38. The total number of carboxylic acid groups in the product P is 38. [2] O C HO 3 O C C O C O C O. Total no. of carboxylic group = 2

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1 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 8. The total number of carboxylic acid groups in the product P is 8. [] C H Heat C Total no. of carboxylic group = C C H 9. A tetrapeptide has CH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine(ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with NH group attached to a chiral center is 9. [4] H H C C Glycine = CH NH CH Alanine = CH CH NH CH Valine (CH ) CH CH NH CH NH phenyl alanine Ph CH CH CH No. of possible sequences (primary structures) with NH group attached to a chiral center is The total number of lone pairs of electrons in melamine is 40. [6].... N NH N.. N. NH... Lone Pairs 6... NH Melamine 6

2 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (7) PART III : MATHEMATICS SECTIN I : (nly ne option correct Type) This section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NLY NE is correct. 4. 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 4. (A) Given equations are ax + by + c = 0 bx + ay + c = 0 Now, x y = = bc ac bc ac a x = ( b a ) c c = a b a+ b (b a)c c y = = a b a+ b Now c c a+ b a+ b (a + b + c) < (a + b) (a + b+ c) < a+ b (a + b + c) < a + b a + b c > 0 b < 4. The area enclosed by the curves y = sin x + cos x and y = cos x sin x over the interval π 0, is (A) 4( ) (B) ( ) (C) ( + ) (D) ( + ) 4. (B) π y = sin x+ 4 π y = sin x 4 Area of shaded part = π/4 π π sin x+ sin x dx π/ π π + sin x+ sin x dx 4 4 π/4 π 4 7

3 (8) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution π/4 π/ π π π π = sin x + + sin x dx + sin x + sin x π/4 = + = 4 4. The number of points in (, ), for which x x sin x cos x = 0, is (A) 6 (B) 4 (C) (D) 0 4. (C) f(x) = x x sin x cos x f (x) = x sin x x cos x + sin x = x ( cos x) > 0 x> 0 So, f (x) = 0 x= 0 For x = 0, f(x) = Hence two solutions < 0 x < 0 (0, ) n 44. The value of cot cot + k is n= k= (A) (B) 5 (C) (D) (B) n cot cot + k n= k= ( n+ ) = cot cot + n = cot n= cot ( n + n + ) n= (n + ) n = cot tan = cot tan n= + n(n+ ) n= + (n + ).n = cot ( tan (n + ) tan n) = cot tan 4 tan n= 4 = cot tan + 4. = cot 5 cot = 5 π 45. A curve passes through the point,. Let the slope of the curve at each point (x, y) 6 be y sec y +, x > 0. Then the equation of the curve is x x y (A) sin y = log x + (B) cosec = log x + x x y y (C) sec = log x + (D) cos = log x + x x 8

4 45. (A) dy dx = y sec y + x x let IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (9) v + x dv dx = v + sec v y x = v y = vx dy dx = v + x dv dx dx cos v dv = x sin v = lnx + lnc π Given it passes (, ) 6 Sin 6 π = ln + lnc lnc = So, curve is sin v = lnx+ sin y x = lnx Let f :, (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)dxlies in the interval (A)( e, e) (B) (e, e ) e (C),e 46. (D) f (x) < f(x) f(x) < f(x) f(x) dx < dx f(x) n{ f(x) } < x+ c f(x) < e x + c f(/) < e + c < e + c.(f(/)) = ) e 0 < e + c 0 < + c c >.(i) (D) 0, e 9

5 (0) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution / / / / / x + c f(x)dx< e dx x + c f(x)dx< e / + + f(x)dx< e e + c f(x)dx< e [ e ] Maximum value of c c / And minimum value of f(x) dx is greater than zero because it is positive, nonconstant function. f(x)dx is [e ] / 47. Let PR = i ˆ+ ˆj kˆ and SQ = ˆ i j ˆ 4k ˆ determine diagonals of a parallelogram PQRS and PT = ˆi + j ˆ+ kˆ be another vector. Then the volume of the parallelepiped determined by the vectors PT, PQ and PS is (A) 5 (B) 0 (C) 0 (D) (C) PR = i ˆ+ ˆj kˆ and SQ = ˆ i ˆ j 4k ˆ are diagonals of a parallelogram. PT = ˆ i+ ˆ j+ k ˆ is another vector. Volume of the parallelepiped by the vector PT, PQ & PS a+ b= i ˆ+ ˆj kˆ S a R a b= ˆi ˆj 4kˆ b a= i ˆ ˆj kˆ b b = ˆi+ ˆj+ kˆ P Q a Volume of parallelepiped formed by PT, PQ & RS is given by scalar triple product of ˆi+ ˆj+ kˆ, i ˆ ˆj kˆ & ˆ i+ ˆ j+ k ˆ Volume = = (5) + ( 5) + (5) = = Perpendiculars are drawn from points on the line x + y + = = z to the plane x + y + z =. The feet of perpendiculars lie on the line (A) x y z = = (B) x y z = = (C) x y z = = (D) x y z = =

