Paper-1: Hints & Solutions
Paper-1: Hints & Solutions
Paper-1: Hints & Solutions
JEE ADVANCED
TARGET : JEE (ADVANCED)-2013
DATE : 01-05-2013 COURSE NAME : REVISION CLASSES
RESONANCE SOL010513 - 1
15. Sound of frequency 1000 Hz...............................
Ans. 4 (2 sin – 3 )( 3 sin + 1) = 0
V v0 330 30 3
Sol. f' f f' = 1000 330 sin = , = 60º
V vs 2
Ans. 4
330 360 21. A man can swim with a......................................
f " f ' f " 1000 300
330 30 Ans. 6
f " 1200Hz 2d
Sol. Ist case: T1 =
16. When a system is taken from..................................
v m
Ans. 5
2d
Sol. Q acb U acb Wacb IInd case: T2 = 2 2
v m v
Q bda U bda Wbda
d d 2dv m
Q acb Q bda U acb U bda Wacb Wbda
IIIrd case: T3 = = 2 2
v m v v m v v m v
U acb U bda 0 Then, T22 = T1T3 = 3 × 12 T2 = 6 sec.
60 Q bda 30 10 22. The cross–section of a glass.....................................
Ans. 2
Q bda 40J Q adb 40J Sol. PQ & RS are parallel (normal to force)
Then, = 2
17. In the given potentiometer arrangement................................... But, + 2 = 180º
Ans. 5 = 36º
Sol. Amptitude of SHM, A = 5 cm, initial position of jockey is at A/2 and Then, x = 2.
moving towards extreme.
23. A glass bottole when full of........................................
T T Ans. 8
t
6 4 mass expelled 3
Sol. Mass of Hg =
18. A block of mass 5 kg is............................... ( Hg glass )T 15 10 5 20 = 1000 gm
=
An s . 3
Sol. As seen from an inside observer, the forces acting on the Initial mass of Hg = 1000 + 3 = 1003 gm
block are pseudoforce, frictional force and the applied force. Mass of empty bottle = 1123 – 1003 = 120 gm
When the applied force is in the direction of pseudo force x = 8.
(in this case less force will be required to move the block) 24. The work of 147 kilojoule is......................................
10 + pseudoforce = µmg ........(1) Ans. 5
When the applied force is opposite to the pseudoforce,
20 – pseudo force = µmg ........(2) f
Adding (1) & (2) Sol. Q = U + W W = –U = nRT
2
30 = 2 µmg = 2µ50 µ = 0.3
19. The molar heat capacity of a................................ f 147
An s . 3 147 = × 8.4 × 7 f = = 5.
2 7 4.2
Sol. Proces PV2 = constant & f=5
PART-II (Chemistry)
f R 5 3 xR 25 (A)
C= R+ = R–R= R =
2 2 1 2 2 2 R
x = 3. Sol. Cm = Cv + +
1 x
20. An elastic ball thrown with a........................... (A) PV–1 = constant
Ans. 4 X=–1
2v 0 sin
Sol. Total time of flight T = R
g Cm = Cv +
2
v0 (B) PV–2 = constant
Time from A to B t1 = X=–2
g sin
Then, B to C = T – t1 R
Cm = Cv + = Cv – R
2v 0
1 3
But, T – t1 = (maximum possible time)
2 3 g (C) PV2 = constant
X=2
2v 0 sin v0 v0 1
R
g g sin = g 3 Cm = Cv + = Cv – R
1
sin
2 sin2 – 1 = (D) PV3 = constant
3 X=3
2 3 sin2 – sin – 3 =0 R
Cm = Cv +
2
2 3 sin2 – 3 sin + 2 sin – 3 =0
Heat exchanges Cm so highest heat exchange is in (A).
RESONANCE SOL010513 - 2
26. (C) two diastereomeric amides
–
O
Sol. O S O O H
39. (9)
O Sol. Initial moles of H2O2 = 2 × 2 = 4
50%
1FeO + CO Fe + CO
2
Br2 / KOH / 2
100 0.8 3 0.6 0.5
retension
3 = 3.
16
RESONANCE SOL010513 - 3
44. (6) PART-III (Mathematics)
49. (C)
Sol. Using property g1g100 = g2 g99 = g3 g98 = .... = g50 g51 = ab
and a1 + a100 = a2 + a99 = a3 + a98 = .... = a50 + a51 = a + b
(ab)50 (ab) 49 ab
given expression = 50 + 49 + ..... +
(a b ) (a b ) ab
Sol.
h 50
1
50 49
h h h h 2
= + + ..... + =
45. (9) 2 2 2 2 h
1
Sol. H2N–CH2–COOH (Glycine) + 2
h(h50 2 50 )
(1) Gly–Gly (2) Gly–ala(+) (3) Gly-ala (–) =
(4) Ala(+) – Gly
250 (h 2)
(5) Ala (–)–Gly (6) Ala(+) – Ala(+) (7) ala (–)–ala(–) 50. (C)
(8) ala (+)–ala(–) 1/ n
(9) ala(–)–ala(+) n!
