3
Unit 1.
MATHEMATICS
Answer (1)
Z
5.
4 2 Z Z
z
4 4 |Z| Z |Z|
|Z|
4 2 |z|
– 2|Z| – 4 0 6.
1 5 | Z | 1 5
1 1 z 2 2
1 1 2 2 2
z
zmin.
1 2
3 2
Answer (3) ⎛ z1 2z2 ⎞ ⎜ ⎟ 1 ⎝ 2 z1z2 ⎠
Hence maximum value = 1 5 2.
1 2
So, | z |
|Z|2 – 4 – 2|Z| 0 |Z|2
Answer (4)
Answer (2)
⎛ z1 2z2 ⎞ ⎛ z1 2z2 ⎞ ⎜ ⎟⎜ ⎟ 1 ⎝ 2 z1z2 ⎠ ⎝ 2 z1z2 ⎠
(0, 1)
z1z1 2z1z2 2z2 z1 4z2 z2
(–1, 0)
(1, 0)
4 2z1z2 2z1z2 z1z1z2 z2
O
z1z1 4z2 z2 4 z1z1z2 z2
zz1 1 z2 z2 4 1 z2 z2 0
|z + i| We have,
z1z1 4 1 z2z2 0
|z – 1| = |z + 1| = |z – i|
z1z1 4
Clearly z is the circumcentre of the triangle formed by the vertices (1, 0) and (0, 1) and (–1, 0), which is unique. 3.
Answer (4)
4.
Answer (3)
|z| = 2 i.e. z lies on circle of radius 2. 7.
Answer (3)
2 + 3i sin θ 1 + 2i sin θ = purely in imaginary × 1 − 2i sin θ 1 + 2i sin θ
⎛ 1 z ⎞ ⎛ zz z ⎞ arg ⎜ arg ⎜ ⎟ ⎟ ⎝ 1 z ⎠ ⎝ 1 z ⎠ = arg (z) =
2 – 6sin2 = 0 sin2 = sin = ±
1 3
1 3
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ANSWERS AND SOLUTIONS
ARCHIVE - JEE (Main)
11. Answer (3)
Answer (2) bx2 + cx + a = 0
Coeff. of x = – 7
Roots are imaginary c2 – 4ab < 0
Constant term = 6
f(x) = 3b2x2 + 6bcx + 2c2
The quadratic eqaution is x2 – 7x + 6 = 0
D = 36b2c2 – 24b2c2 = 12b2c2
x = 1, 6 12. Answer (4)
∵ 3b2 > 0
Let f(x) = 2x3 + 3x + k
⎛ D ⎞ f (x) ⎜ ⎟ ⎝ 4a ⎠
f ( x ) c
f (x) = 6x2 + 3 > 0, xR f(x) is strictly increasing function for all real values of k.
2
No real k exists such that equation has two distinct roots in [0, 1].
Now c2 – 4ab < 0 c2 < 4ab –c2
9.
13. Answer (1)
> – 4ab
f(x) > – 4ab.
∵ The equation x2 + 2x + 3 = 0 has complex roots and coefficients of both equations are real.
Answer (3)
Both roots are common.
and are roots of the equation x2 – x + 1 = 0. = 1, = 1
x
1 3i 1 3i 1– 3i , , 2 2 2
x = – or
a b c 1 2 3 14. Answer (3)
From equation, +=6
2
= –2
Thus, = –2, then = – = –, then =
–2
3
where
The value of
=1
a10 – 2a8 10 10 (8 8 ) 2a9 2(9 9 )
2009 + 2009 = (–)2009 + (–2)2009 =– =
[(3)669.2
–[2
+
(3)1337.]
+ ] = –(–1) = 1
10. Answer (2) p(x) = 0 (a – a1)x2 + (b – b1)x + (c – c1) Let p(x) = 1x2 + 2x + 3
x = 1, 4
p(–1) = 0 – 21 + 2 = 0
…(ii)
or x2 – 5x + 5 = – 1
p(–2) = 2 41 – 22 + 3 = 2
…(iii)
x = 2, 3
1 = 2 2 = 4 p(x) = 2x2 + 4x + 2 p(2) = 2.22 + 4.2 + 2 =8+8+2 = 18
6 3 2 2
x2 – 5x + 5 = 1 …(i)
3 = 2
9 ( ) 9 ( ) 2(9 9 )
15. Answer (4)
p(–1) = 0 1 –2 + 3 = 0
(ii) 2 + (iii)
or x2 + 4x – 60 = 0 x = –10, 6 x = 3 will be rejected as L.H.S. becomes –1 So, sum of value of x = 1 + 4 + 2 – 10 + 6 = 3 16. Answer (3) Rearranging equation, we get
nx 2 1 3 5 .... (2n 1) x 1 2 2 3 ... (n 1)n 10n
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⇒ nx 2 n 2 x
ANSWERS AND SOLUTIONS
(n 1)n(n 1) 10n 3
⎛ n 2 31 ⎞ ⇒ x 2 nx ⎜ ⎟0 ⎝ 3 ⎠
Given difference of roots = 1
n[log4 – log3] log10 10 = 1 1 n log 4 log 3
20. Answer (1) Put n = 0 Then when 1 – 62 is divided by 9 then remainder is same as when 63–61 is divided by 9 which is 2.
|| = 1 D1 n2
61
4 2 (n 31) 1 3
So, n = 11
21. Answer (3) 10
S2 ∑ j
10
j 1
17. Answer (3)
C j 10.29
Statement-2 is false.
2| x – 3 | x ( x – 6) 6 0 2| x – 3| ( x – 3 3)( x – 3 – 3) 6 0 2
Only choice is (3). 22. Answer (3) Statement 1 : (n + 1)7 – n7 – 1
2| x – 3| ( x – 3) – 3 0
= n7 + 7C1n6 + 7C2n5 + ... + 7C6 n + 7C7 – n7 – 1
( x – 3)2 2| x – 3| – 3 0
= 7C1n6 + 7C2n5 + ... + 7C6n
(| x – 3 | 3)(| x – 3 | –1) 0
Statement 1 is true.
| x – 3| 1, | x – 3| 3 0
Statement 2 : By mathematical induction
= 7m mI.
n7 – n is divisible by 7 (true)
x – 3 1
Let n7 – n = 7p pI
x 4, 2
n7 = 7p + n
x = 16, 4
...(i)
(n + 1)7 – n7 – 1 = (n + 1)7 – (7p + n) – 1 = (n + 1)7 – (n + 1) – 7p
18. Answer (3) x2 – x + 1 = 0
= 7l + 7p
Roots are –, –2
Statement 2 is a correct explanation of statement 1.
Let = –, = –2
23. Answer (4)
101 + 107 = (–)101 + (–2)107
24. Answer (3)
= –(101 + 214)
n(A × B) = 2 × 4 = 8
= –(2 + )
The number of subsets of A × B having 3 or more elements.
=1
= 8C3 + 8C4 + ... + 8C8
19. Answer (4) n
9 ⎛3⎞ 1 ⎜ ⎟ 10 ⎝4⎠ n
9 1 ⎛3⎞ ⎜ ⎟ 1 10 10 ⎝4⎠ n
l,pI
⎛4⎞ ⎜ ⎟ 10 ⎝3⎠
= 28 – 8C0 – 8C1 – 8C2 = 256 – 1 – 8 – 28 = 219 25. Answer (3) Given expression can be written as 10
⎛ x1/2 1 ⎞ ⎪⎫ ⎪⎧ 1/3 ⎨ x 1 ⎜⎜ 1/2 ⎟⎟ ⎬ ⎪⎩ ⎝ x ⎠ ⎪⎭
x1/3 x 1/2
10
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ANSWERS AND SOLUTIONS
General term =
ARCHIVE - JEE (Main)
10
Cr x1/3
10 r
x 1/2
r
Sum of coefficient of integral power of x
From question,
50
C0 20 50 C2 22 50 C4 24 50 C50 250
We know that
10 r r 0 3 3 2
(1 + 2)50 =
r=4
50
C0 50 C1 2 ..... 50 C50 250
Then,
i.e., constant term
10
C4 210
50
26. Answer (2)
X {(1 3)n 3n 1, n N }
C0 50 C2 22 ..... 50 C50 250
350 1 2
30. Answer (3) Number to terms is 2n + 1 which is odd but it is given 28. If we take (x + y + z)n then number of terms is n + 2C2 = 28
= 32 ( n C2 nC3 .3 ... 3n 2 ), n N } = {Divisible by 9}
Hence n = 6
Y = {9(n – 1), n N}
6
= (All multiples of 9} So, X Y
2 4 ⎞ ⎛ 2 6 ⎜ 1 2 ⎟ a0 a1x a2 x ...... a6 x x ⎝ x ⎠
i.e., X Y Y
Sum of coefficients can be obtained by x = 1 (1 – 2 + 4)6 = 36 = 729
27. Answer (2) (1 + ax + bx2) (1 – 2x)18 (1 + ax + bx2)[18C0 – Coeff. of x3 =
18C (2x) + 18C (2x)2 – 1 2 18C (2x)3 + 18C (2x)4 – .......] 3 4 –18C3.8 + a × 4.18C2 – 2b × 18 = 0
18 17 16 4a 18 17 .8 36b 0 6 2 = –51 × 16 × 8 + a × 36 × 17 – 36b = 0
So according to what the examiner is trying to ask option 3 can be correct. 31. Answer (3) 21
C1 21C2 ... 21C10
=
= –34 × 16 + 51a – 3b = 0
28. Answer (1) n(A) = 4, n(B) = 2 Required numbers = 8C3 + 8C4 + ...... + 8C8 = 28 – (8C0 + 8C1 + 8C2) = 219
C0
10
C1 10C2 ... 10C10 210 1
= 220 – 210 32. Answer (4)
5
x x3 1 x x3 1
5
2 ⎡⎣x 5 10( x 6 x 3 ) 5 x ( x 6 2 x 3 1)⎤⎦
2 ⎡⎣5 x 7 10 x 6 x 5 10 x 4 10 x 3 5 x⎤⎦
29. Answer (1) 50
2 ⎡⎣x 5 10 x 6 10 x 3 5 x 7 10 x 4 5 x⎤⎦
= 256 – 37
C0 21C1 ... 21C21 1
2 ⎡⎣ 5C0 x 5 5C2 x 3 ( x 3 1) 5C4 x ( x 3 1)2 ⎤⎦
n(A × B) = 8
50
21
= 220 – 1
... (i)
Only option number (2) satisfies the equation number (i).
1 2 x
Required sum = (220 – 1) – (210 – 1)
= 51a – 3b = 34 × 16 = 544 = 51a – 3b = 544
1 2
50
C1 2 x
1
50 C2 (2 x )2 .....
50 C50 ( 2 x )50
Sum of odd degree terms coefficients = 2(5 + 1 – 10 + 5) =2
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33. Answer (3) The number of ways in which 4 novels can be selected = 6C4 = 15 The number of ways in which 1 dictionary can be selected = 3C1 = 3 4 novels can be arranged in 4! ways. The total number of ways = 15 × 4! × 3 = 15 × 24 × 3 = 1080.
3 red balls
9 distinct blue balls
Urn A
Urn B
Two balls from urn A and two balls from urn B can be selected in 3C2 × 9C2 ways = 3 × 36 = 108 35. Answer (3)
C3 C3 10 C2 10
Words starting with SL = 3! = 6 Next words is SMALL Rank = 12 + 24 + 12 + 3 + 6 + 1 = 58 X(4 L 3 G)
Y(3 L 4 G)
3L0G
0L3G
2L1G
1L2G
1L2G
2L1G
0L3G
3L0G
Required number of ways = 4C3 4C3
4
C2 3C1
2
4
C1 3C2
C 2
3
2
3
= 16 + 324 + 144 + 1 = 485
Atleast 1000
37. Answer (2)
41. Answer (1)
4 digit numbers
6,
7,
8
Let S 1
2 6 10 14 ..... 3 3 2 33 3 4
S 1
2 6 10 14 ..... 3 3 2 33 3 4
678 4
Required arrangements = 6C4 × 3C1 × 4! = 1080
n=5
3
3! 3 2!
Number of ways of selecting 1 dictionary from 3 dictionaries = 3C1
n
5,
Words starting with SA
Number of ways of selecting 4 novels from 6 novels = 6C4
n
3,
4! 12 2!
