Jee Advance 2013 Math i Questions Solutions

JEE Advanced 2013 (02Jun13) Question & Solutions PaperI 1 www. prernaclasses.com

Mathematics Paper I Jee Advance 2013

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PART III MATHEMATICS SECTION 1 : (Only One Option Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of whichONLY ONE is correct.

41. Let f : [1 / 2, 1]→R (the set of all real numbers) be a positive, nonconstant and differentiable

function such that f ' (x) < 2 f (x) and f (1 / 2) = 1. Then the value of ∫ 1

2 / 1 ) ( dx x f lies in the interval

(A) (2e – 1, 2e) (B) (e – 1, 2e – 1) (C)

−

− 1 , 2 1 e e

(D)

−

2 1 , 0 e

41. (D) f ' (x) < 2 f (x) ⇒ 2 ) ( ) (

< ′ x f x f (Q f (x) is (+)ve function)

⇒ ∫ < ⌡ ⌠ ′

dx x f x f 2 ) ( ) (

⇒ ln f (x) < 2x + c ⇒ f (x) < ke 2x

Q f (1 / 2) < ke ⇒ ke > 1 ⇒ k > 1 / e

∫ ⌡ ⌠

−

= <

1

2 / 1

2 1

2 / 1 2 ) 1 ( ) ( e e k dx e k dx x f x ⇒

⌡ ⌠ −

<

1

2 / 1 2

) 1 ( 1 ) ( e e e

dx x f

⇒ ⌡ ⌠ −

< <

1

2 / 1 2 1 ) ( 0 e dx x f .

42. A curve passes through the point (1, 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 + (1 / 2) (B) cosec (y / x) = log x + 2 (C) sec (2y / x) = log x + 2 (D) cos (2y / x) = log x + (1 / 2)

42. (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 ⇒ 1 / 2 = 0 + c ⇒ c = 1 / 2 ⇒ sin (y / x) = log x + (1 / 2).


43. Perpendiculars are drawn from points on the line 3 1 1

2 2 z y x

= − +

= +

to the plane x+ y+ z= 3. The

feet of the perpendiculars lie on the line.

(A) 13 2

8 1

5 − −

= −

= z y x

(B) 5 2

3 1

2 − −

= −

= z y x

(C) 7 2

3 1

4 − −

= −

= z y x

(D) 5 2

7 1

2 −

= −

− =

z y x

43. (D) 3 1 1

2 2 z y x

= − +

= +

= r

(hkw)

P L

P = (2r – 2, – r – 1, 3r) = (x 0 y 0 z 0 ) = Variable point on the line L.

3 3 4

3 6 4

3 ) 3 (

1 1 1 0 0 0 0 0 0 + − =

−

− = − + +

− = −

= −

= − r r z y x z w y k x h

Here (h, k,w) is a variable point on the locus joining the feet of the perpendiculars from L. h = 2r – 2 – (4r / 3) + 2 = 2r / 3 k = – r – 1 – (4r / 3) + 2 = – (7r / 3) + 1

w = 3r – (4r / 3) + 2 = (5r / 3) + 2 ⇒ 5 2

7 1

2 −

= − −

= w k h

. ∴ Answer is (D).

44. Let k j i PR ˆ 2 ˆ ˆ 3 − + = and k j i SQ ˆ 4 ˆ 3 ˆ − − = determine diagonals of a parallelogramPQRSand

k j i PT ˆ 3 ˆ 2 ˆ + + = be another vector. Then the volume of the parallelopiped determined by the

vectors PT , PQ and PS is (A) 5 (B) 20 (C) 10 (D) 30

44. (C)

Q

R

P

S

θ

T

k j i PR ˆ 2 ˆ ˆ 3 − + = , k j i SQ ˆ 4 ˆ 3 ˆ − − = , k j i PT ˆ 3 ˆ 2 ˆ + + =

Volume of parallelopiped = (vector area of parallelogram) . PT

− − − = × = PT k j i

PT d d . 4 3 1 2 1 3

2 1 . ) (

2 1

2 1 r r

{ } 10 } 30 20 10 { 2 1 . ) 10 ( ) 10 ( ) 10 (

2 1

= − + − = − + − − − = PT k j i Q volume is +ve.


