Solutions For Jmet Quant Paper-1

Solutions for JMET Quant paper-1 1.

Since in A ∩ B = {3, 5}, the set B should contain the elements 3 and 5. The element 6, should not be present in B as it does not figure in A ∩ B. Now, the remaining elements of µ i.e., 1, 2, 4 and 7 may or may not belong to B 4 ∴ 2 i.e., 16 such subsets are possible Choice (1)

2.

On solving the determinant, we get a linear equation, Hence the given determinant represents a straight line Choice (4)

3.

x − 2 5x +1 =0

⇒ f(0 + 0) = f(0) × f(0) = f(0) ⇒ f(0) = 1 → (2) f (h) − 1 =9 h→0 h

∴from (1) and (2) lim

⎛ f (h) − 1 ⎞ Now, f '(5) = f(5) lim ⎜⎜ ⎟ = 4 × 9 = 36 h→0 ⎝ h ⎟⎠ Alternate method: The property f(x + y) = f(x) . f(y) is satisfied by x x y x+y exponential functions. Let f(x) = a , such that a . a = a . 5 Given a = 4 and f '(0) = 9 x 0 f '(x) = loga . a , so f '(0) = loga . a = 9 or loga = 9 5 ∴f '(5) = loga × a = 9 × 4 = 36 Choice (2)

2

x=

2 5 ±

20 − 4

7.

2

x = 5 + 2, 5 − 2 o Given, angle between these sides is 60 . b ² + c ² − a² From cosine rule, cos A = 2bc ( 5 + 2)² + ( 5 − 2)² − a² 2(1) 2 ⇒ 1 = 18 − a

∴1/2 =

2

or a = 17 i.e. a =

The powers of 7 end in 1, 7, 9 or 3, the number x and y x y should be selected in such a way that 7 has 1 and 7 x y has 9 in units place (or vice versa) or 7 has 3 and 7 has 7 in units place (or vice versa) 1 5 6 Now the powers 7 , 7 , 7 ………(25 in number) have 7 in units place, similar is a case with the other numbers in units place. 1 25 × 25 ∴ Required probability = 4 × = 100 × 100 4 Choice (3)

Choice (4)

17

π

8.

y=∫

0e

ecos x cos x

ecos(π− x )

π

4.

y=∫ D

→ (2) + e −cos x e adding (1) and (2) we have π 2y = ∫ 1 dx 0 or y = π/2 0

E

h

x

α A Let CD be the cliff. tan α = CD/AC tan β = DE/EB = (CD – x) / AC CD – x = AC tan β tan β × CD CD – x = tan α

C 9.

⎡ tan α − tan β ⎤ CD ⎢ ⎥=x tan α ⎣ ⎦ ⎛ ⎞ tan α x cot β ⎟⎟ = ∴h = x ⎜⎜ ⎝ tan α − tan β ⎠ cot β − cot α 5.

( ) [

a. b×c = a b c = = =

6.

Choice (2)

]

Choice (4)

f ( x ) − f (0 ) =9 x−0 f (0) f (h) − f (0) f (0 + h) − f (0) = 9 or lim =9 ⇒ lim h→0 x →0 h h ( Q f(x + y) = f(x) f(y)) f (h) − 1 = 9 → (1) ⇒ f(0) lim h→0 h Now, f(x + y) = f(x) × f(y) for all x, y ∈ R We have f '(0) = 9 ⇒ lim

x →0

cos x

Choice (3)

2

For the roots to be equal b = 4ac 2 If b = 1, b = 1, ac = ¼ (no integral values of a, c) 2 b = 2, b = 4, ac = 1, a = 1, c = 1 2 b = 3, b = 9, ac = 9/4 (no integral values of a, c) 2 b = 4, b = 16, ac = 4, (a = 1, c = 4), (a = 2, c = 2), (a = 4, c = 1) 2 b = 5, b = 25, ac = 25/24 (no integral values of a, c) 2 b = 6, b = 36 ac = 9, (a = 3, b = 3) ∴ Out of 216 possible equations only 5 satisfy the given condition. ∴ Required probability = 5/216 Choice (2)