40. Answer (1)
36. Answer (2)
Words starting with M
39. Answer (4)
34. Answer (4)
n 1
63
ANSWERS AND SOLUTIONS
5
2
= 72
S 1 2 6 10 14 2 3 4 5 ..... 3 3 3 3 3
5 digit numbers
5 5 × 4 × 3 × 2 × 1 = 120
2 2 4 4 4 (S 1) 2 3 4 ..... 3 3 3 3 3
S 1 1
Total number of integers = 72 + 120 = 192 38. Answer (3)
4! 12 Words starting with A 2! Words starting with L = 4! = 24
S 2
2 2 2 2 3 ..... 3 3 3
2 3 1
1 3
=2+1 =3
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ANSWERS AND SOLUTIONS
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42. Answer (2)
47. Answer (2)
Number of notes person counts in 10 minutes. = 10 × 150 = 1500
∵ p, q, r are in AP 2q = p + r
Since, a10, a11, a12, ....... are in A.P. with common difference = –2 Let n be the time taken to count remaining 3000 notes, then
Also
n [2 148 (n 1) –2] 3000 2 n2 – 149n + 3000 = 0
...(i)
1 1 4 4 q p 4 ⇒ q –4r r p –
(n – 24) (n – 125) = 0 n = 24, 125 Time taken by the person to count all notes = 10 + 24 = 34 minutes
...(ii)
From (i) 2(–4r) = p + r p = – 9r
43. Answer (4) a2 + a4 + a6 + .... + a200 =
...(i)
a1 + a3 + a5 + ... + a199 =
...(ii)
a2 – a1 = a3 – a2 = ... = d common difference.
q = – 4r r=r 2 Now | – | ( ) – 4
subtract (i) & (ii)
2
⎛ –q ⎞ 4r ⎜ ⎟ – p p ⎝ ⎠
100 d = – – 100 44. Answer (1)
d=
q 2 – 4 pr |p|
16r 2 36r 2 | –9r |
45. Answer (3) 46. Answer (3) S = 0.7 + 0.77 + 0.777 + ... upto 20 terms =
7 .9 + .99 + .999 + .... 9
=
7 ⎡1 – 0.1 + 1 – 0.01 + 1 – 0.001 ⎤ ⎥ 9 ⎢⎣ + ...upto 20 terms ⎦
=
7 9
⎡ 1 1 1 ⎞⎤ ⎛ 1 ⎢ 20 – ⎜ 10 + 2 + 3 + ... + 20 ⎟ ⎥ 10 10 10 ⎠ ⎦ ⎝ ⎣
7 = 9
⎛ 1 ⎛ 1 ⎞⎞ ⎜ ⎜ 1 – 20 ⎟ ⎟ 10 10 ⎠ ⎟ ⎝ ⎜ 20 – 1⎞ ⎟ ⎛ ⎜ ⎜ 1 – 10 ⎟ ⎟ ⎜ ⎝ ⎠ ⎠ ⎝
7 = 9
⎡ 1⎛ 1 ⎢ 20 – 9 ⎜ 1 – 1020 ⎝ ⎣
=
7 ⎡ 1 ⎤ 179 + 20 ⎥ ⎢ 81 ⎣ 10 ⎦
=
7 ⎡ 179 +10 –20 ⎤⎦ 81 ⎣
⎞⎤ ⎟⎥ ⎠⎦
=
2 13 9
48. Answer (1) 10 9 + 2(11)(10) 8 + 3(11) 2(10) 7 + ... + 10(11) 9 = k(10)9 x = 109 + 2(11)(10)8 + 3(11)2(10)7+ ... +10(11)9 11 x = 11108 + 2(11)2(10)7 +... + 9(11)9 + 1110 10
11 ⎞ ⎛ 9 8 2 7 9 10 x ⎜1 ⎟ = 10 + 11(10) + 11 ×(10) +... +11 – 11 10 ⎝ ⎠ ⎛ ⎛ 11 ⎞10 ⎞ ⎜ ⎜ ⎟ 1⎟ x 10 ⎟ 1110 109 ⎜ ⎝ ⎠ ⎜ 11 ⎟ 10 1 ⎟ ⎜⎜ ⎟ 10 ⎝ ⎠
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ANSWERS AND SOLUTIONS
x (1110 1010 ) 1110 1010 10 x = 1011 = k109
51. Answer (2)
tn
k = 100 49. Answer (2) a, ar,
a, 2ar,
G.P.
ar2
A.P.
ar2
2 × 2ar = a + ar2 4r = 1 + r2 r2 – 4r + 1 = 0 r=
4 16 4 2 3 2
⎡ n n 1 ⎤ ⎢ ⎥ 2 ⎣ ⎦ n2
2
n 12 4
1⎡ 2 n 2n 1⎤⎦ 4⎣
1 ⎡ n n 1 2n 1 2 n n 1 ⎤ 1⎥ ⎢ 4⎣ 6 2 ⎦
1 ⎡ 9 10 19 ⎤ 9 10 9 ⎥ 4 ⎢⎣ 6 ⎦
= 96
r 2 3
52. Answer (1) a + d, a + 4d, a + 8d are in G.P.
r 2 3 is rejected
(a + 4d)2 = (a + d) (a + 8d)
∵ (r > 1)
a2 + 8ad + 16d2 = a2 + 9ad + 8d2
G.P. is increasing.
8d2 = ad
50. Answer (2) l n m 2
a =8 d
Common ratio =
l + n = 2m
…(i)
a + 4d 8+4 4 = = a+d 8 +1 3
53. Answer (1) 2 2 2 16 ⎛ 3⎞ ⎛ 2⎞ ⎛ 1⎞ 2 1 2 3 ⎜ 5⎟ ⎜ 5⎟ ⎜ 5 ⎟ 4 ....... = 5 m ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1 ⎞4
⎛n G1 l ⎜ ⎟ ⎝l ⎠
2
2
2
16 ⎛ 8 ⎞ ⎛ 12 ⎞ ⎛ 16 ⎞ ⎛ 20 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ....10 tens = m 5 5 5 5 6 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2
⎛ n ⎞4 G2 l ⎜ ⎟ ⎝l ⎠
2
16 ⎛4⎞ 2 2 2 2 m ⎜ ⎟ [2 3 4 5 ...... 10 terms] = 5 ⎝5⎠
3
⎛ n ⎞4 G3 l ⎜ ⎟ ⎝l ⎠
2 16 ⎛4⎞ 2 2 2 2 m ⎜ ⎟ [2 3 4 ....... 11 ] = 5 ⎝5⎠
Now G14 2G24 G33
l4
65
2
n ⎛n⎞ ⎛n⎞ 2 (l 2 ) ⎜ ⎟ l 4 ⎜ ⎟ l l ⎝ ⎠ ⎝l ⎠
= nl3 + 2n2l2 + n3l
2
3
16 ⎛4⎞ 2 2 2 2 2 m ⎜ ⎟ [1 2 3 ....... 11 1 ] 5 5 ⎝ ⎠ 2
⎛ 4 ⎞ ⎡11.12.23 ⎤ 16 1⎥ m (given) ⎜ ⎟ ⎢ 6 5 ⎝5⎠ ⎣ ⎦
16 16 [22.23 1] m 25 5
= nl(n + l)2 = nl (2m)2
1 (505) m 5
= 4 nlm2
m = 101
= 2n2l2 + nl(n2 + l2) = 2n2l2 + nl((n + l)2 – 2nl)
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ANSWERS AND SOLUTIONS
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54. Answer (1)
Solving (i) & (ii), We get d = 1, a = 8
9(25a2 b2 ) 25(c 2 3ac ) 15b (3a c ) (15a)2 (3b)2 (5c )2 45ab 15bc 75ac 0 (15a 3b)2 (3b 5c )2 (15a 5c )2 0 It is possible when 15a 3b 0 and 3b 5c 0 and 15a 5c 0 15a 3b 5c
2 Now, a12 a22 ..... a17 140m
82 92 ..... 242 140m
24 25 49 7 8 15 140m 6 6
m 34
57. Answer (2)
a b c 1 5 3
A 12 2.22 32 .... 2.202
b, c, a are in A.P.
(12 22 32 .... 202 ) 4(12 22 32 .... 102 )
55. Answer (4)
20 21 41 4 10 11 21 6 6
As, f ( x y ) f ( x ) f ( y ) xy
Given, f (1) 3
= 2870 + 1540 = 4410
Putting, x y 1 f (2) 2f (1) 1 7
B 12 2.22 32 .... 2.402
Similarly, x 1, y 2 f (3) f (1) f (2) 2 12
(12 22 32 .... 402 ) 4(12 22 32 .... 202 )
10
Now,
∑ f (n ) =
n 1
f (1) f (2) f (3) ... f (10)
= 3 + 7 + 12 + 18 + ... = S (let) Now, Sn 3 7 12 18 ... t n
58. Answer (4)
⎛1 * * ⎞ ⎜ ⎟ Consider ⎜ * 1 * ⎟ . By placing a1 in any one of ⎜ * * 1⎟ ⎝ ⎠
n (n 5) 2
n
∑ tn = 2 ∑ n 2 5 ∑ n 1
the 6 * position and 0 elsewhere. We get 6 nonsingular matrices.
n 1
=
So, S10 =
B – 2A = 33620 – 8820 = 24800 = 248
We get, t n 3 4 5 ... n terms
i.e., Sn =
= 22140 + 11480 = 33620 100 = 24800
Again, Sn 3 7 12 ... t n 1 tn
=
40 41 81 4 20 21 41 6 6
n (n 1)(n 8) 6
⎛ * * 1⎞ ⎜ ⎟ Similarly ⎜ * 1 * ⎟ gives at least one nonsingular ⎜1 * * ⎟ ⎝ ⎠
10 11 18 330 6
59. Answer (3)
56. Answer (3)
A satisfies A2 – Tr(A). A + (det A)I = 0 comparing with A2 – I = 0, it follows Tr A = 0, |A| = –1.
Let a1 = a and common difference = d
60. Answer (2)
Given, a1 + a5 + a9 + ..... + a49 = 416 a + 24d = 32
...(i)
R is Reflexive
Also, a9 + a43 = 66 a + 25d = 33
...(ii)
Let ARB
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ANSWERS AND SOLUTIONS
i.e., A = P–1BP
67
65. Answer (4)
PA = BP
⎡ 1 2 2 ⎤ ⎡ 1 2 a ⎤ ⎡9 0 0 ⎤ ⎢ 2 1 2⎥ ⎢ 2 1 2⎥⎥ ⎢⎢0 9 0 ⎥⎥ ⎢ ⎥⎢ ⎣⎢a 2 b ⎦⎥ ⎢⎣2 2 b ⎦⎥ ⎣⎢0 0 9 ⎦⎥
PAP–1 = B PAP–1 P–1 AP Hence R is not equivalence Statement 1 is false
a 4 2b 0
Statement 2 is true
2a 2 2b 0
61. Answer (2)
a 1 b 0
For skew-symmetric matrix
2a 2b 2
AT = – A (∵ det (–A) = – det A for matrix of odd order)
det AT = det (–A) det A = – det A 2 det A = 0
det A = 0
Statement 1 is true.
a 2b 4
3a 6 a 2
Statement 2 : For every matrix det (AT ) = det (A) But det (–A) = – det A is true for matrix of odd order. Statement 1 is ture and Statement 2 is false.
2 1 b 0
b=–1 a=–2 (–2, –1) 66. Answer (1)
62. Answer (2) ⎡ 0 ⎤ ⎡ 0 ⎤ H2 ⎢ ⎥⎢ ⎥ ⎣ 0 ⎦ ⎣ 0 ⎦
⎡a b ⎤ Let A = ⎢c d ⎥ ⎣ ⎦
⎡2 H2 ⎢ ⎢⎣ 0
0⎤ ⎥ 2 ⎥⎦
⎡ d Then adj (A) = ⎢ – c ⎣
⎡3 H3 ⎢ ⎢⎣ 0
0⎤ ⎥ 3 ⎥⎦
⎡70 Similarly H 70 ⎢ ⎢⎣ 0
– b⎤ a ⎥⎦
|A| = |adj A| = ad – bc ⎡a b ⎤ Also adj[adj A] = ⎢c d ⎥ = A ⎣ ⎦
0 ⎤ ⎡ 0 ⎤ ⎥⎢ ⎥ 70 ⎥⎦ ⎣ 0 ⎦
Both statements are true but (2) is not correct explanation of (1). 67. Answer (2)
=H
Applying D' = D is first determinant and R2 R3 and R1 R2 in second determinant
is complex cube root of unity
a b c a(1)n2 b(1)n1 c(1)n a 1 b 1 c 1 a 1 b 1 c 1 0 a 1 b 1 c 1 a 1 b 1 c 1
70
∵
63. Answer (3) 64. Answer (4)
BB ' ( A1.A ')( A( A1)') = A–1.A.A'.(A–1)1 =
I(A–1A)'
= I.I = I2 = I
{as AA' = A'A}
a ( 1)n 2 a b ( 1)n 1 b c ( 1)n c Then a 1 b 1 c 1 0 a 1 b 1 c 1
if n is an odd integer.
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ANSWERS AND SOLUTIONS
68. Answer (4)
When k = 1, equation change to
The given system of linear equations can be put in the matrix form as
by R2 R2 2R1 R3 R3 3R1
R3 R3 R2
Subtracting the addition of first two equations from third equation, we get, 0 = –5, which is an absurd result. Hence the given system of equation has no solution. 69. Answer (3) For non-trivial solution
1 k 0 1
1(– 3 + k) + k (– k + 3k) + 1(k – 9) = 0 k – 3 + 2k2 + k – 9 = 0 2k2 + 2k – 12 = 0
k2
x + 4y = 2
k + 2y = 3
and 3x + 6y = 8
and x + 2y =
+k–6=0
(k + 3) (k – 2) = 0 k = – 3, 2 Thus, the set of values of k is R – {– 3, 2} for trivial solution.
No solution
72. Answer (2)
⎡1 3⎤ ⎢1 3 3⎥ ⎢ ⎥ ⎢⎣2 4 4 ⎥⎦ |P| = 1(12 – 12) – (4 – 6) + 3(4 – 6) = 2 – 6 Also, |P| = |A|2 = 16 2 = 22 = 11 73. Answer (1)
1 1 1 1
1 2 2
1
1 2 2 1 3 3
1 2 2 1 3 3 1 4 4 1
1
2 1 1 1 1 2
2 2 1 1 1
1
= [(1 – )(1 – )(1 – )]2 So, k 1 74. Answer (3) x1(2 ) 2 x2 x3 0
70. Answer (2)
2 x1 x2 ( 3) 2 x3 0
71. Answer (2)
x1 2 x2 x3 0
k 1 8 0 k k 3
k2 + 4k + 3 – 8k = 0 k = 1, 3
8 3
One value of k exists for which system of equation has no solution.
