Jmetquant Practice Paper-2 Solutions

Solutions for JMET-2 1.

π

π

0

0

9.

∫ x f (sin x ) dx = ∫ (π − x ) f (sin x ) dx π

or ∫ x f (sin x ) dx = 0

1 π π ∫ f (sin x ) dx 2 0 Choice (1)

cos(A + B )

2.

sin(A + B + C)

tan(B + C)

sin(− B )

=

0

sin B

sin(π / 2 − C)

0

tan A

0

sin B 0

− cos C − tan A

4.

5.

6.

7.

cos C tan A (rearranging rows) 0

n(A) = 4, n(B) = 6 Number of mappings that can be defined from 4 A to B = 6 Number of injective mappings that can be defined from 6 A to B = P4 6P 6×5× 4× 3 5 4 ∴ Required probability = = = 64 6 × 6 × 6 × 6 18 Choice (3)

Standard Result: If R and S are relations such that S ⊆ A × B; R ⊆ B × C; and R ο S ⊆ A × C; –1 –1 –1 then; (R ο S) = S ο R –1 –1 –1 Here A = B = C. Hence (R ο S) = S ο R Interchanging R and S, we have –1 –1 –1 (S ο R) = R ο S Choice (3) Let 5sin θ + 12cos θ = k ⇒ 13(5/13 sin θ + 12/13 cos θ) = k or 13(sin α sin θ + cos α cos θ) = k where sin α = 5/13 and cos α = 12/13 i.e., 13[cos (θ – α)] = k or cos(θ – α) = ± 1 ∴ θ – α = 0 or 180 , θ – α ≠ 360° as θ ≤ 360° and α ≠ 0 ∴Two solutions exist. Choice (3) The maximum possible value is obtained at the point of x intersection of the lines y = 4 – x and y = – 1 i.e., at 2 10 10 10 / 3 2 ∴ Max (f(x))= Min (4 – – 1) = , . x = 3 3 2 3 Choice (2) The statement given in choice (3) is true. (1) is false as we must have l + m + n = 0 (2) is false. Volume = 1/6 [a b c ] (4) is false as in the given case [a b c ] = 0 Choice (3)

8.

4

4

3

3

4

Required Area = ∫ [ y] dy = ∫ 3dy = 3( y ) 3 = 3 (4 – 3) = 3 sq.units

⎛ ⎞ r6 ⎟ (4π2 α4 – α6) ⎜ 9 × 64π 4 ⎟ ⎝ ⎠ 2 4 6 If we denote (4π α – α ) by u; then du 2 3 5 = 16π α – 6α dα

= ⎜

Since, the matrix is a skew-symmetric matrix of odd order the determinant is zero. Choice (4) 3.

Let V represent the volume of the cone. ⎛ 1⎞ 2 V= ⎜⎜ ⎟⎟ π x h where h is the height of the cone. ⎝3⎠

⎛ 1⎞ 2 4 2 2 2 2 2 V = ⎜⎜ ⎟⎟ π x h where h = r – x ⎝9⎠

0 − cos C − tan A sin π sin B cos C 0 tan A − sin B

∆ = − sin B

Let r be the radius of the circle. Note that the slant height of the cone is r And the circumference of the base is r α ⎛rα ⎞ If x is the radius of the base x = ⎜⎜ ⎟⎟ ⎝ 2π ⎠

Choice (2)

d2u dα 2

2

2

= 48π α – 30α

4

du 8 π2 2 = 0 gives α = dα 3 2

For this value of α , it can be seen that

d2u

dα 2 negative. Therefore, the required value of the angle α = 2π

2 3

is

Choice (3)

10. If xy > x + y, then 1 > (x + y) / xy or 1/x + 1/y < 1 → (1) Since A.M. ≥ G.M

1/x + 1/y ≥ 2

1 → (2) xy

combining (1) and (2) 2 <1 xy or 2 <

xy ⇒ xy > 4

Choice (1)