6 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) () 48. (D) Any point on x + = y + = z is (r, r, r). Let P(x, y) be feet of perpendicular then, x x y y z z ax+ by+ cz+ d = = = a b c a + b + c x (r ) y ( r ) ) = = z r = 4r 6 So, x = r 4r 6 = r ; 7r 5r + 6 y = ; z = r 7r 5r + 6 bviously,, satisfy x y z = = Four persons independently solve a certain problem correctly with probabilities,,,. Then the probability that the problem is solved correctly by at least one of them is (A) 5 (B) (C) (D) (A) Required Probability = P(No one solves the problem correctly) = 7 = Let complex numbers α and α lie on circles (x x 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) (D) (C) α Z 0 = r () and Z0 = r () α () (α Z 0 ) ( α Z0 ) = r α α Z0 Z0α+ α Z 0 + Z 0 α = α + () α Z0 = r α ( Z 0 ).( Z0 ) r + = r (, Z 0 = r α α = 4r α α α Z 0 α Z 0 + α r + = 4 r α α Z 0 + α Z 0 = + α () = (4) α + () 7r α (4) r +) r = α + 7 r α α = 7

7 () Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution SECTIN II : (ne 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 NE or MRE are correct. 5. A line passing through the origin is perpendicular to the lines ˆ ˆ ˆ :(+ t)i + ( + t)j + (4+ t)k, < t < ˆ ˆ ˆ :(+ s)i + (+ s)j + ( + s)k, < s < Then, the coordinate (s) of the point (s) on at a distance of 7 from the point of intersection of and is (are) (A),, (B) (,,0) (C) (,,) (D),, (B), (D) ˆ ˆ ˆ ˆ ˆ ˆ : r = (i j+ 4k) + t(i + j+ k) ˆ ˆ ˆ ˆ ˆ ˆ : r = (i + j+ k) + s(i + j+ k) Vector perpendicular to ˆ i+ j ˆ + k ˆ and i ˆ+ j ˆ+ kˆ is ˆi ˆj kˆ = i ˆ+ j ˆ kˆ So, line is r = k( i ˆ+ j ˆ k) ˆ For point of intersection of and k ( i ˆ+ j ˆ k) ˆ = i ˆ ˆj + 4kˆ + t (i ˆ+ j ˆ+ k) ˆ ( k )i ˆ+ (k+ )j ˆ+ ( k 4)kˆ = t(i ˆ+ j ˆ+ k) ˆ k = t (i) and k + = t (ii) k 4 = t (iii) k + (ii) (iii) = k 4 k + = k 4 k = From (i), ( ) = t t =. So, : r = i ˆ j ˆ+ kˆ So, point of intersection of and is A (,, ). The line can be written as x y z = = = s Any point on (s +, s +, s + ) = B Given, AB = 7 (s + ) + (s + 6) + s = 7 9s + 8s + 0 = 0 (s + ) (9s + 0) = 0 s = or s = 0 9 So, point B (,, 0) and 7 7 8,, 9 9 9

8 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) () 5. 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 + ) 5. (B), (C) f(x) = x sinπx f (x) = sinπx + (x) π(cosπx) sinπx + xπ cosπx = 0 sinπx = πx cos πx tanπx = πx y 0 x 5. Let S n = 4n k= k(k+ ) ( ) k. Then S n can take value(s) (A) 056 (B) 088 (C) 0 (D) 5. (A), (D) S n = 4n k= k(k+ ) ( ).k = ( ). + ( ). + ( ) 6. + ( ) upto 4n terms = (4n) = n (n + terms) = n n times = 6n + 4n 54. For matrices M and N, which of the following statement(s) is (are) NT correct? (A) N T M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric (B) M N N M is skew symmetric for all symmetric matrices M and N (C) M N is symmetric for all symmetric matrices M and N (D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N