Sol. nlim
(mn)n
46. (3)
1/ n
CF CH O
3 2 CH CH O
3 2 1 2 3 n
= nlim
Sol. (P) CH3 – CH2 – Br . . .......
(1) ( 2)
mn mn mn mn
In reaction 2nd Nucleophile is stronger.
(Q) (Alkene) 1 2 3 n
lim 1 . . .......
(1) ( 2) = e n n n mn mn mn
C2H5O
(Alkene) mn
( CH3 )3 CBr / ( CH3 )2 CHCH2Br /
n 1
st
In reaction 1 3º alkyl halide 1 r x
n dx
1 1
1>2 =
lim
n = m =
n
=
(R) In reaction 1st stable carbocation is formed
e n n r 1
mn e 0
e em em
Ans. (C) is correct.
51. (B)
(Cyclopropyl methyl C) Sol. Focus of x2 = 4y is (0, 1)
1>2 Mid-point of chord of circle is of the form (t2 , 2t)
(S) In reaction 1st reagent in HI. t4 + 4t2 < r2 .....(i)
and t2x + 2ty = t4 + 4t2
HI is more ionised, gives sufficent concentration of H ion
It passes through (0, 1) 2t = t4 + 4t2
and I is a good Nucleophile also. t = 0 or t3 + 4t – 2 = 0
1>2 t = 0 gives axis of parabola x2 = 4y
t3 + 4t – 2 = 0 gives exactly one real root ( f(t) = 3t2 + 4 > 0)
47. (6) If this value of t satisfies (i), then one line is obtained.
Sol. 3 O NH OH / H
2
52. (B)
excess
7 6 x 2 13
2 x 2 13 2
Sol. =0
2
x 13 3 7
R1 R1 + R2 + R3
cis - cis
trans - trans x2 4 x2 4 x2 4
cis R trans 2
trans S cis 2 x 13 2
=0
Total isomeric product = 4 x 2 13 3 7
Optical active = 2
48. (6) 1 0 0
C 2 2
x 15 0
| (x2 – 4) =0
Sol. C–C–C–Cl x 13 x 16 20 x 2
2 2
|
C (x2 – 4) (x2 – 15) (20 – x2) = 0
roots are ±2, ± 15 , ± 20
sum of roots other than 2 is –2.
RESONANCE SOL010513 - 4
53. (D)
Sol. Required area is
1
[( x
2
– 2x – 1) – (– e x – 1)] dx = 2 e – 1
–1
3 e
graph of region is
54. (C)
Sol. f(x) = f(x) × f(x)
Let f(x) be a polynomial of degree n x = c cuts the graph at two different points if c > 0.
degree of f(x) = n – 1 & degree of f(x) = (n – 3) i.e c (0,)
n = (n – 1) + (n – 3) n = 4
also f(x) = 0 satisfies x = 1, 2, 3 only 60. (A)
f(x) = (x – 1)2 (x – 2) (x – 3) or f(x) = (x – 1) (x – 2)2 (x – 3)
or f(x) = (x – 1) (x – 2) (x – 3)2 x 2 mx 1
This implies that either f(1) = 0 or f(2) = 0 or f(3) = 0 Sol. –1 1, x R
f(1) f(2) f(3) = 0 (C) is correct. 3( x 2 x 1)
–3x2 – 3x – 3 x2 + mx + 1 3x2 + 3x + 3
55. (B) 4x2 + (m + 3) x + 4 0 and 2x2 + (3 – m) x + 2 0, x R
Sol. Let center be (– r, – r) (m + 3)2 – 64 0 and (3 – m)2 – 16 0
m2 + 6m – 55 0 and m2 – 6m – 7 0
2r r 1 (m + 11) (m – 5) 0 and (m – 7) (m + 1) 0
=r –11 m 5 and –1 m 7
5 –1 m 5
(3r – 1)2 = 5r2
9r2 – 6r + 1 = 5r2 61. (C)
4r2 – 6r + 1 = 0 Sol. f(x) = x2 + 2x – 3 = (x + 1)2 – 4 is increasing in [–1, 5]
f(–1) = – 4 and f(5) = 32 D = [– 4, 32]
1 5 1
r= . t n 1
4
=
–I
n 10
n
also let center be (r, – r)
(r + 1)2 = 5r2
4r2 – 2r – 1 = 5r 2 1
In + In+2 =
1 5 n 1
r=
4 1
nn+1 In+3 + In+1 =
3 5 n2
radius of larger circle is
4 n n
1
56. (A)
Sol. 1000C2 – 250C2 – (250C2 – 1) = 437251
Sn = (
n0
n n 2 ) n 3 n n 2 =
n 1
n1 n3
n 0
57. (B) n n
Sol. Let (x,y) be any point whose distance from (0,0) is 1. 1 1 1 1
the region is | x | + | y | = 1. It intersects y = 2x at two = n 1 n 2 n 1 – n 2
n 0 n0
points sum of whose absissae is 0.