11. Given equation: a sinx + b cosx = c

Dividing above equation with a a2 + b2

. sin x +

sin ( x + φ)=

b a2 + b2

c 2

a + b

2

a2 + b2 ; we have . cos x =

where φ = sin

–1

As 0 ≤ sin ( x + φ) ≤ 1; we have 0 ≤ c ≤ Hence, if c >

c a2 + b2

⎞ ⎛ a ⎟ ⎜ ⎟ ⎜ 2 2 ⎜ a + b ⎟ ⎠ ⎝ a2 + b2

a2 + b2 ; no solution exists. Choice (3)

12. Let AB = 2x be the vertical pole and C be the midpoint of AB such that AC = BC = x. B Given AP = Nab = 2nx ∠CPA = α and ∠BPA = β. le x Fro, ∆ CAP, C AC x 1 = = tan α = AP 2nx 2n x β le From ∆ BAP α A P

1

–1

16. We have 0 ≤ cos x ≤ π –1 –1 –1 ∴ cos x = π, cos y = π, cos z = π ⇒ x = –1, y = –1, z = –1 ⇒ xy + yz + xz = 3.

AB x 1 = = AP 2nx 2n ∴tan β = tan ((α+ β) – α) tan (α + β ) − tan α = 1 + tan(α + β ) tan α tan(α+ β) =

⎛ 1⎞ ⎛ 1 ⎞ ⎜⎜ n ⎟⎟ − ⎜⎜ 2n ⎟⎟ n ⇒ tan β = ⎝ ⎠ ⎝ ⎠ = 2 ⎛ 1 ⎞ ⎛ 1 ⎞ 2n + 1 1 + ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎝ n ⎠ ⎝ 2n ⎠

Choice (1)

13. Given: A = 0; b = z; C = zi ; D = z(1 + i) Taking z = x + iy; we have

A

2

⇒c=±

D •

•

Now AB = BC = =

(x

2

x + y ;

+ y ) + (y − x ) 2

2

⎞ ⎛ 2 ⎜ x2 + y2 ⎟ ⎠ ⎝

CD =

x2 + y2 ;

AD =

(x

− y )2 + (y + x )2

x 2 + y 2 ; BD =

x2 + y2

Here AB = CD; BC = AD and AC = BD ∴ABCD form a square.

Choice (3)

14. Given lines: 2x + y = 3 y = 3 – 2x 2 Area A = xy = 3x – 2x

dA = 3 – 4x = 0 dx 3 x= 4 3 3 9 Area = × = . 4 2 8 15. A1 = =

Since c ∈ [0, 2], c =

S2 =

1 ∞ ⎛a+4⎞ ⎟ ∑⎜ 5 n=0⎝ 5 ⎠

2 3

Choice (1)

1 1− a

n

2 ⎤ 1 ⎡ ⎛a+ 4⎞ ⎛a+ 4⎞ ⎢1 + ⎜ ⎟ + .....∞⎥ ⎟+⎜ 5 ⎢ ⎜⎝ 5 ⎟⎠ ⎜⎝ 5 ⎟⎠ ⎥ ⎣ ⎦ 1 1 1 ⋅ = = S1 = 5 ⎛ a + 4 ⎞ 1− a 1 − ⎜⎜ ⎟⎟ ⎝ 5 ⎠

=

Choice (2)

19. Let f: A → B and f(x) = ⏐x⏐ If x1, x2 ∈ A then f(x1), f(x2) ∈ B Now if f(x1) = f(x2) ⇒ ⏐x1⏐ = ⏐x2⏐ ⇒ x1 = ± x2 ∴ f is not injective. Also f(1) = ⏐1⏐ = 1 and f(–1) = ⏐–1⏐ = 1 If y ∈ B, such that f(x) = y then ⏐x⏐ = y, i.e., y is always positive, which is not ture since –1 ∈ B ∴There is no x in A for every y in B. ∴ f(x) is not surjective Choice (4) n–1

( ) ( ) ) log (A R log A 2R 2n

Choice (3)

1 OA × OB 2

A 2 = OA × OC = 12 (b × a ) = 12 a × b

1 1 A1 = ⇒ A1 = A2 4 4 A2 1 4

3

20. xn = AR log xn = log A + (n – 1) log R Given determinant, on applying C1 + C3,

1 6 a × b = 3 a× b 2

∴k =

2

∴ S1 = S2.