9 (4) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 54. (C), (D) (C) is wrong as (MN) T = N T M T = NM MN (D) is wrong as adj (MN) = adj (N) adj (M) for all invertible M and N. 55. 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) (C) 45 (D) (A), (C) Let the sides be 8C and 5C where C is a parameter given (8C + 5 C) = K (constant) C = K 46 Now, 5C x Volume is given by V = (8C x) (5C x) x x x x x or V = 4x 46x C + 0C x dv dx = x 9 xc + 0 C 8C x () area of all squares = 4x = 00 x = 5 dv = 0 for maximum volume at x = 5 dx using x = 5 in () we get 0 C 460 C + 00 = 0 6C C + 5 = 0 C = or 5/6 but C = 5/6 and x = 5 makes sides negative. Thus C = is the only value of the parameter. Hence sides are 4, 45. SECTIN III : (Integer value correct Type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). { } 56. Consider the set of eight vectors V = ai + bj + ck:a, b,c {,}. Three non coplanar vectors can be chosen from V in p ways. Then p is, 56. [5] Total number of vectors are, ˆi+ ˆj+ kˆ ˆi+ ˆj kˆ ˆ i+ ˆ j+ k ˆ ˆ i+ ˆ j k ˆ ˆi ˆj+ kˆ ˆi ˆj kˆ ˆ i ˆ j+ k ˆ ˆ i ˆ j k ˆ Total no. of ways choosing non coplanar vectors = ( 4 C ) ( 4 ) ( 4 C + C C + C) = = P = P = 5 4

10 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (5) 57. f the three independent events E, E and E, the probability that only E occurs is α, only E occurs is β and only E occurs is γ. Let the probability p that none of events E, E or E occurs satisfy the equations (α β) p = αβ and (β γ) p = βγ. All the given probabilities are assumed to lie in the interval (0, ). Pr obability of occurrence of E Then Pr obability of occurrence of E = 57. [6] P (E ) = x P (E ) = y P (E ) = z Then x ( y) ( z) = α y ( z) ( x) = β z ( x) ( y) = γ ( x) ( y) ( z) = P Now, (α β) P = αβ [x ( y) ( z) y ( z) ( x)] ( x) ( y) ( z) = xy ( y) ( z) ( x) x ( y) y ( x) = xy x xy y + xy = xy x = y x y = () Also, [y ( z) ( x) z ( x) ( y)] ( x) ( y) ( z) = yz ( z) ( y) ( x) y ( z) z ( y) = yz y y z = 0 z = x z = The coefficients of three consecutive terms of ( + x) n + 5 are in the ratio 5 : 0 : 4. Then n = 58. [6] Let the coefficients of three consecutive terms be n + 5 Cr, n + 5 Cr+ and n + 5 Cr+ Given n + 5 Cr : n + 5 Cr+ : n + 5 Cr+ : : 5 : 0 : 4 Solving n + 5 Cr : n + 5 Cr+ : : 5 : 0 We get r = n + () likewise solving n+ 5 Cr+ = 0 we get n+ 5 Cr+ 4 r = 5n + 6 () solving () & () we get n = 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 = 5

11 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 59. [5] n(n + ) The sum of first n cards is, S =. if two consecutive cards numbered k and k + are removed then s = n(n + ) (k + k + ) = 4 (given) n(n + ) = k, which is satisfied for n = 50 and k = 5 in the given limits k = 5 and k 0 = 5. x y 60. A vertical line passing through the point (h, 0) intersects the ellipse + = at the 4 points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. max min If Δ(h) = area of the triangle PQR, Δ = Δ( h) and, ( h) / h Δ = / h Δ then Δ Δ = 60. [9] Let, line x = h For P and Q points h y + = P h, 4 h 4 h y = 4 (h, 0) R y = (4 h ) 4 Q h, 4 h y= ± 4 h Tangent at P, y 4 h xh + 4 =. So, R, 0 4 h / 4 Δ P Q R = ( 4 h ) h h = 4 h (4 h ) 4 h = h h Δ = 6 4 h < 0 h h Δ is decreasing / 4 / 4 5 So, Δ = max Δ (h) at h = = = h 4 Δ = / min Δ (h) (at h = ) = So, / Δ Δ = 5 / = = 9 6

12 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (7) 7