4n (4n + ), n N, satisfied by (A) and (D) when n = 8, 9 respectively. 55. Let f (x) = x sin πx, x > 0. Then for all natural numbers n, f ' (x) vanishes at (A) a unique point in the interval n, n + (B) a unique point in the interval + n, n + (C) a unique point in the interval (n, n + ) (D) two points in the interval (n, n + ) 55. (BC) f ' (x) = πx cos πx + sin πx = 0 sin πx = πx cos πx tan πx = πx πx tan πx / 0 / 3/ Hence, x + n, n + and also x (n, n + ). JEE Advanced 03 (0 Jun 3) Question & Solutions Paper I 6 www. prernaclasses.com
SECTION 3 (Integer value correct Type) This section contains5 questions. The answer to each question is asingle digit integer,ranging from 0 to 9. (both inclusive). 56. A pack contains n cards numbered from to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 4. If the smaller of the numbers on the removed cards is k, then k 0 = n ( n + ) 56. (5) ( k + ) = 4 Q k n n + n 450 n 4 Hence,k= 5. n + n 4k = 450 49 < n < 5 n = 50 57. Consider the set of eight vectors V = { a i ˆ + b ˆ j + c k ˆ : a, b, c {, } }. Three non coplanar vectors can be chosen from V in p ways. Then pis 57. (5) Considering the origin at the centre of a cube, then the given vectors are the vectors joining centre and eight vertices of this cube. A set of coplanar vectors (or three vertices in this case) can be selected iff out of three vertices we select two vertices which are situated at the two ends of a body diagonal and the third vertex can be any point out of remaining six vertices. There are 4 6 = 4 coplanar vectors (as there are four body diagonals). No. of non coplanar vectors = 8 C 3 4 = 3. p = 5. 58. A vertical line passing through the point (h, 0) intersects the ellipse x + y = at the points Pand 4 3 Q. Let the tangents to the ellipse at Pand Q meet at the pointr. If (h) = area of the triangle PQR, = max ( h ) and min ( ) / h h 8 =, then 8 = / h 5 58. (9) Taking R (k, 0) and Chord of contact with respect to R, i.e. PQ is x = (4 / k) = h. 4 3/ R, 0 3 ( 4 h ) d ( h ) =. < 0 h h dh (h) min. = () = (9 / ) = h, & (h) max. = ( /) = 8 8 5 3. 5 8 = 45 36 = 9 5 = JEE Advanced 03 (0 Jun 3) Question & Solutions Paper I 7 www. prernaclasses.com
59. The coefficients of three consecutive terms of ( + x) n+ 5 are in the ratio 5 : 0 : 4. Then n= n + 5 n + 5 59. (6) Since, C C = 3r = n + 6 and n + 5 r : r n + 5 7 C r + : C r = r = 5n + 8. n = 6. 5 60. Of the three independent events E, E and E 3, the probability that only E occurs is α, only E occurs is β and only E 3 occurs is γ. Let the probability p that none of events E, E or E 3 occurs satisfy the equations (α β) p =αβ and (β 3γ) p = βγ. All the given probabilities are assumed Probabilit y of occurrence of E to lie in the interval (0, ). Then = Probabilit y of occurrence of E 60. (6) Let x, y, zbe the probabilities of occurrence of E, E and E 3 respectively. px Then, α= x ( y) ( z) = (Q p = ( x) ( y) ( z)) ( x ) 3 Similarly, py β = ( y ) pz and γ = ( z ) px py px py (α β) p = αβ p = x y x y x = y and (β 3γ) p = βγ x = 6z x / z = 6. py y 3 pz py pz p = z y z y = 3z JEE Advanced 03 (0 Jun 3) Question & Solutions Paper I 8 www. prernaclasses.com