Alter
3 1
x + 4y = 2
4x + 8y = 12
Clearly the given system of equations has no solution.
k 3
and x + 4y = 2
When k = 3
⎡ 1 2 1 ⎤ ⎡ x1 ⎤ ⎡ 3 ⎤ ~ ⎢⎢0 –1 –1⎥⎥ ⎢⎢ x2 ⎥⎥ ⎢⎢ –3 ⎥⎥ ⎢⎣0 –1 –1⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ –8 ⎥⎦
1 k
2x + 8y = 4
Infinitely many solutions
⎡ 1 2 1⎤ ⎡ x1 ⎤ ⎡3 ⎤ ⎢ 2 3 1⎥ ⎢ x ⎥ ⎢ 3 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣3 5 2⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ 1⎥⎦
⎡ 1 2 1⎤ ⎡ x1 ⎤ ⎡3 ⎤ ~ ⎢⎢0 1 1⎥⎥ ⎢⎢ x2 ⎥⎥ ⎢⎢3 ⎥⎥ ⎢⎣0 0 0 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣5 ⎥⎦
ARCHIVE - JEE (Main)
2 2 1 2 3 2 0 1 2
(2 )(2 3 4) 2( 2 2) (4 3) 0
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2 2 6 8 3 3 2 4 4 4 1 0
= 3 (2 – 4)
3 2 5 3 0
⎡⎛ 1 3i ⎞ ⎛ 1 3i ⎞ ⎤ ⎟⎜ ⎟⎥ = 3 ⎢⎜ 2 ⎠ ⎝ 2 ⎠⎦ ⎣⎝
3 2 5 3 0
= 3 3i
3 2 2 2 2 3 3 0
= –3z
2
( 1) 2 ( 1) 3( 1) 0
k = –z 78. Answer (1)
( 1)( 2 2 3) 0
⎡ 2 3 ⎤ A⎢ ⎥ ⎣ 4 1 ⎦
( 1)( 3)( 1) 0
1, 1, 3 A I
Two elements.
2
3
4
1
75. Answer (3)
1
69
= (2 – 2– + 2) – 12
1
f ( ) 2 3 10
1 1 0 1 1 1( + 1) – (–
2
∵ A satisfies f ( )
+1) + 1(– –1) = 0
A2 – 3A –10I = 0
3 –+ + 1 – – 1 = 0
A2 – 3A = 10I
3 – = 0
3A2 – 9A = 30I
(2 –
3A2 + 12A = 30I + 21A
1) = 0
= 0, = ±1
⎡30 0 ⎤ ⎡ 42 63 ⎤ ⎢ ⎥⎢ ⎥ ⎣⎢ 0 30 ⎦⎥ ⎣⎢ 84 21 ⎦⎥
Exactly three values of 76. Answer (1)
⎡ 72 63 ⎤ ⎢ ⎥ ⎢⎣ 84 51 ⎥⎦
A - adj A = IAI = A.AT adj A = AT ⎡ 2 b ⎤ ⎡ 5a 3 ⎤ ⎢ 3 5a ⎥ ⎢ b 2 ⎥ ⎣ ⎦ ⎣ ⎦
⎡ 51 63 ⎤ adj(3 A2 12 A) ⎢ ⎥ ⎢⎣84 72 ⎥⎦
5a = 2, b = 3
79. Answer (3)
So, 5a + b = 5
1 1 1
77. Answer (4)
1 a 10
2 + 1 = z, z 3 i
a b 1
1 3i Cube root of unity. 2
–(1 – a)2 = 0
C1 C1 + C2 + C3 1
1
1 2
1 1 1
2
1 1
2
1
3
1
2
1 2
a=1 For a = 1 Eq. (1) & (2) are identical i.e.,x + y + z = 1
1 0
To have no solution with x + by + z = 0.
7
b=1
1 2
0 2
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ANSWERS AND SOLUTIONS
ARCHIVE - JEE (Main)
83. Answer (2)
80. Answer (3) x 4 2x 2x 2x x 4 2x 2x
cos( )
5 – 1st quadrant 13 2 = ( + ) + ( – )
x4
2x
sin( )
x = –4 makes all three row identical hence (x + 4)2 will be factor
Also, C1 C1 C2 C2 5 x 4 2x 5x 4 x 4 5x 4
2x
4 + 1st quadrant 5
tan2
tan( ) tan( ) 1 tan( )tan( )
3 5 56 = 4 12 3 5 33 1 . 4 12
2x 2x x4
84. Answer (3)
5x – 4 is a factor
(5 x 4)( x 4)2 B = 5, A = –4
R 2 n
81. Answer (2) ∵ System of equation has non-zero solution.
r n
a sin 2R n
1 k 3 3 k –2 0 2 4 –3
44 – 4k = 0 k = 11 Let z = x + 11y = –3
a tan 2r n
r cos R n
n=3
gives
r 1 R 2
n=4
gives
r 1 R 2
n=6
gives
r 3 R 2
and 3x + 11y = 2
x
5 ,y – ,z 2 2
85. Answer (2) sin + sin4 + sin7 = 0
xz y2
5 · 2 10 2 ⎛ ⎞ ⎜– 2⎟ ⎝ ⎠
2sin4 cos3 + sin4 = 0 sin4 (2cos3 + 1) = 0 sin4 = 0 or cos3θ = –
82. Answer (2) 2(cos cos + sin sin) +2(cos cos + sin sin) +2(cos cos + sin sin) +
sin2
+
cos2
+
sin2
+
cos2
+
sin2
+
cos2
=0
(sin + sin + sin)2 + (cos + cos + cos)2 = 0 sin + sin + sin = 0 = cos + cos + cos Both A and B are true.
1 2
⎛ 2 ⎞ 3 = 2n ± ⎜ ⎟ ⎝ 3 ⎠ =
2n 2 3 9
or 4 = n n 4 =
3 2 4 8 , , , , , 4 2 4 9 9 9
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86. Answer (1)
89. Answer (2)
87. Answer (1)
A
q +
2
p
D
q
B
f4 ( x ) f6 ( x )
p
C
AB BD sin sin( ) AB
1 (sink x cosk x ) k
fk ( x )
2
=
1 1 (sin4 x cos4 x ) (sin6 x cos6 x ) 4 6
=
1 1 1 2 sin2 x cos2 x 1 3 sin2 x cos2 x 4 6
=
1 1 1 = 4 6 12
90. Answer (2)
p 2 q 2 ·sin sin ·cos cos ·sin
p2 q 2 ·sin sin ·
q p2 q 2
( p q 2 )sin p cos q sin
cos ·
tan A cot A 1 cot A 1 tan A
tan2 A cot A tan A 1 1 tan A tan2 A cot A tan A 1
tan3 A 1 tan A(tan A 1)
tan2 A tan A 1 tan A
p2 q 2
sin A cos A 1 cos A sin A
1 sin A cos A sin A ·cos A = 1 + secA· cosecA
20 y
t=1s From figure tan 45 and tan30
20 x
20 xy
so, y 20( 3 1) i.e., speed = 20( 3 1) m/s.
P
91. Answer (1) AO = h cot30º
h 3
h
BO = h
CO
= tanA + 1 + cotA
20
45° 30° x
p
B
A
88. Answer (2)
71
h 3
A
30º B
60º 45º C
O
AB AO BO h 3 h 3 h BC BO CO h 3
92. Answer (2) cosx + cos2x + cos3x + cos4x = 0 2cos
5x 3c 5x x ⋅ cos + 2cos ⋅ cos = 0 2 2 2 2
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ANSWERS AND SOLUTIONS
2cos
x=
ARCHIVE - JEE (Main)
5x x × 2cos⋅ x cos = 0 2 2
(2n + 1) π (2k + 1) π ,
5
2
95. Answer (2)
tan
, (2r + 1) π ,
tan
0 x < 2
where n, k,
B
1 4
π 3 π 2π 9 π π 3 π Hence x = , , π, , , , 5 5 5 5 2 2
93. Answer (3)
1 2
h
30°
A
10v
B
x
h tan 60 x
cos 3 x 1
So, time = 5 minutes.
3x
94. Answer (3)
5 tan2x = 9 cos2x + 7
x
5 sec2x – 5 = 9 cos2x + 7
5 7 , , 9 9 9
Sum
5 9t 12 t
k
9t2 + 12t – 5 = 0
1 2
5 7 , , 3 3 3
Let cos2x = t
cos2 x
2 9
⎛3 1 ⎞ 8cos x ⎜ 1 cos2 x ⎟ 1 ⎝4 2 ⎠
cos3 x
t
A
96. Answer (2)
x 1 ⇒ x 5v 10v x 3
as
4x
⎛ 3 4cos2 x ⎞ 8cos x ⎜ ⎟ 1 ⎜ ⎟ 4 ⎝ ⎠
h tan30 10v x
1 3
P
1⎞ ⎛ 8cos x ⎜ cos2 sin2 x ⎟ 1 6 2⎠ ⎝
60°
let speed = v units/min
t
1 tan 1 4 1 2 1 tan 4
Solving tan
x C x
13 9
13 9
97. Answer (1) 5 3
P
1 , cos 2x = 2cos2x – 1 3
=
45º
1 3
T
cos4x = 2 cos2 2x – 1
2 = 1 9 =
7 9
Q
30º
M
30º
R
Let height of tower TM be h PM = h
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tan30º
In TQM,
h QM
⎛ x ⎞⎞ 1 ⎛ = tan ⎜ tan ⎜ ⎟ ⎟ ⎝ 4 2 ⎠⎠ ⎝
QM 3 h In PMQ,
=
PM 2 QM 2 PQ 2 2
2
h ( 3h ) 200
4h 2 2002
h = 100 m
2y xz 2 1 xy 1 y
y 2 xz
y
xz ⎞ = tan ⎜ ⎟ ⎝ 1 xz ⎠ –1 ⎛
2tan–1y
102. Answer (1) Lines perpendicular to same line are parallel to each other. p = –1 There is exactly one value of p. 103. Answer (1) Let (x, y) denote the coordinates of A, B and C.
99. Answer (1)
1 x 2
Then, 1 x2
8x2 + 8y2 – 20x + 8 = 0
x2 y 2
1 ⎛ dy ⎞ ⎜ dx ⎟ 2 ⎝ ⎠ x 1
5 x 1 0 2
⎛5 ⎞ A, B, C lie on a circle with C ⎜ , 0⎟ . ⎝4 ⎠
100. Answer (1)
104. Answer (4)
⎛ 2x ⎞ tan1 y tan1 x tan1 ⎜ ⎟ ⎝ 1 x2 ⎠ 3 1 ⎛ 3 x x ⎞ 3tan–1 x = tan ⎜⎜ 2 ⎟ ⎟ ⎝ 1 3x ⎠
3x x3 1 3x 2
13 32 32 8 1⇒ b 20 5 b b 5 The line K must have equation
x y x y 1 a or 5 a 20 a 5 20 Comparing with
101. Answer (1)
f ( x ) tan
( x 1)2 y 2 1 ( x 1)2 y 2 9
9x2 + 9y2 – 18x + 9 = x2 + y2 + 2x + 1
dy x dx 1 x2
1
⎞ 2 ⎛ 2 ⎜ x ⎟ ⇒ y 2x 3 6⎠ 3 ⎝
–p(p2 + 1) = p2 + 1
x=y=z
y
1 and at x , f ( x ) 6 3 2
So, equation of normal is
x, y, z are in GP
y sec sec –1
x 4 2
f ( x )
2
98. Answer (1) ∵ 2y = x + z and
73
x y 1 c 3
3⎞ ⎛ ⎜ Given 20a 3, c 5a 4 ⎟ ⎝ ⎠
1 sin x 1 sin x
Distance between lines is 2
x x⎞ ⎛ ⎜ cos 2 sin 2 ⎟ ⎝ ⎠ = tan1 2 x x⎞ ⎛ cos sin ⎜ 2 2 ⎟⎠ ⎝
=
3 1 23 20 1 1 17 17 25 400 400 a 1
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ANSWERS AND SOLUTIONS
ARCHIVE - JEE (Main)
105. Answer (4)
107. Answer (2)
Given
108. Answer (2)
x + y = |a|
(1, a – 1)
ax – y = 1
109. Answer (2) y
Reflected ray
x y 1 1 1 a
a 30°
x + y = a.
0
...(i) for Ist quadrant
ax – y = 1.
x
(3, 0)
...(ii)
1
Slope of incident ray = After solving (i) & (ii)
3
= 150°
x=1
Slope of reflected ray = tan30°
y=a–1 Clearly a – 1 0
=
a1 a [1, )
Reflected ray is y =
106. Answer (4)
⎛ 9 – 2t ⎞ Let vertex of C be ⎜ t, 3 ⎟⎠ ⎝ C
A (2, – 3)
(t,
1 3
1 3 ( x 3)
3y x 3
110. Answer (2)
9 – 2t ) 3 2x + 3y = 9
(0, 2) B
B (– 2, 1)
(0, 1)
(1, 1)
Let (h, k) be centroid
–3 1 t2– 2 , k= h= 3 3 h= k =
t 3
...(i)
–6 9 – 2t 9
from (i) and (ii) k =
9 – 2t 3
...(ii)
O (0, 0)
So, x co-ordinate of incentre
9k = 3 – 6h 6h + 9k = 3 2h + 3k = 1
A (2, 0)
Required triangle is OAB
3 – 2 3h 9
(1, 0)
20 2 2 2 2 0 222 2
4 42 2 2 2 2
Required locus is 2x + 3y = 1
2 2
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75
111. Answer (4)
P(2,2)
(x, y)
Q(6,– 1)
R(7,3)
S
2(ad – bc ) –(5bc – 4ad ) –2ab –2ab
S is mid-point of QR
⎛ 7 6 3 – 1 ⎞ ⎛ 13 , So S ⎜ , ⎜ 2 ⎟⎠ ⎝ 2 ⎝ 2
2ad – 2bc = – 5bc + 4ad
⎞ 1⎟ ⎠
3bc – 2ad = 0
...(i)
113. Answer (4) 2–1 2 Slope of PS = – 13 9 2– 2
(0, 41) A39
2 Equation of line y – (–1) – ( x – 1) 9 9y + 9 = – 2x + 2 2x + 9y + 7 = 0 112. Answer (1) Let (, -) be the point of intersection 4a – 2a + c = 0 ⇒
c 2a
and 5b – 2b + d = 0 ⇒
d 3b
⇒
3bc = 2ad
⇒
3bc – 2ad = 0
A2
B2 B1
A1
(41, 0)
(0, 0)
Total number of integral coordinates as required = 39 + 38 + 37 + ....... + 1
39 40 780 2 114. Answer (3)
A
Alternative method : The point of intersection will be
N (1, 2)
x –y 1 = = 2ad – 2bc 4ad – 5bc 8ab – 10ab 2(ad – bc ) –2ab
x
5bc – 4ad y –2ab
∵ Point of intersection is in fourth quadrant so x is positive and y is negative.