45. The value of

+ ∑ ∑

= =

− 23

1 1

1 2 1 cot cot n

n

k k is

(A) 23 / 25 (B) 25 / 23 (C) 23 / 24 (D) 24 / 23

45. (B) ∑ ∑ =

− − − −

=

− = − =

+ + − +

= + + 23

1

1 1 1 1 23

1

1 25 23 tan ) 1 ( tan ) 24 ( tan

) 1 ( 1 ) 1 ( tan )) 1 ( 1 ( cot

n n n n n n n n

∴ Exp.= 23 / 25 23 25 cot cot 1 =

− .

46. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than 2√2. Then (A) a + b – c > 0 (B) a – b + c < 0 (C) a – b + c > 0 (D) a + b – c < 0

46. (AC) Point of intersection is

+ −

+ −

b a c

b a c ,

∴ 2 2 2 2

<

+ + + b a c b a

⇒ 2 0 < +

+ + <

b a c b a

⇒ a + b > c ⇒ a + b – c > 0. Since, option (C) is correct as a > b > c > 0.

47. Let complex numbers α and α 1 lie on circles (x – x 0 ) 2 + (y – y 0 ) 2 = r 2 and

(x – x 0 ) 2 + (y – y 0 ) 2 = 4r 2 , respectively. If z 0 = x 0 + iy 0 satisfies the equation 2 | z 0 | 2 = r 2 + 2, then | α | = (A) 1 / √2 (B) 1 / 2 (C) 1 / √7 (D) 1 / 3

47. (C) z 0 = α

− α 1 2

2 1 2 2 2 2

+ = α

− α r ⇒ 2 1 1 2 2 2 2

+ α

− α = α

− α

⇒ 0 8 1 7 = − + 2 2

α α ⇒ 1 =

2 α or 1 / 7 ⇒ 7 1

= α or 1.

48. The number of points in (– ∞, ∞), for which x 2 – x sin x – cos x = 0, is (A) 6 (B) 4 (C) 2 (D) 0

48. (C) f(x) = x 2 – x sin x – cos x ⇒ f '(x) = x (2 – 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) = – 1. ⇒ f (x) = 0 is satisfied by two points in (–∞,∞)


49. The area enclosed by the curves y = sin x + cos x and y = | cos x – sin x | over the interval [0, π / 2] is (A) 4(√2 – 1) (B) 2√2(√2 – 1) (C) 2(√2 + 1) (D) 2√2(√2 + 1)

49. (B) Area = dx x x x x dx x x x x )) cos (sin ) cos ((sin )) sin (cos ) cos ((sin 2 /

4 /

4 /

0

− − + + − − + ∫ ∫ π

π

π

= ∫ ∫ π π

π

+ 4 /

0

2 /

4 / cos 2 sin 2 xdx xdx = ( ) 1 2 2 2 −

50. Four persons independently solve a certain problem correctly with probabilities 1 / 2, 3 / 4, 1 / 4, 1 / 8. Then the probability that the problem is solved correctly by at least one of them is (A) 235 / 256 (B) 21 / 256 (C) 3 / 256 (D) 253 / 256

50. (A) Probability of solving the problem correctly by at least one of them

256 235

8 1 1

4 1 1

4 3 1

2 1 1 1 =

−

−

−

− −


SECTION 2 (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 whichONE or MORE are correct.

51. For 3 × 3 matricesM andN, 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 matricesM andN (C) MN is symmetric for all symmetric matricesM andN (D) (adjM) (adjN) = adj (MN) for all invertible matricesM andN

51. (CD) (A) (N T MN) T = N T M T (N T ) T = N T MN (since M T =M) (B) (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) = adjM. adjN ∴ statements (C) and (D) are incorrect.