⎛ ⎞ = 2 ⎜ x2 + y2 ⎟ . ⎝ ⎠ AC =

f ( 2) − f (0 ) 2 2 = 3c – 5 = –1 ⇒ 3c = 4 2−0

18. S1 = ∑ an = 1 + a + a2 + .....∞ =

• C A = (0,0); B (x, y) C = (–y, x) and D = (x – y, x + y) 2

17. In this case, we need to identify a point 'c' in the interval f (b) − f (a) (a, b) such that f '(c) = (slope of line joining b−a 'a' and 'b' the end points). 3 2 Given f(x) = x – 5x ⇒ f '(x) = 3x – 5 f (b) − f (a) taking, f '(c) = b−a

3c – 5 =

B •

Choice (4)

Choice (4)

( ) ( ) ) log (AR log ARn

log A 2R 2n + 6

log ARn + 3

2 2n +12

n+6

log xn log xn + 3 log xn + 6

Observe that the first column in twice the second column. Hence, the value of the determinant equals zero. Choice (4) 21. We know that the number of non-negative integral solutions of the equation (n + k – 1) x1 + x2 + x3 + …… + xk = n (n∈N) is C k–1 So, the number of non-negative integral solutions of 10 ≤ x1 + x2 + x3 + x4 ≤ 20 13 14 15 16 17 18 19 20 = C3 + C3 + C3 + C3 + C3 + C3 + C3 + C3 + 21 21 22 23 C3 + C3 + C3+ C3 13 13 14 23 13 = ( C4 + C3) + C3 + …….+ C3 – C4 n n (n + 1) [ ∵ Cr + Cr–1 = Cr] 24 13 = C4 – C4 (applying the formula successively) 24 × 23 × 22 × 21 13 ×12 ×11×10 = − 4 × 3 × 2 ×1 4 × 3 × 2 ×1

= 10626 – 715 = 9911 .

Choice (2)

2

22. The number of friends of Anand from city Y is x and the number of friends of Bimal from city Z is x + 3. Let the number of common friends in IIT be y. ∴total number of friends = x + (x + 3) + y = 12 (given) or 2x + y = 9 ; x, y > 0. Also the total number of games played, as per the given conditions is, N = x(x + 3) + xy + (x + 3)y The possible conditions are listed:

X 1 2 3 4 Y 7 5 3 1 N 39 45 45 39 ∴ the maximum number of games that could have been played is 45. Choice (4) 23. In option 2: lim x sin 12 = 0 θ→0 x But, given that f(0) = 3 ∴f(x) is not continuous at x = 0

Choice (2)

24. Given f(a) = 3; f '(a) = –3; g(a) = 2, g '(a) = 3 Using L' Hospital rule we get f (a ) g (x ) − f (x ) g (a ) f (a ) g' (a ) − f ' (a ) g (a ) = lim x →a a−x −1 3(3 ) + 3(2) = –15 Choice (1) = (− 1) 25. There are 43 numbers between 101 and 400 which are multiples of 7; 23 numbers between 101 and 400 which are multiples of 13 and 3 numbers between 101 and 400 which are multiplies of 7 and 13. Therefore, (43 − 3 ) = 2 . P (player wins Rs. 100/–) = 300 15 ( 23 − 3 ) 1 P (Player wins Rs. 150/–) = = 300 15 3 1 = . and P (player wins Rs. 250/–) = 300 100 The expected value in the game 2 1 1 ⎞ ⎛ = ⎜⎜100 × + 150 × + 250 × ⎟ rupees. 15 15 100 ⎟⎠ ⎝

= Rs. 25.83 Since the admission fee is Rs. 25/– gain of for the player = (25.83 – 25) rupees. = Rs. 0.83. Choice (1)

27.