Also distance from axes is same So x = – y (∵ distance from x-axis is – y as y is negative)
(2, 3) ⎛ a 2 b 3⎞ M⎜ , ⎟ 2 ⎠ ⎝ 2
(a, b) B (Image of A)
After solving equation (i) & (ii) 2x – 3y + 4 = 0
...(i)
2x – 4y + 6 = 0
...(ii)
x = 1 and y = 2 Slope of AB × Slope of MN = – 1
b3 2 b3 2 1 a 2 a 2 1 2
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ARCHIVE - JEE (Main)
(y – 3)(y – 1) = –(x – 2)x – 4y + 3 =
A (2, –6)
+ 2x
– 2x – 4y + 3 = 0
Circle of radius =
8
+
y2
For k = 2
=
x2
–x2
2
E
m
y2
115. Answer (2)
(5, 2) B
Point of intersection of sides
D
D
C M (–1, –2)
m= –2 C (–2, 2)
m=0 Equation of AD, x=2
...(i)
Also equation of BE,
A
y 2
B
(1, 2)
1 ( x 5) 2
x–y+1=0
2y 4 x 5
and 7x – y – 5 = 0
x 2y 1 0
x = 1, y = 2 Slope of AM =
Solving (i) & (ii), 2y = 1
4 =2 2
y
1 Equation of BD : y + 2 = − ( x + 1) 2 x + 2y + 5 = 0 Solving x +2y + 5 = 0 and 7x – y – 5 = 0 x=
...(ii)
⎛1 8⎞ 1 8 ,y= − ⎜⎝ , − ⎟⎠ 3 3 3 3
⎛ 1⎞ Orthocentre is ⎜ 2, ⎟ ⎝ 2⎠ 117. Answer (3) x y 1 a b (i) passes through the fixed point (2, 3)
Let the equation of line be
116. Answer (3) k 3 k 1 1 5 k 1 28 Area = 2 k 2 1
1 2
2 3 1 a b
...(i)
...(ii)
P(a, 0), Q(0, b), O(0, 0), Let R(h, k),
k 5 4k 0 5 k k 2 0 56 k 2 1
(k 2 7k 10) 4k 2 20k 56 5k 2 13k 10 56 5k 2 13k 46 0
5k 2 13k 46 0 k =
13 169 920 10
= 2, –4.6 reject
5K 2 13K 66 0
⎛h k ⎞ Midpoint of OR is ⎜ , ⎟ ⎝2 2⎠ ⎛a b⎞ Midpoint of PQ is ⎜ , ⎟ ⇒ h a, k b ... (iii) ⎝2 2⎠
From (ii) & (iii), 2 3 1 h k
locus of R(h, k)
2 3 1 x y
3x + 2y = xy
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k = 2
118. Answer (1) x2
(x2
y2
y2
+ + 3x + 7y + 2p – 5 + + + 2x + 2y – p 2 ) = 0, –1 passes through point of intersection of given circles. Since it passes through (1, 1), hence 7 – 2p + (6 – p2) = 0 7 – 2p + 6 – p2 = 0 If
= –1, then 7 – 2p – 6 + p2 = 0 p2 – 2p + 1 = 0 p=1
The circle is (x –3)2 + (y + 2)2 = 4 Clearly the point (5, –2) lies on it. 123. Answer (1)
x2 y 2 1 16 9 9 = 16 (1 – e2)
e2
7 16
e
7 4
∵ –1 hence p 1 All values of p are possible except p = 1 119. Answer (2) Centre (2, 4)
r2 = 4 + 16 + 5 = 25
Distance of (2, 4) from 3x – 4y = m must be less than radius
77
| 6 16 m | 5 5
foci ( 7, 0) Equation of required circle is (x – 0)2 + (y – 3)2 = 7 + 9 x2 + y2 – 6y – 7 = 0 124. Answer (2)
–25 < 10 + m < 25
C
–35 < m < 15 120. Answer (4)
(0, y)
(0, 1)
(1, 1)
T ⎛ 1 1⎞ ⎜ , ⎟ ⎝ 2 2⎠
(1, 0)
C ( x 1)2 ( y 1)2 1 Radius of T = |y|
Equation of a circle is
(x – 0) (x – 1) + (y – 1) (y – 0) = 0
T touches C externally
x2 + y2 – x – y = 0
(0 – 1)2 + (y – 1)2 = (1 + |y|)2
121. Answer (4)
1 + y2 + 1 – 2y = 1 + y2 + 2|y|
122. Answer (3)
If y > 0,
Let the circle be (x – 3)2 + (y + k)2 = k 2
y2 + 2 – 2y = y2 + 1 + 2y 4y = 1 y=
1 4
If y < 0, (3, –k)
y2 + 2 – 2y = y2 + 1 – 2y 1 = 2 (Not possible)
It passes through (1, –2) 4 + (4 + k2 –4k) = k2
y
1 4
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ANSWERS AND SOLUTIONS
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125. Answer (3)
128. Answer (2)
y
x2 + y2 – 4x – 6y – 12 = 0
P
22 32 12 5
C1(center) = (2, 3), r =
x2 + y2 + 6x + 18y + 26 = 0
64 8
x
O
C2(center) (– 3, –9), r 9 81 26
x 2 ( y 4)
C1C2 = 13, C1C2 = r1 + r2
⎛t t2 Let a point on the parabola P ⎜⎜ , 4 4 ⎝2
Number of common tangent is 3. 126. Answer (3)
⎞ ⎟⎟ ⎠
Equation of normal at P is
k
6 (4, 4)
y
(h1 k)
(r = k)
Radius =
x axis
16 16 4 6
t2 1⎛ t⎞ 4 ⎜x ⎟ 4 t⎝ 2⎠
x ty
t3 7 t 0 4 2
It passes through centre of circle, say (0, k)
(6 + k)2 = (h – 4)2 + (k – 4)2
tk
Replace h x, k y (y + 6)2 – (y – 4)2 = x2 – 8x + 16 (2y + 2) (10) = x2 – 8x + 16 20y + 20 = x2 – 8x + 16
t3 7 t 0 4 2
t = 0, t 2 14 4k Radius = r
x2 – 8x – 20y – 4 = 0
0k 2
(Length of perpendicular from (0, k) to y = x)
Centre lies on parabola 127. Answer (1)
r
(+2, –3) 50
...(i)
C1 5
k 2
A
Equation of circle is x 2 ( y k )2
k2 2
It passes through point P
C2(–3, 2)
Eq. x2 + y2 – 4x + 6y – 12 = 0 C1; (2, –3), r1 4 9 12 5 C2 = (– 3, 2)
2
⎞ t2 ⎛ t2 k2 ⎜⎜ 4 k ⎟⎟ 4 ⎝ 4 2 ⎠
t 4 t 2 (8k 28) 8k 2 128k 256 0 For t = 0
k 2 16k 32 0
k 84 2
C1C2 52 52 50 2
...(ii)
2
Then, C2 A 5 ( 50) 75 5 3
r
k 2
4( 2 1)
(discarding 4( 2 1) ) ...(iii)
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∵ Line (i) touches the circle
For t 14 4k 2
2
(14 4k ) (14 4k )(8k 28) 8k 128k 256 0
2k 2 4k 15 0
k
2 34 2
r
k 2
2(–8) – (–6) 5 4 1
100 – c
5 100 – c
c = 95 131. Answer (1) 17 2 2
(Ignoring negative
...(iv)
A(2, 1) (0, 1)
value of r)
(2, 0)
From (iii) & (iv),
rmin
79
ANSWERS AND SOLUTIONS
2
y x2 + =1 4 1
(4, 0)
17 2 2 x2 y 2 1 16 b 2
But from options, r 4( 2 1) y (0, 4)
Let the equation of the required ellipse is But the ellipse passes through (2, 1) x
(0, 0)
1 1 1 4 b2
1 3 b2 4
129. Answer (3) A (–3, 5) B (3, 3)
2 b
A
B
Hence equation is
C
x2 y 2 3 1 16 4
So, AB 2 10 Now, as, AC
So, radius =
x2 + 12y2 = 16
3 AB 2
132. Answer (3) Locus of P from which two perpendicular tangents are drawn to the parabola is the directrix of the parabola
3 3 5 AB 10 3 4 2 2
130. Answer (4)
Hence locus is, x = –1
Equation of tangent at (1, 7) to curve x –1
4 3
x2
= y – 6 is
Co-ordinates of focus are ( 2, 0)
1 ( y 7) – 6 2
2x – y + 5 = 0
133. Answer (4)
( a e, 0) …(i)
Centre of circle = (–8, –6) Radius of circle 64 36 – c 100 – c
ae=2 e=2 a=1
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ANSWERS AND SOLUTIONS
ARCHIVE - JEE (Main)
b2 = a2(e2 – 1)
For hyperbola
y
b2 = 4 – 1 =3
( x 2 y 2 )2 a2 x 2 b2 y 2
x2 y 2 1 a2 b2
y2 = 4x …(1) x2 = –32y
⇒ 3x2 – y2 = 3
…(2)
m be slope of common tangent
134. Answer (1)
Equation of tangent (1)
135. Answer (4) 136. Answer (2)
1 m
y = mx +
5 2
…(i)
Equation of tangent (2) y = mx + 8m2 …(ii)
2
y 4 5x
(i) and (ii) are identical
Equation of tangent to parabola is
y mx
( x 2 y 2 )2 6 x 2 2y 2
138. Answer (3)
x2 y 2 1 1 3
x2 y 2
...(ii)
Eliminating m, we get
Standard equation of hyperbola is
⇒
x m
5 m
…(i)
1 8
m3 =
For circle, y mx
1 = 8m2 m
5 1 m2 2
…(ii)
(i) and (ii) are identical,
1 2
m
Alternative method :
5 5 (1 m 2 ) 2 2 m
Let tangent to y 2 4 x be
2 = m4 + m2
y mx
m4 + m2 – 2 = 0
as this is also tangent to x 2 32y
m = ±1 which satisfy given equation Statement (1) is true and statement (2) is true. 137. Answer (1) Here ellipse is
1 m
Solving x 2 32mx
32 0 m
Since roots are equal x2 y 2 1 , where a2 = 6, b2 = 2 a2 b2
Now, equation of any variable tangent is
y mx a2m2 b2
...(i)
D=0 ⇒ (32)2 4
⇒ m3
where m is slope of the tangent So, equation of perpendicular line drawn from centre to tangent is
⇒ m
32 0 m
4 32
1 2
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ANSWERS AND SOLUTIONS
142. Answer (2)
139. Answer (4)
x2 y 2 1 Ellipse is 9 5
b e a 2
2 3
We know that b 2 a 2 (e 2 1)
As, required area
29 2a2 27 (2/3) e
b2 a
140. Answer (4) x2
2
= 8y
⎛ t2 ⎞ P ⎜⎜ t , ⎟⎟ ⎝ 2⎠
h = t, k
e
y
2 3
2
1
Let P be (h, k)
e2 1
4 e2 2 e2 1 , e 3 4
Let Q be (4t, 2t2)
2b 2 8, 2b ae a
Given
i.e., a2 = 9, b2 = 5 So, e
81
O
t2 2
P
3
Q(4t, 2t )
143. Answer (1)
x
x = –4
2k h2
Locus of (h, k) is x2 = 2y. 141. Answer (4)
e
1 2
a 4 e
Let the normal of parabola be
a 4 e
a2 P(2m , –4m) 2
(0, –6)
C
y = mx – 4m – 2m3 (0, – 6) lies on it
Now, b2 a2 (1 e2 ) 3 Equation to ellipse
x2 y 2 1 4 3
– 6 = – 4m – 2m3
Equation of normal is
m3 + 2m – 3 = 0
3 y x 1 2 4 x 2y 1 0 1 3 4 23
(m – 1)
(m2
+ m + 3) = 0
m=1 Point P: (2m2, –4m)
144. Answer (1)
= (2, – 4) Equation of circle is (x – 2)2 + (y + 4)2 = (4 + 4) x2 + y2 – 4x + 8y + 12 = 0
x2 y 2 1 a2 b2 a2 b2 4
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ANSWERS AND SOLUTIONS
and
ARCHIVE - JEE (Main)
2 3 2 1 2 a b
2 4b
2
3 b2
4 2 i.e., tan 3 2 8 1 3
1
P(16, 16)
b 3 2
a2 1 x2
C(4, 0)
A
B(24, 0)
y2 1 3 147. Answer (1)
Tangent at P ( 2, 3) is
2x
y 3
1
Clearly PQ is a chord of contact, i.e., equation of PQ is T 0 y = –12
Clearly it passes through (2 2, 3 3)
Solving with the curve, 4x2 – y2 = 36
145. Answer (2)
As, |a – 5| < 1 and |b – 5| < 1 4 < a, b < 6 and
(a 6)2 (b 5)2 1 9 4
i.e., P (3 5, 12); Q( 3 5, 12); T (0,3) Area of PQT is
Taking axes as a-axis and b-axis (a 6)2 (b 5)2 1 9 4
a=6 (6, 7)
b (3, 5)
(6, 5)
⇑
P Q (6, 6)
y T (0, 3)
1 6 5 15 2
x
= 45 5
Q
P
148. Answer (2)
(9, 5)
b=5
(4, 5) (6, 4) S R (6, 3) (0, 0)
x 3 5, y 12
a
The set A represents square PQRS inside set B representing ellipse and hence A B.
p
q
~q
p (~q)
~[p (~q)]
p q
T
T
F
F
T
T
T
F
T
T
F
F
F
T
F
T
F
F
F
F
T
F
T
T
Statement (1) is true and statement (2) is false.