52. A line l passing through the origin is perpendicular to the lines

∞ < < ∞ − + + + − + + t k t j t i t l , ˆ ) 2 4 ( ˆ ) 2 1 ( ˆ ) 3 ( : 1

∞ < < ∞ − + + + + + s k s j s i s l , ˆ ) 2 ( ˆ ) 2 3 ( ˆ ) 2 3 ( : 2

Then, the coordinate(s) of the point(s) on l 2 at a distance of 17 from the point of intersection of l and l 1 is (are) (A) (7 / 3, 7 / 3, 5 / 3) (B) (– 1, – 1, 0) (C) (1, 1, 1) (D) (7 / 9, 7 / 9, 8 / 9)

52. (BD) L 1 : ) ˆ 2 ˆ 2 ˆ ( ) ˆ 4 ˆ ˆ 3 ( k j i t k j i + + + + −

L 2 : ) ˆ ˆ 2 ˆ 2 ( ) ˆ 2 ˆ 3 ˆ 3 ( k j i s k j i + + + + + Let a, b, c be DR of req. line L 1

∴ a + 2b + 2c = 0 2a + 2b + c = 0 ⇒ a : b : c = 2 : – 3 : 2

Eq. of L : ) ˆ 2 ˆ 3 ˆ 2 ( ) ˆ 0 ˆ 0 ˆ 0 ( k j i λ k j i + − + + + Point of int. of L 1 and L Any point of L: (2λ, – 3λ, +2λ) and L 1 : ( λ 1 + 3, 2λ 1 , –1, 2λ 1 + 4) For point of int

2λ = λ 1 + 3, – 3λ = 2λ – 1, 2λ = 2λ 1 + 4 λ = 1, λ 1 = – 1

∴ point of int ≡ (2, – 3, 2) Any point of L 2 : (2k 1 + 3, 2k 1 + 3, k 1 + 2)

⇒ (2k 1 + 3 – 2) 2 + (2k 1 + 3 + 3) 2 + (k 1 + 2 – 2) 2 = ( ) 2 17 k 1 = – 2, k 1 = – 10 / 9

∴ points are ≡ (–1 , – 1, 0) , ( 7 / 9, 7 / 9, 8 / 9).


53. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 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 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are (A) 24 (B) 32 (C) 45 (D) 60

53. (AC) V = (15x – 2y) (8x – 2y) y [y side of square, clearly y = 5 for max. volume] ⇒ V = 120 x 2 y – 46xy 2 + 4y 3 dV / dy = 120x 2 – 46 × 2xy + 12y 2 = 0 when y = 5, 120x 2 – 92 × 5x + 300 = 0 ⇒ 12x 2 – 46x + 30 = 0 ⇒ x = 3, 5 / 6. If x = 3 sides : 45, 24. If x = 5 / 6 : (Not possible since 2y > 8x).

54. Let ∑ =

+

− = n

k

k k

n k S 4

1

2 2 ) 1 (

) 1 ( . ThenS n can take value(s)

(A) 1056 (B) 1088 (C) 1120 (D) 1332

54. (AD) S n = – 1 2 – 2 2 + 3 2 + 4 2 – 5 2 – 6 2 + 7 2 + 8 2 ..... + (4n – 1) 2 + (4n) 2 = 3 2 + 4 2 – 1 2 – 2 2 + 7 2 + 8 2 – 5 2 – 6 2 ..... + (4n – 1) 2 + (4n – 3) 2 + 4n 2 – (4n – 2) 2 = 4 [2 + 3 + 6 + 7 + 10 + 11 + ........ (4n – 2) + (4n – 1)] = 8(1 + 3 + 5 + ........ + (2n – 1)) + 4(3 + 7 + 11 + ........ + (4n – 1)) = 16n 2 + 4n = 4n (4n + 1), 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

+

2 1 , n n

(B) a unique point in the interval

+ + 1 ,

2 1 n n

(C) a unique point in the interval (n, n + 1) (D) two points in the interval (n, n + 1)

55. (BC) f ' (x) = πx cos πx + sin πx = 0 ⇒ sin πx = – πx cos πx ⇒ tan πx = – πx

tan πx

−πx

−1/2 0 1/2 1 3/2

Hence, x∈

+ + 1 ,

2 1 n n and also x∈ (n, n + 1).