10. Initially we select five places for the vowels out of 9, in which they can be arranged only in 1 way. Now the remaining 4 places can be occupied by the four consonants in 4! ways. 9 ∴ Number of such words = C5 × 4! 9×8×7 ×6 = × 4! 4 × 3 × 2 ×1

l m n ⎡∧ ∧ ∧ ⎤ 1 1 1 ⎢ i j k⎥ ⎦ m n l ⎣

l (l − n) − m (l − m) + n (n − m) 2 2 2 l + m + n − lm − mn − nl

dx

e − cos x dx

π

y=∫

β

dx → (1)

cos(π − x ) + e−cos(π − x ) 0e

Consider the given figure:

B

+ e − cos x

= 3024 –

Choice (2) –

–

2

11. cot (2x + 1) + cot (4x + 1) = cot (x /2) 2 ⎛ 1 ⎞ ⎛ 1 ⎞ ⇒ tan − ⎜ ⎟ = tan − ⎟ + tan − ⎜ x2 ⎝ 4x + 1 ⎠ ⎝ 2x + 1 ⎠

∴

tan −

⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟+⎜ ⎟ ⎝ 2x + 1 ⎠ ⎝ 4 x + 1 ⎠ = tan − 2 1 1 x2 1− 2x + 1 4 x + 1 1

⇒ x (3x + 1) = 8x + 6x 2 i.e., x = 0 or 3x – 7x – 6 = 0 ⇒ x = 0, 3 or –2/3, x = –2/3 is rejected as it makes LHS negative where as RHS is positive. ∴Number of values that x can take is 2. Choice (3) 2

2

12. Given equation: 2 2 25x – 100x + 16y – 144y – 44 =0 81 2 2 25(x – 4x + 4) + 16(y – 9y + ) = 468 4

18. y = f(x), slope of the tangent at the point (x, f(x)) is dy = f ' ( x ) = 6x + 2, integrating this equation, we have dx 2 y = 3x + 2x + k, since this curve passes through (2, 16), 2 16 = 3 × (2) + 2 × 2 + k, or k = 0 2 Now, the area enclosed by y = 3x + 2x, the x-axis and 2

the line x = 2 is given by A = ∫ (3x2 + 2x ) dx = [ x3 + x2 ]20 0

= 12 sq.units.

2

9⎞ ⎛ 2 25(x – 2) + 16 ⎜ y − ⎟ = 468, this reduces to the form 2⎠ ⎝

(x − h)2 + (y − k )2

= 1, on dividing by 468, which is a a2 b2 general form of an ellipse. Choice (2) 3

5

7

13. We know that sin x = x – x /3! + x /5! – x /7! + ….. ⎞ d ⎛ x3 x5 d ⎜x − ∴ (sin x ) = cos x + − ..... ⎟⎟ = ⎜ dx ⎝ 3! 5! ⎠ dx Choice (2)

Choice (2) n

n

19. Given f(x) = (1 – x) , then f(0) = (1 – 0) = 1 1 n – 1 2 n – 2 also, f (x) = n(1 – x) , f (x) = n(n –1) x ,…., 1 f (x) = n(n – 1)(n – 2)…..2.1 n – 1 2 hence, f′(0) = n = n(1 – 0) = n, f (0) = n(n –1), 3 f (0) = n(n – 1) (n – 2) and so on. f 1(0) f 2 (0) f n (0 ) + + ....... + 1! 2! n! n(n − 1) n(n − 1) (n − 2) n(n − 1) (n − 2)...2.1 = 1+ n + + + .... + 2! 3! n! n n n n 6 = C0 + C1 +…. Cn = 2 = 2 = 64 Choice (3)

∴ f(0) =

2

14. Let R and S be the relations defined on set A = {a, b,c} such that R = {(a, a), (a, b)} and S = {(b, b), (b, c)}. Now, R and S are transitive in nature but R ∪ S = {(a, a) (b, b), (a, b) (b, c)}, is not transitive, as (a, b), (b, c) ∈ R ∪ S but (a, c) ∉ (R ∪ S) ∴R ∪ S is not transitive Choice (3) 15. The given graphs intersect at only one point (1, 0)

a<1 a>1 (1, 0) Choice (2)