146. Answer (2)
149. Answer (3)
y2 = 16x Tangent at P(16, 16) is 2y = x + 16
... (1)
150. Answer (1)
Normal at P(16, 16) is y = –2x + 48
... (2)
A T
B T
T F
F T
T T
T F
F F
T F
F
F
F
F
F
F
i.e., A is (–16, 0); B is (24, 0) Now, Centre of circle is (4, 0) Now, mPC mPB = –2
4 3
A B A ( A B) ( A B ) A ( A B ) T T T T
(3) and (4) are not tautology according to above table.
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ANSWERS AND SOLUTIONS
A B AB A (AB) T T T T T F F F F T T F F F T F
A ( A B ) B
A(AB) B T T T T
B (A (AB)) T T F T
is tautology according to above
table
= ( p q ) ( p q ) = ( p q p) ( p q q ) = t (p q) = pq 156. Answer (4)
151. Answer (4)
p
q
152. Answer (2)
T T F
T F T
F
F
p q ~ p ~ q a = (p q) b = (~ q ~ p) T T F F T T T F F T F F F T
T
F
T
T
F
T
T
T
T
F
a b (p ~ q) (~ p q) T F T T
F F
T
F
83
p q (~p q) (~p q) q (p q) [(~p q) q] T F T
T T T
T F T
T T T
T
F
T
T
(a tautology)
157. Answer (1) ∼ ( p q ) (∼ p q )
By property, ( ∼ p ∼ q ) ( ∼ p q ) = ~p 158. Answer (2) A B = A C and A B = A C B=C
153. Answer (3) ~(p ~ q)
159. Answer (2)
p
q
~q
p ~q
~ (p ~q)
F
F
T
F
T
F
T
F
T
F
T
F
T
T
F
T
T
F
F
T
160. Answer (3) Statement (2) is true.
var x =
Clearly equivalent to p q 154. Answer (4) ∼ ( ∼ s ( ∼ r s ))
=
2 (n + 1) (2n + 1) – (n + 1)2 3
=
(n 1) {4n + 2 – 3n – 3) 3
=
(n 1) (n – 1) 3
=
n2 – 1 3
( p ∼ q ) q (∼ p q )
= ( p q ) ( ∼ q q ) ( ∼ p q ) = ( p q ) t (∼ p q )
2
4 n (n 1) (2n 1) – (n + 1)2 6n
= sr 155. Answer (2)
n
⎛ ∑ xi ⎞ –⎜ ⎟ ⎝ n ⎠
=
= s (r ∼ s ) = (s r ) (s ∼ s )
∑ xi2
Statement (1) is false. Statement (2) is true.
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ANSWERS AND SOLUTIONS
ARCHIVE - JEE (Main)
161. Answer (2)
x
1 (1 d ) (1 2d ) .....(1 100d ) 101
164. Answer (3) 165. Answer (4)
101 d (1 2 3 .....100) x 101
x
But standard deviation will remain unaffected as each data has been decreased by a constant.
With increase in data, mean will also increase by the same, hence variance will remain unchanged.
100 101 2 101
101 d
166. Answer (4) Variance =
x 1 50d
xi2 ( x )2 N
Mean deviation =
|1 50d 1| |1 50d 1 d | .....|1 50d 1 100d | 101
=
50d 49d 48d .....d 0 d 2d .....50d 101
2
=
2 = 833 167. Answer (4)
d = 10.1
Mean = 16
162. Answer (2)
–
(E(X))2
E(X2)
Sum = 16 × 16 = 256 =4
New sum = 256 – 16 + 3 + 4 + 5 = 252
=4+4=8 Mean =
∑ X i2 40 E(Y2) = 5 + 16 = 21
Var = 2 =
∑Yi 2 105
2
2
22 32 a 2 112 ⎛ 2 3 a 11 ⎞ ⎜ ⎟ 3.5 4 4 ⎝ ⎠
∑ ( X i Yi ) 30 ∑ ( X i2 Yi 2 ) 145
Variance(combined data) =
x12 x n
Standard Deviation =
∑ X i 10, ∑Yi 20
252 = 14 18
168. Answer (1)
E(Y2) – (E(Y))2 = 5
4(12 22 32 .... 502 ) (51)2 50
= 3434 – 2601
50 51 d 255 101
E(X2)
2
⎛ 50 51 101 ⎞ 2 = 4⎜ ⎟ (51) 50 6 ⎝ ⎠
⎛ 50 51⎞ 2d ⎜ ⎝ 2 ⎟⎠ = 101
22 42 ... 1002 ⎛ 2 4 ... 100 ⎞ ⎜ ⎟ 50 50 ⎝ ⎠
2
145 55 11 9 10 10 2
163. Answer (3)
134 a2 ⎛ 16 a ⎞ ⎜ ⎟ = (3.5)2 4 ⎝ 4 ⎠ 4 (134 + a 2 ) 16
−
(16
2
+ a 2 + 32a ) 16
= (3.5)2
Since weight of each fish is measured 2 gm lesser
536 + 4a2 – 256 – a2 – 32a = 196
actual mean : 30 + 2 = 32
3a2 – 32a + 84 = 0
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ANSWERS AND SOLUTIONS
169. Answer (3)
Differentiating,
Standard deviation of xi – 5 is 9
∑ ( xi 5)2
i 1
9
⎛ 9 ⎞ ⎜ ∑ ( xi 5) ⎟ ⎜ i 1 ⎟ ⎜ ⎟ 9 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠
2
2.x x .x x (1 loge x ) dy ⎡ ⎤ 2 ⎢ x x cosec 2 y cot y .x x (1 log x ) ⎥ 0 dx ⎣ ⎦
Put x = 1 and y = 2 2.
5 1 2 As, standard deviation remains constant if observations are added/subtracted by a fixed quantity.
2
dy 20 0 dx
dy 1 dx
174. Answer (3)
So, of xi is 2
Given f(x) = (x – 1)2 + 1
170. Answer (3) R is not an equivalence relation because 0 R 1 but 1 R 0 , S is an equivalence relation.
y = (x – 1)2 + 1 (x – 1)2 = y – 1
⇒
171. Answer (2)
We have, f(x) = (x +
1)2
– 1, x – 1
f(x) = 2 (x + 1) 0 for x – 1 f(x) is one-one Since co-domain of the given function is not given, hence it can be considered as R, the set of reals and consequently R is not onto. Hence f is not bijective statement-2 is false. 1)2
Also
f(x) = (x +
Rf = [–1, )
– 1 –1 for x – 1
x 1 y 1
f–1(x) = 1 x 1 Statement-1 : f(x) = f–1(x) (x – 1)2 + 1 = 1 x 1 (x – 1)4 = (x – 1) (x – 1) ((x – 1)3 – 1) = 0 After solving x = 1, 2 Statement-1 is true.
Clearly f(x) = f –1(x) at x = 0 and x = – 1.
Statement-2 :
Statement-1 is true.
f–1(x) = 1 x 1
172. Answer (2)
Statement-2 is also true.
f(x) = x3 + 5x + 1 f (x) =
85
3x2
+5>0xR
Hence f(x) is monotonic increasing. Therefore it is one-one.
But statement-2 is a correct explanation of statement 1. 175. Answer (1) 176. Answer (3)
Also it onto on R
–3(x – [x])2 + 2[x – [x]) + a2 = 0
Hence it one-one and onto R.
3 {x}2 – 2{x} – a2 = 0
173. Answer (4)
∵ ( x x )2 2.x x cot y 1 , when x = 1, y =
2
2 ⎛ ⎞ a 0, 3 ⎜ { x } 2 { x } ⎟ a2 3 ⎝ ⎠ 2
1⎞ 1 ⎛ a2 3 ⎜ { x } ⎟ 3⎠ 3 ⎝
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ANSWERS AND SOLUTIONS
0 { x } 1and
ARCHIVE - JEE (Main)
1 1 2 {x} 3 3 3
f(x) = f(–x)
2
1⎞ 4 ⎛ 0 3 ⎜ {x} ⎟ 3 3 ⎝ ⎠
2 2 −x = − +x x x 4 =0 x
2x −
2
1 1⎞ 1 ⎛ 3 ⎜ {x} ⎟ 1 3 3⎠ 3 ⎝
x= ± 2 179. Answer (2)
For non-integral solution 0 < a2 < 1 and a (–1, 0) (0, 1)
f (x)
Alternative –3{x}2 + 2{x} + a2 = 0 Now,
–3{x}2
f ( x )
+ 2{x}
x 1 x2
(1 x 2 ) 1 x 2 x 2 2
(1 x )
1 x2 (1 x 2 )2
f(x) changes sign in different intervals. Not injective.
2/3
1
1
y
x 1 x2
yx 2 x y 0 For y 0
to have no integral roots 0 < a2 < 1 a(–1, 0) (0, 1) 177. Answer (2)
1 f '( x ) f (g ( x )) x f '(g ( x )) g '( x ) 1 1 x5
g '( x )
1 1 (g ( x ))5 f '(g ( x ))
178. Answer (2)
For, y = 0 x = 0 Part of range ⎡ 1 1⎤ Range : ⎢ , ⎥ ⎣ 2 2⎦
Surjective but not injective. 180. Answer (1) We have,
⎛ 1⎞ f ( x ) + 2f ⎜ ⎟ = 3 x ⎝x⎠ 3 ⎛ 1⎞ f ⎜⎝ ⎟⎠ + 2f ( x ) = x x
3f(x) =
⎡ 1 1⎤ D 1 4 y 2 0 ⇒ y ⎢ , ⎥ {0} ⎣ 2 2⎦
6 − 3x x
⎛2 ⎞ f(x) = ⎜⎝ − x ⎟⎠ x
f(–x) = −
2 +x x
f:RR
lim
x
f (3 x ) 1 f (x)
⎛2 ⎞ f ⎜ x⎟ f (2 x ) f (2 x ) ⎝ 3 ⎠ . f (x) ⎛ 2 ⎞ f (x) f ⎜ x⎟ ⎝3 ⎠ ⎛x⎞ f⎜ ⎟ 1 f (2 x ) 3 . . ⎝ ⎠ ( ) f x 2 2 ⎛ ⎞ ⎛ x⎞ f ⎜ x⎟ f⎜ ⎟ ⎝3 ⎠ f ⎛ x⎞ ⎝ 3 ⎠ ⎜3⎟ ⎝ ⎠
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ANSWERS AND SOLUTIONS
f (2 x ) l x f ( x )
Taking limit x and lim
lim sin
( sin2 x ) x2
lim sin
( sin2 x ) sin2 x x2 ( sin2 x )
x 0
We find that,
1 1 l 1 1 l
x 0
2
181. Answer (2)
184. Answer (3)
f : R (0, ) (f ( x ))2 9 x 5 | x 5|
2 sin2 x 3 cos x x 2 =2 tan 4 x x x x2 4x 4x lim
lim
⎛0⎞ For existance of limit above form must be ⎜ ⎟ ⎝0⎠ This is possible if and only if
lim f ( x ) 3
185. Answer (2)
p = lim 1 tan2 x x 0
x 5
| x |⎞ ⎛ d ∵⎜ | x | x ⎟⎠ ⎝ dx
4f ( x ).f ( x ) | x 5 | lim | x 5| x 5 x 5 4f ( x ).f ( x ) Now R.H.L = lim x 5 0 x 5 1
L.H.L = lim x 5
4f ( x )f ( x ) 5 x 0 1
= e
lim 1 x 0 2
logp =
2 sin x (3 cos x ) tan 4 x x 0 4x2 4x lim
2 4 2 4
183. Answer (2) sin( cos2 x ) x 0 x2 lim
sin( (1– sin2 x ) x 0 x2
lim
lim sin x 0
( – sin2 x ) x2
1 2x
tan2 x
⎛ tan x ⎜⎜ x ⎝
⎞ ⎟⎟ ⎠
2
1
= e2
1 2
186. Answer (1) 1
⎡ (n 1)(n 2)(n 3)....(n 2n ) ⎤ n p lim ⎢ ⎥ n. n. .....n n ⎣ ⎦
1 2n ⎛nr ⎞ ∑ log ⎜⎝ n ⎟⎠ n n r 1
log p lim
182. Answer (4) 2
1
lim
= e x 0 2 x
Applying L' Hospital rule. 2.f ( x ).f ( x ) 1 | x 5| x 5 1 . 2 | x 5| x 5
[∵ sin( ) sin ]
⎛ sin x ⎞ lim 1 ⎜ ⎟ x 0 ⎝ x ⎠
l 2 = 1 l = 1.
lim
87
2
=
∫ log(1 x )dx
0
2
=
log(1 x )dx 02 ∫
0
1.x dx 1 x
⎡2 ⎛ 1 ⎞ ⎤ ⎢ 2log3 1 ⎟ dx ⎥ ∫⎜ = ⎢⎣ 0 ⎝ 1 x ⎠ ⎦⎥ 2
= 2log3 x log(1 x )0 = 2log3 – (2 – log3] logp = 3 log3 – 2
p e3log32
elog27 e
2
27 e2
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ANSWERS AND SOLUTIONS
ARCHIVE - JEE (Main)
gof is differentiable at x = 0.