(1 + cos 10 x )−1 (1 − cos 10 x )−1

=

2

n

n

1

[ Q 1 − x + x + ….. + (−1) x +….. = (1 + x)− and 2 n –1 1 + x + x + …. + x + ……= (1 − x) ] 1 − cos 10 x 1 − sin 5 x = ⇒ 1 + cos 10 x 1 + sin 5 x By method of elimination of choices we find that only choice (4) satisfies the equation. Choice (4) 4 5

28. Given focus = (0, ±4) = (0, ± be) and e =

⇒ be = ± 4 4b 2 2 2 = ± 4 ⇒ b = ± 5 and a = b (1 − e ) ⇒ 5 ∴ Required ellipse is

x2 y2 = 1. + 9 25

Choice (3)

5 1/3

29. Given f(x) = (5 – (x – 8) ) let f(x) = y 5 1/3 ⇒ (5 – (x – 8) ) = y 5 3 ⇒ (x – 8) = 5 – y 3 1/5 ⇒ x = 8 + (5 – y ) 3 1/5 Now, let z = g(x) = (8 + (5 – x ) , then 3 1/5 5 1/3 f(g(x))[5 – [8 + (5 – x ) – 8] ] ] 3 1/3 = (5 – (5 – x )) =x Similarly, we can show that g(f(x)) = x. 3 1/5 Hence, g(x) = 8 + (5 – x ) is the inverse of f(x). 30. S = 1 +

Choice (2)

1. 3 . 5 1. 3 1 + ………∞ + + 7 . 14 7 7 .14 . 21 2

3

1. 3 . 5 ⎛ 1 ⎞ ⎛ 1 ⎞ 1. 3 ⎛ 1 ⎞ S = 1 + 1 ⎜⎜ ⎟⎟ + ⎜ ⎟ + ⎜ ⎟ + ………∞ 2! ⎜⎝ 7 ⎟⎠ 3! ⎜⎝ 7 ⎟⎠ ⎝7⎠ The expansion is in the form ⎛ x ⎞ p (p + q) (1 – x)−p/q = 1 + p ⎜⎜ ⎟⎟ + 2! ⎝q⎠ Here p = 1, p + q = 3, x/q = 1/7 S = (1 – x)

–p/q

= (1 – 2/7)

–1/2

2

⎛x⎞ ⎜⎜ ⎟⎟ + .......... ⎝q⎠ ⇒ q = 2, x = 2/7

= (5/7)

–1/2

= 7/5 Choice (2)

2

31. Given |x – 3x + 1| = x – 3, Hence (x – 3) ≥ 0 2 2 If x – 3x + 1 ≥ 0; x – 3x + 1 = x – 3 2 2 x – 4x + 4 = 0 or (x – 2) = 0 ⇒ x = 2, since x – 3 ≥ 0, x = 2 is not a valid solution. 2 2 also, if (x – 3x + 1) < 0 then –(x – 3x + 1) = x – 3 or

26.

B

2

x – 2x – 2 = 0 ⇒ x = 1 ±

30° 30°

A

D

As (1 ± 3 ) – 3 ≱ 0, x = 1 ± 3 is not valid. Hence, the number of solutions = 0. Choice (1) 2

Let AB = ℓ, then l 3 l and BD = ℓ sin 30 = 2 2

⎛ 3 l⎞ ∴ The coordinates of B are ⎜ l, ⎟ ⎜ 2 2⎟ ⎝ ⎠ ⎛ 3 ⎞ ⎛l⎞ 2 l⎟ since, B lies on y = 8x, ⎜⎜ ⎟⎟ = 8⎜ ⎜ 2 ⎟ ⎝2⎠ ⎝ ⎠ 2

or ℓ = 16 3 units

2logk

32. Let α and β be the roots of the equation x –3kx + 2e then αβ = 7 (given) 2logk ⇒ (2 e –1) = 7

C

AD = ℓ cos30 =

3

–1

2

⇒ 2 elog k = 8 2 ⇒k =4 i.e., k = ± 2.