SECTION 3 (Integer value correct Type) This section contains5 questions. The answer to each question is asingledigit integer,ranging from 0 to 9. (both inclusive).

56. A pack containsn cards numbered from 1 ton. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards isk, then k– 20 =

56. (5) 1224 ) 1 2 ( 2

) 1 ( = + −

+ k n n ⇒ n 2 + n – 4k= 2450

Q 1 ≤ k ≤ n – 1

1 4 2450 1

2 − ≤

− + ≤ n n n

⇒ 49 <n< 51 ⇒ n= 50

Hence,k= 25.

57. Consider the set of eight vectors } } 1 , 1 { , , : ˆ ˆ ˆ { − ∈ + + = c b a k c j b i a V . Three non coplanar vectors can be chosen fromV in 2 p ways. Thenp is

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 = 24 coplanar vectors (as there are four body diagonals). ∴ No. of non coplanar vectors = 8 C 3 – 24 = 32. ∴ p= 5.

58. A vertical line passing through the point (h, 0) intersects the ellipse 1 3 4

2 2 = +

y x at the pointsPand

Q. Let the tangents to the ellipse atPandQmeet at the pointR. If∆(h) = area of the trianglePQR,

) ( max 1 2 / 1

1 h h

∆ = ∆ ≤ ≤

and ) ( min 1 2 / 1

2 h h

∆ = ∆ ≤ ≤

, then = ∆ − ∆ 2 1 8 5 8

58. (9) Taking R (k, 0) and Chord of contact with respect toR, i.e. PQ is x = (4 / k) = h.

∴

0 , 4 h

R h h h

3/2 2 ) 4 ( . 2 3 ) ( −

= ∆ ∴ 0 < ∆

⇒ dh d

∈ ∀ 1 ,2 1 h

∴ ∆(h) min. = ∆(1) = (9 / 2) = ∆ 2

& ∆(h) max. = ∆(1 /2) = 8

15 15 . 3 = ∆ 1

⇒ 2 1 ∆ − ∆ 8 5 8

= 45 – 36 = 9


59. The coefficients of three consecutive terms of (1 +x) n+ 5 are in the ratio 5 : 10 : 14. Thenn=

59. (6) Since, 2 : 1 5 5 = −

+ + r

n r

n C C ⇒ 3r = n + 6 and

5 7 : 5

1 5 = +

+ +

r n

r n C C ⇒ 12r = 5n + 18. ⇒ n = 6.

60. Of the three independent eventsE 1 ,E 2 andE 3 , the probability that onlyE 1 occurs isα, onlyE 2 occurs is β and onlyE 3 occurs is γ. Let the probabilityp that none of eventsE 1 ,E 2 orE 3 occurs satisfy the equations (α– 2β)p=αβand (β– 3γ)p= 2βγ. All the given probabilities are assumed

to lie in the interval (0, 1). Then = 3

1 of occurrence of y Probabilit of occurrence of y Probabilit E E

60. (6) Letx,y,zbe the probabilities of occurrence ofE 1 ,E 2 andE 3 respectively.

Then, α= x (1 – y) (1 – z) ) 1 ( x

px−

= (Q p= (1 – x) (1 – y) (1 – z))

Similarly, ) 1 ( y py−

= β and ) 1 ( z pz−

= γ

(α – 2β) p = αβ ⇒

−

− =

−

− − y

py x

px p y

py x

px 1 1 1

2 1 ⇒ x= 2y

and (β – 3γ) p = 2βγ ⇒

−

−

=

−

− − z

pz y

py p z

pz y

py 1 1

2 1 3

1 ⇒ y= 3z

⇒ x= 6z ⇒ x / z= 6.

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Jee Advance 2013 Math i Questions Solutions