187. Answer (1)
lim
cot x cos x
x 2
⎧ – 2 x cos x 2 ⎪ (gof) (x) = ⎨ 0 ⎪ 2 ⎩ 2 x cos x
3
( 2x )
x t Put, 2 lim
= lim
8t 3
LHD = lim –
– 2 x cos x 2 x
RHD = lim
2 x cos x 2 x
x0
sin t 2 sin2 8t
t 0
3
t 2
x0
=2
190. Answer (2) We have,
188. Answer (3)
f : ( 1, 1) R
1 ⎡ 1⎤ 1 1 ⎢ ⎥ As x ⎣x⎦ x
f (0) 1
g(x) = 2[ f(2f(x) + 2)] × f (2f(x) + 2) × 2f (x) g(0) = 2[ f(2 f(0) + 2)] × f (2 f(0) + 2) × 2f (0)
⎞ 15 ⎛ r ⎞
15
= 2[f(0)] × f (0) × 2f (0)
r
∑ ⎜⎝ x 1⎟⎠ ∑ ⎜⎝ x ⎟⎠ ∑ x
r 1
r 1
f (0) 1
g(x) = [ f(2f(x) + 2)]2
2 ⎡2⎤ 2 1 ⎢ ⎥ x ⎣x⎦ x ⎛r
=–2
(gof) is not twice differentiable at x = 2.
1 = . 16
15
x0
For second derivative,
tan t sin t
t 0
x0 x0
= 2 × –1 × 1 × 2 × 1 = –4
r 1
191. Answer (1)
⎛ 15 ⎡ r ⎤ ⎞ 120 lim x ⎜ ∑ ⎢ ⎥ ⎟ 120 ⎜ ⎟ x 0 ⎝ r 1 ⎣ x ⎦ ⎠
f (0)
Statement-1 is true.
⎛ ⎡ 1⎤ ⎡2⎤ ⎡ 15 ⎤ ⎞ ⇒ lim x ⎜ ⎢ ⎥ ⎢ ⎥ ...... ⎢ ⎥ ⎟ 120 x x ⎣ x ⎦⎠ ⎝⎣ ⎦ ⎣ ⎦ x 0
f (x)
189. Answer (2) f(x) = x |x| and g(x) = sin x ⎧ – sin x 2 ⎪ (gof) (x) = ⎨ 0 ⎪ 2 ⎩ sin x
ex ex e x e x 4 2 2
x0
For first derivative LHD = lim
x0
– sin x x
=
lim
x 0–
=0 RHD =
– x sin x 2 x2
0
0 f (x)
x0
sin x 2 x =0 x x
1 3/4
4
4
1 43/4 4
1 2 2
Equality holds if ex = 2e–x e2x = 2. Since
lim
1 ex ex e x e x 2 2
By AM – GM
x0 x0
2
1 3
f (c )
1 1 by intermediate value theorem 3 2 2 1 same c R. 3
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192. Answer (1)
89
197. Answer (1)
⎧ ⎛ 1⎞ ⎪ x sin ⎜ ⎟ , x 0 F(x) ⎨ ⎝x⎠ ⎪ 0 , x 0 ⎩ Statement-1
⎛ 1⎞ lim F ( x ) lim x sin ⎜ ⎟ 0 x 0 x 0 ⎝x⎠ Also, F(0) = 0
ANSWERS AND SOLUTIONS
lim F ( x ) F (0)
⎪⎧k x 1 , 0 x 3 g( x ) ⎨ ⎪⎩ mx 2 , 3 x 5
R.H.D.
g (3 h ) g (3) h
lim
h 0
m(3 h ) 2 2k h 0 h
= lim
x 0
F(x) is countinuous at x = 0 F(x) is countinuous x R Statement-2
lim
h 0
(3m 2k ) mh 2 m h
and 3m – 2k + 2 = 0 L.H.D.
f1(x) = x It is continuous on R
⎧ ⎛ 1⎞ , x0 ⎪sin f2 ( x ) ⎨ ⎜⎝ x ⎟⎠ ⎪ 0 , x 0 ⎩ lim sin
x 0
1 does not exist x
It is not countinuous at x = 0 f2(x) is discontinuous on R Thus statement-2 is false. 193. Answer (1) x 2f (a ) a 2f ( x ) ⎛ 0 ⎞ lim ⎜0⎟ x a x a ⎝ ⎠
k (3 h ) 1 2k h 0 h lim
k [ 4 h 2] h 0 h lim
lim k
h 0
4h4 h( 4 h 2)
k m and 3m – 2k + 2 = 0 4 m
2 8 and k 5 5 8 2 10 2 5 5 5
k m
2 x f (a ) a2f ( x ) x a 1
Alternative Answer
= 2a f(a) – a2 f '(a) 194. Answer (3)
⎪⎧k x 1 , 0 x 3 g( x ) ⎨ ⎪⎩ mx 2 , 3 x 5
195. Answer (1)
g is constant at x = 3
196. Answer (2)
k 4 3m 2
Using, mean value theorem
f (c )
f (1) f (0) 4 1 0
g (c )
g (1) g (0) 2 1 0
so, f (c ) 2g (c )
k 4
From above,
Applying L' hospital rule
lim
2k = 3m + 2
…(i)
k ⎛ ⎞ m Also ⎜ ⎟ ⎝ 2 x 1 ⎠ x 3 k m 4
k = 4 m
…(ii)
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ANSWERS AND SOLUTIONS
Now P(–1) < P(1)
8m=3m+2 m
ARCHIVE - JEE (Main)
P(–1) cannot be minimum in [–1, 1] as minima in this interval is at x = 0.
2 8 ,k 5 5
Hence in [–1, 1] maxima is at x = 1
2 8 mk 2 5 5
Hence P(–1) is not minimum but P(1) is the maximum of P.
198. Answer (1)
202. Answer (4)
g ( x ) f (f ( x )) | log2 sin | log2 sin x ||
Let there be a point P(t2, t) on x = y2 Its distance from x – y + 1 = 0 is
g(x) = f(f(x)) = log2 – sin(log2 – sinx)
t2 t 1
g(x) = cos(log2 – sinx)x – cosx
2
g (0) = cos(log2) 199. Answer (4)
f ( x ) 2 tan1(3 x x )
Min (t2 – t + 1) is
1⎞ ⎛ For x ⎜ 0, ⎟ ⎝ 4⎠
3 4
Shortest distance = f ( x )
9 x
3 4 2
3 2 8
203. Answer (4)
1 9x3
We have,
g( x )
9
y x
1 9x3
200. Answer (1)
f ( x ) | x | (e|x| 1)sin| x | x = , 0 are repeated roots and also continuous. Hence, 'f' is differentiable at all x. 201. Answer (1)
4 x2
dy 8 1 3 dx x The tangent is parallel to x-axis, hence
dy 0 dx x3 = 8
x=2 and y = 3 The equation of the tangent to the given curve at (2, 3) is
(0, d) –1
O
1
We have P(x) = x4 + ax3 + bx2 + cx + d
⎛ dy ⎞ y 3 ⎜ ( x 2) 0 ⎟ ⎝ dx ⎠(2, 3) y=3 204. Answer (4)
P (x) = 4x3 + 3ax2 + 2bx + c P (0) = 0 c = 0 Also P (x) = 0 only at x = 0
2x + 3 k – 2x
P (x) is a cubic polynomial changing its sign from (–)ve to (+)ve and passing through O. P (x) < 0 x < 0 P (x) > 0 x > 0 Hence the graph of P(x) is upward concave, where P (x) = 0
1
k+2 –1 k+21 k –1
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ARCHIVE - JEE (Main)
ANSWERS AND SOLUTIONS
205. Answer (4)
210. Answer (4)
P is point of contact
Curve is x2 + 2xy – 3y2 = 0
P is mid point of AB.
dy 2y dx – 2x
dy ⎡ dy ⎤ Differentiate wr.t. x, 2 x 2 ⎢ x y ⎥ 6y 0 dx ⎣ dx ⎦
(0, 2y)A
P
(2, 3)
(x, y)
dy y – dx x
B (2x, 0)
dy – dy y x
⎛ dy ⎞ 1 ⎜ ⎟ ⎝ dx ⎠(1, 1) So equation of normal at (1, 1) is y – 1 = – 1 (x – 1)
ln y = – ln x + c
...(i)
(i) passes through (2, 3)
y=2–x Solving it with the curve, we get x2 + 2x(2 – x) – 3(2 – x)2 = 0
ln3 = – ln2 + c c = ln 6
–4x2 + 16x – 12 = 0
Equation of curve is xy = 6
x2 – 4x + 3 = 0 x = 1, 3
206. Answer (4)
So points of intersections are (1, 1) & (3, –1) i.e. normal cuts the curve again in fourth quadrant.
tan x is even function. f(x) = x
211. Answer (3)
lim f x 1
Let f(x) = a0 + a1x + a2x2 + a3x3 + a4x4
x 0
f(0 + h) f(0)
⎡ f (x) ⎤ Using lim ⎢1 2 ⎥ 3 x 0 ⎣ x ⎦
and f(0 – h) f(0) x = 0 is point of local minima
f(x) has local minima at x = 0 also, f (x) = 0 at x = 0 but statement 2 is not correct explanation of statement 1 207. Answer (2)
lim
x 0
lim
f (x) 2 x2
a0 a1x a2 x 2 a3 x 3 a4 x 4 x2
x 0
2
So, a0 = 0, a1 = 0, a2 = 2
208. Answer (1)
i.e., f(x) = 2x2 + a3x3 + a4x4
209. Answer (1)
Now, f (x) = 4x + 3a3x2 + 4a4x3
2
f ( x ) log | x | x x f ( x )
91
= x[4 + 3a3x + 4a4x2] Given, f (1) = 0 and f (2) = 0
2x 1 0 at x = –1, 2 x
3a3 + 4a4 + 4 = 0
…(i)
2 1 0 2 1
...(i)
and 6a3 + 16a4 + 4 = 0
4 1 0 2
...(ii)
Solving, a4
8 2
6 3 2
1 2
…(ii)
1 , a = –2 2 3
i.e., f ( x ) 2 x 2 – 2 x 3
1 4 x 2
i.e., f (2) 0
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ANSWERS AND SOLUTIONS
ARCHIVE - JEE (Main)
212. Answer (2) Length of wire = 2
r
r
r 2 ⎛ 20 2r ⎞ ⎜ ⎟ 2⎝ r ⎠
A
r
⎛ 20r 2r 2 ⎞ 2 A⎜ ⎟ 10r r ⎝ ⎠ 2
r
A to be maximum
x Given
dA 10 2r 0 ⇒ r 5 dr
4x + 2r = 2
2x + r = 1
...(i) d2A
2
⎛ 1 − πr ⎞ + πr 2 A = x2 + r2 = ⎜ ⎝ 2 ⎟⎠
Hence for r = 5, A is maximum
dA ⎛ 1 − πr ⎞ ⎛ π ⎞ − + 2πr = 2⎜ ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ dr
Now, 10 + ·5 = 20 = 2 (radian) 2 5 2 25 sq m 2
Area =
dA For max and min =0 dr
215. Answer (4)
(1 – r) = 4r 1 = 4r + r
2 0
dr 2
...(ii)
from (i) and (ii) 2x + r = 4r + r
y2 = 6x ; slope of tangent at (x1, y1) is m1
also 9 x 2 by 2 16; slope of tangent at (x1, y1) is
x = 2r 213. Answer (1)
m2
y ( x 2)( x 3) x 6
9x1 by1
As m1m2 1
At y-axis, x = 0, y = 1 Now, on differentiation.
dy ( x 2)( x 3) y (2 x 5) 1 dx
27 x1
9 as y12 6 x1 2
216. Answer (4)
dy 6 1 dx 6
hx
Now slope of normal = –1 Equation of normal y – 1 = –1(x – 0) ... (i)
⎛ 1 1⎞ Line (i) passes through ⎜ , ⎟ ⎝2 2⎠
214. Answer (2) 2r r 20
A = area =
... (i) r 2 r 2 2 2
1
by12
b
dy (6) 1( 5) 1 dx
y+x–1=0
3 y1
... (ii)
x2
1
x2 x1 x
x
2 x1
2 x1
x1
x
1 0, x
x 1x
x
1 0, x
x 1x
2 x1
x
x
(2 2 , ]
x
( , 2 2]
Local minimum is 2 2
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ARCHIVE - JEE (Main)
ANSWERS AND SOLUTIONS
221. Answer (1)
217. Answer (3) 218. Answer (3)
∫x
5
2 x 12 + 5 x 9
∫ (x
f ( x 3 )dx
1 3 x f ( x 3 ) 3 x 2dx 3∫
=
1 1 x 3 ∫ f ( x 3 ) 3 x 2dx ∫ 3 x 2 ∫ f ( x 3 ) 3 x 2dx dx 3 3
1 3 x ( x 3 ) ∫ x 2 ( x 3 )dx C 3
dp(t ) 1 p(t ) 200 dt 2
⎞ ⎜ 2 p(t ) 200 ⎟ ⎝ ⎠
∫⎛
1 1⎞ ⎜⎝1 + 2 + 5 ⎟⎠ x x
3
1 1 + =t x2 x5
∫
−dt 1 +C = 3 t 2t 2
∫ dt
=
⎛ p(t ) ⎞ 2log⎜ 200⎟ t cx ⎝ 2 ⎠ p(t ) 200 2
dx
)
+ x3 + 1
5 ⎞ ⎛ 2 ⎜⎝ 3 + 6 ⎟⎠ dx x x
1+
= d ( p(t ))
2
5 ⎞ ⎛ 2 ⎜⎝ − 3 − 6 ⎟⎠ dx = dt x x
219. Answer (3)
∫⎛1
1 1 1⎞ ⎛ 2 ⎜1 + 2 + 5 ⎟ ⎝ x x ⎠
3
+C =
(
x10
)
2 x5 + x3 + 1
2
+C
222. Answer (1) In ∫ tann xdx, n 1
t e2k
Using given condition p(t) = 400 – 300 et/2
I4 I6 ∫ (tan4 x tan6 x )dx ∫ tan4 x sec 2 xdx
220. Answer (4) dx I∫ 2 4 ∫ x ( x 1)3/4
Let tanx = t
dx 1 ⎞ ⎛ x 5 ⎜1 4 ⎟ x ⎠ ⎝
sec2x dx = dt
3/4
∫ t 4 dt
Let 1
4 t ⇒ 5 dx dt x4 x
t5 C 5
So, I
1 dt 1 3/ 4 t dt 4 ∫ t 3/ 4 4 ∫
1 tan5 x C 5
93
1 ⎛ t 1/4 ⎞ ⎜ ⎟c 4 ⎜⎝ 1/ 4 ⎟⎠
a
1 ,b 0 5
223. Answer (2) 1/4
1 ⎞ ⎛ = ⎜1 4 ⎟ x ⎠ ⎝
c
where c is an arbitrary constant. So, option (4) is right answer.