16. Let z = x + iy.

z + ⏐z⏐ = x – y + 2xyi + 2

2

2

x2 + y2 = 0

⇒x –y + 2

x2 + y2 = 0

2

⇒ 2xy = 0 ⇒ x = 0 or y = 0 when x = 0 y = 0, ± 1 ⇒ z = 0, i, –i ∴ Three such complex numbers exists. 17. We have x = Xcos30° − Ysin30°, y = X sin30° + Ycos30° substituting these values in x + y = 2, we get ⎞ ⎛ ⎛ 3⎞ ⎟ − Y⎛⎜ 1 ⎞⎟ + X⎛⎜ 1 ⎞⎟ + Y⎜ 3 ⎟ = 2 X ⎜ ⎜2⎟ ⎜2⎟ ⎟ ⎜ ⎜ 2 ⎟ 2 ⎝ ⎠ ⎝ ⎠ ⎠ ⎝ ⎠ ⎝

⎞ ⎛ ⎛ 3 + 1⎞ ⎟ + Y⎜ 3 − 1 ⎟ = 2 ⇒ X⎜ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎠ ⎝ ⎠ ⎝ Y X = 1, ⇒ + ⎛ 4 ⎞ ⎛ 4 ⎞ ⎟ ⎟ ⎜ ⎜ ⎜ 3 + 1⎟ ⎜ 3 − 1⎟ ⎠ ⎠ ⎝ ⎝ 4 4 where h = ,k = 3 +1 3 −1 Now,

1 2

+

1 2

=

h k 2(1 + 3) 1 = = . 16 2

21. K: Sonia know the answer. G: Sonia guesses. C: Question being answered correct. 2 3 P(G) = 1 − = 5 5 1 and P(C/K) P(C/G) = 5 We now want P(G) ⋅ P(C / G) P(G/C) = P(G) ⋅ P(C / G) + P(k ) ⋅ P(C / K )

= Choice (2)

2

20. Given inequality is log9x + (log3x) < 6 2 2 ⇒ 2log9x + (log3x) < 6 ⇒ log3 x + (log3 x )2 < 6 2 put log3x = t 2 ⇒ t + t − 6 > 0 ⇒ (t + 3) (t − 2) < 0 ⇒ t ∈ (−3, 2) ⇒ log3x ∈ (−3, 2) ⎛ 1 ⎞ , 9 ⎟⎟ . ⇒ x ∈ ⎜⎜ Choice (4) ⎝ 27 ⎠

3 1 ⋅ 3 5 5 . = 3 1 2 13 ⋅ + ⋅1 5 5 5

Choice (2)

22. Given equation is nx − (2n + 1) x + (n + 1) = 0 2 ⇒ nx − nx − (n + 1) x + (n + 1) = 0 ⇒ nx (x − 1) − (n + 1) (x − 1) = 0 ⇒ [nx − (n + 1)] (x − 1) = 0 n +1 ⇒ x = 1, n 6 Now both the roots are less then 5 n +1 6 < ⇒ 5n + 5 < 6n ⇒ n > 5 ⇒ n 5 ∴ n = 6. Choice (2) 2

( x + 1) = 90 2 ⇒ x = 179, which is odd. x If f ( x ) = + 1 = 90 3 ⇒ x = 267, which is not even. ∴x = 179

23. For f ( x ) =

( 3 + 1)2 ( 3 − 1)2 + 16 16 Choice (1)

Choice (2) 2

24. We know that cos−1x + sin−1x =

π 2

Hence maximum speed is when log x =

π , then y = x 2

if sin−1 x + cos−1y =

x=

1

−1 or when 2

Choice (3)

e

∴ we have 2x 2 3 x 3 4 x 4 2 x 4 3 x 9 4 x16 + − + ..... = x 2 − + − ..... 3 4 5 3 4 5 By method of elimination of choices we can see that only x = 1 satisfies above equation. Choice (3) x−