Choice (2)

33. Given y = x 2 − 35 − 16 , for the curve to meet the x-

axis, y = 0. x 2 − 35 − 16 = 0

Choice (4)

3

x = 31; 2y + z = 7 ; y = 1,2,3……; No of values = 3 x = 32; 2y + z = 4 ; y = 1……; No of values = 1 ∴Number of solutions = (1+ 2 + 3 + ……+ 48) – (2 + 5 + …… + 47)

∴ x 2 − 35 = 16 or x – 35 = ± 16 2

2

2

∴x = 51 or x = 19 51 , − 15 ,

⇒x=

19 , − 19 .

∴ the curve intersects the x-axis at 4 points. Choice (4) 34. Consider the given figure: let AB and CD be the towers and P the points selected. We need to minimize (AP + PC).

=

= 24 × 49 – 8 × 49 = 784.

15

21 P 20

B

• A

• P

• Q

• B

Let AP = x and AQ = y, given x − y ≤ 20 Plotting the region required on the graph paper, we hav y

D

(100,120) R

C (0,120)

15

Choice (1)

37. Consider the line segment AB

C

A

48 × 49 16 (2 + 47) − 2 2

(x– y) = 0 B (120, 100)

Q

A'

(0,20) S

Now , to minimize (AP + PC), A’PC should be a straight line. 2 2 2 In ∆A'CQ : (A'C) = (A'Q) + (CQ) 2

2

∴ (A'C) = (48) + (36)

A (120, 0)

P

O

2

⇒ A'C = 60. Hence, the minimum value of S is 60 m. Choice (3)

Q

(20, 0)

x

The region PQRS which represents x − y ≤ 20 is the favourable region.

x + 1 + y = 4 and

35. The regions represented by

∴required probability =

x − 1 + y = 4 are shown in the given figure. =

A

–5

–4

1

–3

–2 –1

–3

120

2

2

⇒ π r h = 1 or r =

2 D

120 2 − 2 × (1/ 2 × 100 2 )

B 1 2 –1 –2

2

11 ⎛ 100 ⎞ = 1 – ⎜⎜ ⎟⎟ = 36 ⎝ 120 ⎠ Choice (4)

38. Let ‘h’ and ‘r’ be the height and the radius of the cylinder. 2 Hence, the volume V= π r h, given V= 1

4 3

Ar. of PQRS Ar. of OABC

3

4

5

1 πh

Let K be the cost per square metre for the side, then the cost for top and bottom will be 4k per square metre. 2 Total cost C = (2 π r ) × (4K) + (2π rh) × k

⎛ 1 ⎞ ⎟ + 2π k ⎝ πh ⎠

C= 8π k ⎜

C –4

The shaded region gives the required area Area of ABCD = 4 × (1/2 OA × OB) = 18 sq.units Choice (4)

⎛ ⎜h × ⎜ ⎝

1 ⎞ ⎟ πh ⎟⎠

⇒ C = 8k/ h + 2k

πh

For minimizing C ;

8k 2k π dC = 0⇒ 2 = dh h 2 h

h=

4 3

π

Choice (2)

+

36. Given 3x + 2y + z = 100; x, y x ∈ Z for x = 1; 2y + z = 97 ; y = 1,2……48; No of values = 48 x = 2; 2y + z = 94 ; y = 1,2……46; No of values = 46 x = 3; 2y + z = 91 ; y = 1,2……45; No of values = 45 x = 4; 2y + z = 88 ; y = 1,2……43; No of values = 43 x = 5; 2y + z = 85 ; y = 1,2……42; No of values = 42 . . . x = 30; 2y + z = 10 ; y = 1,2……4; No of values = 4

n

n–1

39. Sum of the roots of an equation of form a0 x + a1 x + ….. + an = 0 is given by –a1/a0 In this case the sum of the roots = –(–4)/1 = 4 Choice (1) 3

40. Taking x = y = z =1 , we have (x + 2y + 3z) 3 3 = (1 + 2 + 3) = (6) = 216, which the sum of the coefficients Choice (2)

4