I∫
(sin
sin2 x.cos2 x dx 2
x cos2 x ) (sin3 x cos3 x )
2
Dividing the numerator and denominator by cos6x
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ANSWERS AND SOLUTIONS
I∫
ARCHIVE - JEE (Main) 1
tan2 x sec 2 x dx (1 tan3 x )2
2
= ∫ x.0dx ∫ 0
1
=0
Let, tan3x = z 3tan2x.sec2xdx = dz 1 dz 1 I ∫ 2 C 3 z 3z
1 C = 3(1 tan3 x )
x2 2
=
1 1 2 4
=
3 4
2 1
xdx 2∫
x2
1.5 2
xdx
1.5 2
227. Answer (1) 228. Answer (3)
224. Answer (3)
P
0
2000
I ∫ [cot x ]dx
∫
dp
25
∫ 100 – 12 0
x dx
I ∫ [cot( x )]dx
P – 2000 = 2500 – 12 ×
0
P = 3500
2I ∫ ([cot x ] [ cot x ])dx
229. Answer (4)
0
Statement (1)
2I ∫ ( 1)dx 0
I
2 × 125 = 1500 3
I
2
3
dx tan x
∫ 1 6
225. Answer (2) I
We have, p(x) = p(1 – x), x [0, 1], p(0) = 1, p(1) = 41
3
∫ 1
dx cot x
6
p(x) = –p(1 – x) + C 3
1 = –41 + C
2I ∫ dx
C = 42
6
p(x) + p(1 – x) = 42 1
1
0
0
2I
I ∫ p( x )dx ∫ p(1 x )dx 1
1
0
0
I
2I ∫ ( p( x ) p(1 x ))dx ∫ 42.dx 42
y ∫ t dt 0
x[ x 2 ]dx
1
= ∫ x[ x 2 ] dx ∫ 0
Statement (1) is false, Statement (2) is true.
x
226. Answer (1) 1.5
12
230. Answer (1)
I = 21
∫0
6
1
2
x[ x 2 ] dx ∫
1.5 2
x[ x 2 ] dx
dy |x| 2 dx x = ±2
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ARCHIVE - JEE (Main)
ANSWERS AND SOLUTIONS
Case-1, x = 2
4
2I ∫ 1dx
2
y ∫ t dt 2
2
0
2I = 2
Equation of tangent is y – 2 = 2(x – 2) y x 1 2 1
3 4
∫
When x = –2 2
∫ 0
I=1 233. Answer (1)
x-intercept = 1
y
4
2
⎡ t2 ⎤ t dt ⎢ ⎥ = –2 ⎣⎢ 2 ⎦⎥ 0
1 dx x 2 2cos2 2 dx
x-intercepts = ± 1
tan
231. Answer (2)
0
/3
∫ 0
x 1 ⎡ ⎤ ⎢sin 2 2 ⎥ ⎢ ⎥ ⎢⇒ x x ⎥ ⎢ 2 6 3⎥ ⎢ ⎥ x 5 5 ⎢ ⎥ x 6 3 ⎦⎥ ⎣⎢ 2
x 1 dx 2
x⎞ ⎛ ⎜ 1 2sin 2 ⎟ dx ∫ ⎝ ⎠ /3 /3
x⎤ ⎡ ⎢ x 4cos ⎥ 2 ⎦0 ⎣
x ⎛ ⎞ ⎜ 2 sin 2 1⎟ dx ⎝ ⎠
x ⎡ ⎤ ⎢ 4 cos x ⎥ 2 ⎣ ⎦ /3
⎛ 3 3 ⎞ 4 4 ⎜0 4 ⎟ ⎜ 3 2 2 3 ⎟⎠ ⎝ = 4 34 3
3 tan 8
4
log x 2 dx 2 2 2 log x log(36 – 12 x x )
4
log(6 – x )2 dx I∫ 2 2 2 log x log(6 – x )
3 4 3 1 cos 4 1 cos
2 1 2 1
2 1 1
⎤ ⎥ 2 1⎥ 2 1 ⎥ ⎥⎦
2 1
( 2 1) ( 2 1) 2 234. Answer (4)
I
2
∫
2
sin2 xdx
... (i)
1 2x
Also, I
2
∫
2 x sin2 xdx 1 2x
2
232. Answer (3) I∫
4
⎡ 1 cos ⎢ 4 ⎢ tan 8 ⎢ 1 cos ⎢⎣ 4
x x 1 4 sin 4 sin dx 2 2
x dx 2
3 tan 8 8
2
∫ 2sin
2
4
Hence, here x-intercept is –1
0
∫ sec
3
y = 2x + 2
∫
3 4
x⎤4 ⎡ tan ⎥ 1⎢ 2 ⎢ ⎥ 2⎢ 1 ⎥ ⎣ 2 ⎦
y + 2 = 2(x + y)
95
... (ii)
Adding (i) and (ii)
2I
2
2
∫ sin
xdx
2
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ANSWERS AND SOLUTIONS 2
2
0
0
ARCHIVE - JEE (Main)
237. Answer (4) The area loaded by the curves y2 = 4x and x2 = 4y
2I 2 ∫ sin2 xdx ⇒ I ∫ sin2 xdx ... (iii)
Y
2
x = 4y
2
I ∫ cos2 xdx
(1, 1)
... (iv)
0
2
2I ∫ dx 0
X
(0, 0)
Adding (iii) & (iv)
⇒I 2 4
2
y = 4x
1 ⎛ x2 ⎞ A ∫ ⎜⎜ 2 x ⎟ dx 4 ⎟⎠ 0⎝
235. Answer (2) The equation of tangent at (2, 3) to the given parabola is x = 2y – 4
16 square units. 3
238. Answer (1)
(2, 3)
239. Answer (1)
(9, 3) ( –4, 0) 2
(y – 2) = (x – 1) Required area =
3
∫0
(3, 0)
{( y 2)2 1 2 y 4 }dy
9
3
⎡ ( y 2)3 ⎤ y 2 5y ⎥ = ⎢ 3 ⎣⎢ ⎦⎥ 0 1 8 9 15 = 3 3
= 9 sq. units.
Required area ∫ x dx 0
18 – 9 9 240. Answer (4)
I=
236. Answer (1)
O
2
3 2
1 63 2
∫
1⎫ ⎧ ⎛⎜ x 1 ⎞⎟ 1 ⎞ x ⎪ ⎪ ⎝ x⎠ ⎛ x ⎜ 1 2 ⎟ e x ⎬ dx ⎨e x ⎠ ⎝ ⎪⎩ ⎪⎭
= x.e As
∫
x
1 x
c
( xf ( x ) f ( x ))dx xf ( x ) c
241. Answer (3) Required area
/4
∫
(cos x sin x )dx
0
5 /4
∫
(sin x cos x )dx
/4
3 /2
∫
(cos x sin x )dx
2
2
x +y =1
5 /4
(4 2 2) sq. units
Shaded area
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ARCHIVE - JEE (Main)
243. Answer (1)
1
(1)2 2∫ (1 x ) dx 2 0
97
ANSWERS AND SOLUTIONS
2(1 x )3/2 ( 1) 2 3/2
4 (0 ( 1)) 2 3
4 2 3
(2, 2)
1
0
(2, 0)
2
Area =
242. Answer (4)
y=1
1
π ⋅ 22 − ∫ 2 xdx 4 0
y=1 = π− 2⋅
1 2
= π−
2 32 x 3
2
0
8 3
244. Answer (3) y
y2
After solving y = 4x – 1 and
y 4
= 2x ) ,2 (1
2
y 1 2
(0, 1)
x
2y2 – y – 1 = 0
1 y 1, 2
1 1 8 1 3 y 4 4 1
1
y2 ⎛ y 1⎞ A ∫ ⎜ dy ∫ dy ⎟ 4 ⎠ 2 1/2 ⎝ 1/2 1
1
⎤ 1 ⎡y2 1 ⎡y3 ⎤ ⎢ y⎥ ⎢ ⎥ 4 ⎣⎢ 2 ⎥⎦ 1/2 2 ⎣⎢ 3 ⎦⎥ 1/2
1 ⎡ 4 8 1 4 ⎤ 1 ⎡ 8 1⎤ ⎥ 2 ⎢ 24 ⎥ 4 ⎢⎣ 8 ⎦ ⎣ ⎦ 1 ⎡15 ⎤ 9 4 ⎢⎣ 8 ⎥⎦ 48
=
15 6 32 32
9 32
O
(2, 1)
x=1 x=2
x+
x y=
3
x=0 y
Area of shaded region 1 2 ⎛ ⎛ x2 ⎞ x2 ⎞ ∫ ⎜ x 1 ⎟ dx ∫ ⎜ (3 x ) ⎟ dx ⎝ 4 ⎠ ⎝ 4 ⎠ 0 1
5 sq. unit 2
245. Answer (1) 18 x 2 9x 2 0 (6 x )(3 x ) 0 x , 6 3 , 6 3
y (gof )( x ) cos x
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ANSWERS AND SOLUTIONS
Area =
=
=
3 cos x dx 6
∫
sin x
3 6
3 1 2 2 1 2
ARCHIVE - JEE (Main) 1
I.F. = e
1 y
–
Put –
246. Answer (3) Put e c2 k Then y = c1.kx loge y = loge c1 + x loge k
= ∫e
1 y
which is linear in z
1 dy dt y2
t t = – te – e c
1 1 y 2 ( y )2 0 y y
dz 1 tan x z sec x, z dx y
1 dy y3
= e
1 dy 1 tan x sec x y 2 dx y
1 y
t = – ∫ e . t dt
t t = e – te c
The given differential equation can be put in the form
–
1 t y
–
1 y
1
1 –y e c y 1
x = 1
247. Answer (1)
⇒
–
1 y loge k y
yy = (y )2
= e
The required solution is x. e
3 1 sq. units
∫ y 2 dy
1 ce y y
Put x = 1, y = 1 1 = 1 + 1 + ce ce = – 1
1 e
c=–
1
1 ey x = 1 – y e
249. Answer (4) tan x dx I.F e ∫ eln sec x sec x
The solution is z.sec x ∫ sec 2 xdx tan x c
where c is a constant of integration
⇒ sec x y tan x c 248. Answer (1) dy y3 The given diff. equation reduces to du 1 – xy
dx 1 – xy 1 x 3 – 2 dy y3 y y
dx x 1 dy y 2 y 3
250. Answer (3)* It is best option. Theoretically question is wrong, because initial condition is not given. x log x
dy + y = 2x logx dx
If x = 1 then y = 0
dy y 2 dx x log x 1
I.F. e
∫ x log x dx
eloglog x log x
Solution is y log x ∫ 2log x dx c y log x 2( x log x x ) c
x = 1, y = 0 Then, c = 2, y(e) = 2
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251. Answer (3) ydx – xdy =
ANSWERS AND SOLUTIONS
99
253. Answer (3) –y2xdx
ydx xdy y2
xdx
⎛x⎞ d ⎜ ⎟ xdx ⎝y⎠
sin x
dy y cos x 4 x , x (0, ) dx
dy 4x y cot x dx sin x
I.F. e ∫
cot x dx
sin x
Solution is given by
On integrating both sides 2
x x c y 2
y sin x
y·sinx = 2x2 + c
it passes through (1, –1) 1
1 1 c ⇒ c 2 2
x x2 1 So, y 2 2
y
c–
2 2
Equation is y sin x 2 x 2 –
2 2
when x
when x
2 x
⎛ 1⎞ 4 i.e., f ⎜ ⎟ x 1 ⎝ 2⎠ 5 2
252. Answer (4) (2 sin x )
dy ( y 1)cos x 0 dx
ln| y 1| ln(2 sin x ) ln C ( y 1)(2 sin x ) C
Put x = 0, y = 1 (1 1) 2 C C = 4
Now, ( y 1)(2 sin x ) 4 For, x
2
( y 1)(2 1) 4 y 1 y
4 3
4 1 1 3 3
y–
,y=0 2
1 2 2 – then y · 2· 6 2 36 2
8 2 9
254. Answer (2) Direction ratios are a = 6, b = –3 and c = 2 Then direction cosines are
6
⎛⎞ y (0) 1, y ⎜ ⎟ ? ⎝2⎠ 1 cos x dy dx 0 2 sin x y 1
4x
∫ sin x ·sin x dx
36 9 4 =
,
3 36 9 4
,
2 36 9 4
6 3 2 , , 7 7 7
255. Answer (4) [3u pv pw ] [ pv w qu ] [2w qv qu ] = 3p2 [u .(v w )] pq[v .(w u )] 2q 2 [w .(v u )] (3 p2 pq 2q 2 )[u .(v w )] 0 But u .(v w ) 0 3p2 – pq + 2q2 = 0 p=q=0 256. Answer (1) We have ab c 0 ⇒ a ab ac 0
⇒ a .b a a.a b a c 0
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
ANSWERS AND SOLUTIONS
2 i 2 j 4k ⇒ b i j 2k 257. Answer (1) We have a.b 2 4 2 0 a . c 1 2 0 b . c 2 4 0
C 4k
3 j 3k 2i j k
262. Answer (3) AB + AC AD = 2
2j +
⇒ 3a 2b a c 0 ⇒ 2 b 3a a c ; a c 2i j k
ARCHIVE - JEE (Main)
A
=
(, ) = (–3, 2) 258. Answer (1)
c (a 3b ) b 2c a b 2(a 3b ) a
1 6 b 2 – a 0 6 + 1 = 0, 2 =
3iˆ + 4kˆ + 5iˆ
| AD | =
16 + 1+ 16 =
33
263. Answer (3) l+m+n=0 l2 = m2 + n2 Now, (–m – n)2 = m2 + n2 mn = 0 m = 0 or n = 0 If n = 0
If m = 0 then l = –n l2
+
m2
+
then l = –m
n2
=1
1
1 Now, c (a 3b ) 6c a 3b 0 6
i.e. (l1, m1, n1)
2
1 ⎞ ⎛ 1 , 0, = ⎜ ⎟ 2 2⎠ ⎝
259. Answer (2) Given vectors piˆ ˆj kˆ, iˆ qjˆ kˆ , iˆ ˆj rkˆ to be coplanar
l2 + m2 + n2 = 1 2m2 = 1
Gives n
m2
1 2
1
m Let m
l
p 1 1 1 q 1 0 1 1 r
2 1 2
1 2
n=0 (l2, m2, n2) ⎛ 1 1 ⎞ , ,0 ⎟ = ⎜ 2 2 ⎠ ⎝
p (qr – 1) – 1 (r – 1) + (1 – q) = 0 pqr – p – r + 1 – q + 1 = 0
cos
260. Answer (4) 261. Answer (1)
2
1 1 , 6 3
pqr – p – q – r = – 2
– 2ˆj + 4kˆ
= 4iˆ – ˆj + 4kˆ
and 2 – 4 + 4 + = 0 = –3
B
3i + 4k
Thus = 1 – 2 3 = 6, = 2
D
5i –
100
1 2
3
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ANSWERS AND SOLUTIONS
264. Answer (2)
268. Answer (1)
Clearly, u (a (a b ))
L.H.S.