2

2 2 2 ∫ π [f ( x )] dx = ∫ π y dx = ∫ π [x ( x − 2)] dx

b

a

25. If r is the radius of the circumscribing circle, →

→ ⎞ ⎛ → = (1 − n) ⎜⎜ OA 2 × OA1 ⎟⎟ ⎠ ⎝

Choice (2)

26. Required sample space is, S = {00, 100, 0100, 0101, 0110,0111, 1010, 1100, 1011, 1101, 1110, 1111} Choice (4) 27. The number of terms of the sequence 3,7,11, …up to th 2 3 6 7 group = 1 + 2 + 2 + 2 + …... + 2

[∵Sn =

(

)

a rn − 1 ] r −1

= 127 th th ⇒ the first number in the 8 group = 128 term of the sequence. Now, 3, 7, 11, ….. is an arithmetic progression with common difference 4. th ∴ required number = 128 term = 3 + (128 – 1)4 [∵tn = a + (n + 1)d] = 511. Choice (3) 28.

0

2 0

−

( )

4 3 x 3

2 0

( )

4 2 2⎤ x 0⎥ 2 ⎦

+

32 4π ⎛ ⎞ + 8 ⎟⎟ = = π ⎜⎜ 4 − 3 3 ⎝ ⎠

→

)

a 2

( )

Where n a unit vector normal to place of the polygon. 2π ⎞ → ⎛ = ∑⎜ r 2 sin ⎟n n ⎠ ⎝

(

2

= ∫ π ( x − 4 x + 4 x ) dx 3

⎡1 = π ⎢ x4 ⎣4

⎛ → ⎞ ⎧⎪ 2π ⎫⎪ → ∑⎜ OA i × OA i+1 ⎟ = ∑ ⎨ OAi OA i+1 sin ⎬ n ⎜ ⎟ ⎪ n ⎪⎭ ⎝ ⎠ ⎩

1 27 − 1 2−1

2

b

0

OA i =r for i = 1,2,..,n

=

2

30. Given curve is y = x(x – 2) . The curve meets y-axis at the points (0, 0) and (0,2). ∴Required volume bounded by the curve =

31. Given function is f(x) =

Choice (3)

x +1 − x −1 x

, x > 0.

Case (i): when 0 < x < 1, x + 1 = x + 1 and x − 1 = – (x – 1) (as x + 1 > 0 and x – 1 < 0) (x + 1) + (x − 1) = 2 x Case (ii): when x ≥ 1, x + 1 = x + 1 and x − 1 = x – 1

∴f (x) =

(as x + 1 > 0 and x – 1 ≥ 0) (x + 1) + (x − 1) = 2 ∴f (x) = x x Hence, the graph of the given function is y

•

2

O

•

1

x

dθ = 2π dt

Area of the triangle AOP = y =

1 2 r sin θ 2

dy 1 2 dθ = r cos θ dt dt 2 o

When θ = 60 , dy 1 2 1 πr 2 = r × × 2π = dt 2 2 2

Choice (4)

29. Speed S = −kx log x where k is a cons tan t 2

dS = −2kx log x − kx dx

d2 S

= −2k (1 + log x ) − k dx 2 dS −1 = 0 gives x = 0 or log x = and dx 2

for log x =

−1 d S , is negative . 2 dx 2 2

Choice (2) 32. We know that, if α and β are the roots of f(x) = 0, then the α + h and β + h will be the roots of f(x – h) = 0. ∴ – 3 < α < –1 and β > –1 ⇒ –1 < α + 2 < 1 and β + 2 > 1 2 ∴the roots of a(x – 2) + b (x – 2) + c = 0 are both greater than –1. Choice (1) 33. Let A (x, 2), B (2, x) and C (3,4) be the given vertices Given, Area of ∆ABC is less than 5. 1 x1 − x 2 y1 − y 2 ⇒ <5 2 x 2 − x3 y 2 − y 3

i.e.,

1 x−2 2−x <5 2 −1 x−4

⇒ (x – 2) (x –4) + 1 (2 – x) < 10 2 ⇒ x – 7x < 0 ⇒ x (x – 7) < 0 ⇒ x ∈ (0, 7).

Choice (1)

3

3

34. Given f(x) = min {x, x } 3 and g(x) = max {x, x } Choice (1): when x = 1, f(x) = 1 and g(x) = 1 3 ⇒f(x) + g(x) = 1 + 1 = 2 ≠(1) ∴ choice (1) is false. 3 Choice (2): f(x) = g(x) ⇒x = x 2 ⇒ x (x – 1) = 0 ⇒ x = 0,1, –1. ∴ choice (2) is false. 3 Choice (3): If f(x) = x and g(x) = x ⇒ then (fog) x = f(g(x)) 3 3 3 = f(x ) = x ⇒ and (gof)x = g(f(x)) = g(x) = x 3 if f(x) = x and g(x) = x 3 then (gof)x = g(f(x)) = g(x) = x and 3 (fog)x = f(g(x) = f(x) = x Hence, choice (3) is true. Choice (3) 35. Let the identical twins be A, A and B, B the identical triplets be C, C, C; D, D, D and E, E, E and the other two children be F and G. A, A and B, B should occupy the extreme ends as a set. ∴A A ………………..BB and BB ……………………AA are the two favourable cases. In each of these cases the remaining children can be 11! arranged in ways. 3! 3! 3!

∴total number of ways = 2.

11 ! (3 ! )3

.

⇒ y = ± 10. Hence the required points are (2,10) and (–2, –10). Choice (2) 38. Let y = f(x) be the required curve.

Given, gradient of the curve is

2x − x 2 2

dy 2x − x 2 = dx 2 Integrating w.r.t x, we get ⇒

x2 x3 − +c 2 6 But the curve is passing through the point (0,1) ⇒ 1 = 0 + c i.e c = 1 y=

Hence, the required curve is y =

x2 x3 − +1 2 6 Choice (2)

39.

•B h 15°

C

Choice (3)

•A

75°

2500

F

D 36.

h B(0, 1)

• M2 O(0, 0)

•

E• Let DF be the surface of the take and AC = d cm AB From ∆ACB, tan 15° = AC h − 2500 h − 2500 i.e, d = ------ (1) 2– 3 = d 2 − 3

P(1/2,1/2)

M1 • • P2 P1

A(1, 0)

From ∆ACE, tan 75° =

h + 2500 d h + 2500 i.e, d = ------ (2) 2+ 3 From (1) and (2), we get h − 2500 h + 2500 = 2− 3 2+ 3

From the figure: OA = OB =1 OP1 = P1 P = 1/2 OP2 = P2 M1 = 1/4 and so on. ∴required sum of the areas 1 1 1 1 1 1 1 )+ = ) + …… × (1 × 1) + ×( × × ( × 2 2 2 2 2 4 4 2 4 6 ⎤ 1 ⎡ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎢1 + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + .....⎥ 2 ⎢ ⎝2⎠ ⎥ ⎝2⎠ ⎝2⎠ ⎦ ⎣ 1 a 1 [∵S∞ = = × ] 1 2 1− r 1− 4 2 units. Choice (1) = 3

AE AC

⇒2+

=

3 =

(2 + 3 ) (h – 2500) = (2 – 3 )(h + 2500) ⇒ 2 3 h = 10000 ⇒h=

5000

i.e., h =

3

5000 3 cm. 3

Choice (4)

→ → →

40.

x . y + x ,z=0

→ → → →

37. Given curve is y =

20 x

dy −20 = dx x2 Tangent at P(x, y) is parallel to the line y = 2 –5x −20 ⇒ = –5 (∵slopes of parallel lines are equal) x2 2 ⇒x =4⇒x=±2 ⇒ slope of the tangent at P(x, y) =

y . z + z , x =0

→ → → →

z . x + z , y =0 2

⎛ → → → ⎞ ⎛ → →→ ⎞ x + y + z =⎜⎜ x + y + z ⎟⎟ ⋅ ⎜⎜ x + y z ⎟⎟ ⎝ ⎠ ⎝ ⎠

→ → →

→

2

→

2

→

2

= x + y + z = 4 + 9 + 16 = 29 Answer =

29

Choice (1) 4