= (a b ) [(b c ) (c a )]
u ((a . b )a | a |2 b )
= (a b ) [(b c a )c – (b c c )a] = (a b ) [[b c a ]c ]
u (2a 14b ) 2 (2iˆ 3 jˆ kˆ ) 7( jˆ kˆ )
[∵ b c . c 0]
= [a b c ] (a b c ) [a b c ]
as, u b 24
[a b b c c a ] [a b c ]2
4(iˆ 2 jˆ 4kˆ ) ( jˆ kˆ ) 24
So = 1
= –1
265. Answer (1)
So, u 4(iˆ 2 jˆ 4kˆ )
1 (a c ) b ( b c ) a | b | | c | a 3
1 (b c ) | b | | c | 3
cos
sin
| u |2 336 269. Answer (1) The point (2, 1, –2) is on the plane x + 3y – z + =0
1 3
2 + 3 + 2 + = 0
Hence
2 2 3
2 + = –5 Also
266. Answer (3)
(
) = (aic ) b − (aib) c
=
3 – 15 – 2 = 0
3 b + c and 2
(
)
2 = –12 = –6 Put = –6 in (i)
3 a⋅b = − 2 cos = −
... (i)
1(3) + 3(–5) + –(2) = 0
after comparing
=
u 2 (2iˆ 4 ˆj 8kˆ )
2
a× b×c
101
= 12 – 5 = 7 (, ) (–6, 7)
3 2
270. Answer (3) cos245° + cos2120° + cos2 = 1
5π 6
267. Answer (1) | (a b ) c | 3
a b 2iˆ 2 ˆj kˆ
⇒ | a b | | c | sin 30 3 ⇒ |c | 2
|c a | 3
⇒ | c |2 | a |2 2(a c ) 9 932 a c 2 2
|a | 3 ab
1 1 cos2 1 2 4 cos2
cos
1 4
1 ⇒ 60o or 120o 2
271. Answer (1) The image of the point (3, 1, 6) w.r.t. the plane x – y + z = 5 is x 3 y 1 z 6 2(3 1 6 5) 1 1 1 1 1 1
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ANSWERS AND SOLUTIONS
x 3 y 1 z 6 2 1 1 1
x=3–2=1
274. Answer (3) Any point on a line parallel to the given line x = y = z and passing through (1, – 5, 9) is ( + 1, – 5, + 9)
y=1+2=3
It lies on given plane
z=6–2=4 which shows that statement-1 is true. We observe that the line segment joining the points A(3, 1, 6) and B(1, 3, 4) has direction ratios 2, –2, 2 which one proportional to 1, –1, 1 the direction ratios of the normal to the plane. Hence statement2 is true. 272. Answer (3)
( + 1) – ( – 5) + (+ 9) = 5 + 15 = 5 = – 10 Point is (– 9, – 15, – 1)
102 102 102 = 10 3
Required distance = 275. Answer (4)
Maximum number of triangle = 10C3 – 6C3 =
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276. Answer (2) 277. Answer (3)
10 9 8 6 5 4 0 32
5 8 7 units Distance between the planes 2 3 2
= 100 273. Answer (1)
278. Answer (3)
Let co-ordinates of Q be
P(3,–1, 11)
Q R x = y – 2= z – 3 = t 4 3 2 x = 2t y = 2 + 3t
1
1
–k
1 =0 Given lines are coplanar if k 2 1 –1 –1 1 (– 2 + 1) – 1 (– k – 1) – k (– k – 2) = 0 – 1 + k + 1 + k2 + 2k = 0 k = 0 or – 3 Exactly two values of k. 279. Answer (3)
z = 3 + 4t
A (1, 3, 4)
Direction ratios of PQ are (2t – 3, 3 + 3t, 4t – 8)
3iˆ + jˆ + kˆ
Direction ratios of . Q.R. are (2, 3, 4) PQ QR 2(2t – 3) + 3(3 + 3t) + 4(4t – 8) = 0 29t – 29t = 0
P
t=1
A
Co-ordinates of Q are
3iˆ + jˆ + 5kˆ
(a, b, c)
x = 2, y = 5, z = 7 The length of the perpendicular PQ
a 1 b 3 c 4 1 2 1
=
(3 2)2 ( 1 5)2 (11 7)2
=
12 62 4
a = 2 + 1
=
53
c=4+
b=3–
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⎞ ⎛ P ⎜ 1, 3 , 4 ⎟ 2 2⎠ ⎝ ⎞ ⎛ ⎞ ⎛ 2( 1) ⎜ 3 ⎟ ⎜ 4 ⎟ 3 0 2⎠ ⎝ 2⎠ ⎝ 2 2 3 +
4 3 0 2 2
3 + 6 = 0 = – 2
ANSWERS AND SOLUTIONS
2l – m – 3 = 0
103
...(i)
(3, –2, –4) lies on the plane 3l – 2m + 4 = 9 3l – 2m = 5
...(ii)
Solving (i) and (ii) l = 1, m= –1 l2 + m2 = 2 283. Answer (1)
a = – 3, b = 5, c = 2 So the equation of the required line is
x 3 y 5 z2 3 1 5
(1, –5, 9) P
280. Answer (4)
x 2 y 1 z 2 3 4 12
P 3 2, 4 1, 12 2 Lies on plane x – y + z = 16
Q L: x = y = z Equation of line PQ:
Then,
An x point Q on the line PQ is ( + 1, – 5. + 9)
3 2 4 1 12 2 16
∵ Point Q lies on the plane : x – y + z = 5 (+ 1) – (– 5) + + 9 = 5
11 5 16 1
Distance =
P 5, 3, 14 16 9 144 169 13
281. Answer (3) Required plane is (2x – 5y + z – 3) + (x + y + 4z – 5) = 0 It is parallel to x + 3y + 6z = 1
2 5 1 4 1 3 6
11 Solving = 2 Required plane is
11 (x + y + 4z – 5) = 0 (2x – 5y + z – 3) – 2 x + 3y + 6z – 7 = 0
+ 10 = 0 = – 10 Point Q is (– 9, – 15, – 1) PQ =
(1 + 9)2 + ( −5 + 15)2 + (9 + 1)2 = 10 3
284. Answer (1) Let the plane be
a( x 1) b( y 1) c ( z 1) 0 It is perpendicular to the given lines a – 2b + 3c = 0 2a – b – c = 0 Solving, a : b : c = 5 : 7 : 3 The plane is 5x + 7y + 3z + 5 = 0 Distance of (1, 3, –7) from this plane =
2i j 3k l i m j k 0
83
285. Answer (1)
282. Answer (3) Line is perpendicular to normal of plane
10
Equation of PQ,
x 1 y 2 z 3 1 4 5
Let M be ( 1, 4 2, 5 3)
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ANSWERS AND SOLUTIONS
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Length of projection of the line segment on the plane is AC
P
AC 2 AB 2 BC 2 2
M
AC 2
Q
4 2 3 3
2 3
288. Answer (4)
As it lies on 2x + 3y – 4z + 22 = 0
Restricting sample space as S = {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40}.
=1 For Q, = 2 Distance PQ 2 12 42 52 2 42
P(sum of digits is 8) =
286. Answer (2)
L1 is parallel to
289. Answer (2)
iˆ jˆ kˆ 2 –2 3 iˆ jˆ
Total number of cases = 9C3 = 84 Favourable cases = 3C1.4C1.2C1 = 24
1 –1 1 iˆ jˆ kˆ L2 is parallel to 1 2 –1 3iˆ – 5 jˆ – 7kˆ 3 –1 2
So, required plane is
1 3
5 7
y–
8 7
p
290. Answer (3) The outcomes 2, 8, 14, 20 is an AP with common difference 6. 291. Answer (2)
z
1 –5
0 0 –7
⎛ Ac B c P ⎜⎜ C ⎝
3
1
7x – 7y + 8z + 3 = 0 Now, perpendicular distance
24 2 84 7
Statement-2 is false.
⎛5 8 ⎞ Also, L2 passes through ⎜ , , 0 ⎟ ⎝7 7 ⎠ x–
162
1 . 14
3 2
287. Answer (4)
B (4, –1, 3) n=i+j+k
⎞ P ( Ac Bc C ) ⎟⎟ P (C ) ⎠
P (C ) P (C A) P (C B ) P ( A B C ) P (C )
Let A, B, C be pairwise independent events
P (C ) P (C ).P ( A) P (C ).P (B ) 0 P (C )
= 1 – P(A) – P(B)
(∵ P(C) 0)
= P(Ac ) – P(B)
C
A (5, –1, 4)
292. Answer (1) 293. Answer (3)
Normal to the plane x + y + z = 7 is n iˆ ˆj kˆ AB iˆ kˆ ⇒ | AB | AB 2 BC = Length of projection of AB on n | AB nˆ |
iˆ kˆ
iˆ ˆj kˆ 3
2 3
4
Required probability
⎛ 1⎞ ⎛ 2⎞ ⎛ 1⎞ 5C4 ⎜ ⎟ ⎜ ⎟ 5C5 ⎜ ⎟ ⎝3⎠ ⎝3⎠ ⎝3⎠ 5
5
1 2 1 81 3 35
10 1 11 35 35 35
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294. Answer (1)
P( A B)
1 1 5 ⇒ P ( A B ) 1– 6 6 6
3 2 , q 5 5
var(X) = n.p.q
1 1 3 ⇒ P ( A) 1– 4 4 4
P ( A) ∵
p
= 10
6 12 25 5
298. Answer (1)
P ( A B ) P ( A) P (B ) – P ( A B )
5 3 1 P (B ) – 6 4 4
P ( A ) P (B ) P ( A B )
1 4
1 3
P (B ) P (C ) P (B C )
1 4
P (C ) P ( A) P ( A C )
1 4
P (B )
∵ P(A) P(B) so they are not equally likely. Also P(A) × P(B) =
3 1 1 4 3 4
P ( A) P (B ) P (C ) P ( A B ) P (B C )
= P(A B) ∵
P( A C )
P ( A B ) P ( A) P (B ) so A & B are independent.
∵ P( A B C )
295. Answer (1)* Question is wrong but the best suitable option is (1). Required probability =
12
C3
P( A B C )
11
29 55 ⎛ 2 ⎞ = 3 ⎜⎝ 3 ⎟⎠ 312
P (E2 )
3 8
1 16 3 1 7 8 16 16
299. Answer (4) Total number of ways =
296. Answer (3) P (E1)
105
11C 2
= 55
1 6
Favourable ways are
1 6
Probability =
(0, 4), (0, 8), (4, 8), (2, 6), (2, 10), (6, 10)
P(E1 E2) = P(A shows 4 and B shows 2) =
300. Answer (2) E1 : Event that first ball drawn is red.
1 P (E1 ).P (E2 ) 36
E2 : Event that first ball drawn is black. E : Event that second ball drawn is red.
So E1, E2 are independent Also as E1 E2 E3 =
⎛E ⎞ ⎛ E ⎞ P (E ) P (E1 ).P ⎜ ⎟ P (E2 ).P ⎜ ⎟ ⎝ E1 ⎠ ⎝ E2 ⎠
So P(E1 E2 E3) P(E1. P(E2). P(E3) So E1, E2, E3 are not independent. 297. Answer (4)
4 6 6 4 10 12 10 12
2 5
n = 10 p(Probability of drawing a green ball) =
6 55
15 25
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