58. ( A)
1 1 1 1 1 1
Sol. The region intersects x-axis at two points (–1, 0) and (1,0) 1 – – ..... – 1–
=
2 2 3 n 1 n 2 n2
the required difference between the maximum and mini-
mum distance of region from the point (, 0) = distance lim Sn = 1 – 0 = 1
between the points (–1, 0) and (1,0) = 2. n
59. (D)
64. (2)
Sol. Let (x,y) be point on the locus, then | x + 1 | = | x – 1 | + | y |
Sol. 2 cos 3B + 3 cos 4A = 3
2 x if 0x 1 2 cos 3B = 3 (1 – cos 4A) = 6 sin22A
|y|=|x+1|–|x–1|= and 2 sin 3B – 3 sin 4A = 0
2 if 1 x 2 sin 3B = 6 sin 2A cos 2A
then cot 3B = tan 2A
RESONANCE SOL010513 - 5
69. (9)
cos 3B sin 2 A Sol. There are two sets of five parallel lines at equal distance. Clearly,
i.e. = i.e. cos (2A + 3B) = 0
sin 3B cos 2 A lines 1, 2, m1, m3 form a squares whose diagonal's length is 2
So, the number of required squares = 3 × 3 = 9
since 0 < A, B <
4
5
0 < 2A + 3B < 2A + 3B =
4 2
65. (3)
3 1 3 1
Sol. Point B –
, – , A (0, 1), D
,
2 2 2 2
70. (2)
Sol. If the roots are x1, x2, x3, x4
then x1 + x2 + x3 + x4 = 4
Area (quad. z1 z2 z3 (–z2)) = AB × AD = AB x1 x 2 x 3 x 4
{ AD = 1} =1
4
= 3 square of the area = 3 and x1x2x3x4 = 1
since A.M. = G.M. then
66. (3) x1 = x2 = x3 = x4 = 1
Sol. g(x) = 2 – x1/3 Let x1/3 = t so by theory of equations we get a = 6 and b = –4 only values
g(t) = 2 – t possible so a + b = 2.
Now f(2 – t) = –t3 + 5t – t2
71. (1)
= (2 – t)3 – 7(2 – t)2 + 11(2 – t) – 2
f(t) = t3 – 7t2 + 11t – 2 x
Now f(x) = x – 7x2/3 + 11 x1/3 – 2 Sol. Put x = 1 in 2f(x) = f(xy) + f .............(1)
y
14 11 1
f(x) = 1 – x–1/3 + x–2/3 = 0
3 3 2f(1) = f(y) + f ............(2)
x–1/3 = 1, 3/11
y
Replace x by y and y by x in (1)
14 22
f(x) = x–4/3 – x–5/3 x=1 y
9 9 2f(y) = f(yx) + f ..........(3)
f(x) < 0 x
x = 1 is the point of local maxima (1) – (3)
Local maximum value of f(x) = 3
x x x
67. (0) 2{f(x) – f(y)} = f – – f 2f ..........(4)
Sol. f(x) = ax3 + bx2 + cx + d y
y y
As eccentricities of a parabola and a rectangular hyperbola are
the roots of f(x) = 0 x
f(x) – f(y) = f ...........(5)
y
f (1 h) – f (1)
lim = f(1) = 1
as a, b, c, d h 0 h
f(1) must be equal to 0.
f(1) = a + b + c + d = 0 f (1 h)
lim = 1 as f(1) = 0
68. (1)
h0 h
Sol. 1x + 2x + 3x + ...... + nx = (n + 1)x
f ( x h) – f ( x )
x x x f(x) = lim
1 2 n h 0 h
+ +......... + =1
n 1 n 1 n 1 h
Now drawing the graph of
f 1 1
= lim x =
x x x
h 0 h x
1 2 n
f(x) = + +.........+ f(x) = log|x| + c
n 1 n 1 n 1
f(1) = 0 c = 0
Which is stricity decreasing and g(x) = 1
f(e) = 1
72. (7)
Sol. n (S) = 6 × 6 × 6 = 216
also to get sum of '8' x1+ x2 + x3 = 8
where 1 xi 6
xi = ti + 1 0 t1 5
t1 + t2 + t3 = 5
RESONANCE SOL010513 - 6
by fictious partition method number of solution of this equation is 3f1 = 4f2 = 12Hz f1 = 4Hz & f2 = 3Hz.
7
C2 (A) |5f1 – 4f2| = |20 – 12| = 8 (r)
21 7 (B) 3f1 3f 2 12 9 3 (q)
n (E) = 21 P = =
216 72
(C) 9f1 8f 2 36 24 12 (s)
PAPER-2 (D) 5f1 5f 2 20 15 5 (t)
PART-I (Physics)
1. Two point charges q1 and q2 ................................. 8. A train is moving with a constant.............................
Sol. (A,B) The charge q3 must lie on the line (3x + 4y – 12 = 0) joining Ans. 3
charges q1 and q2 to be in electrostatic equilibrium.
V VO
2. Two conducting plates A and B............................... Sol. f 'f
Sol. (C,D) Let a charge q’ be present on the inner surface of plate V VS
A so that on its outer surface the charge is (q 1 – q’).
Obviously a charge -q’ will get induced on the inner surface of V 15 cos 45
the plate B and a charge (q2 + q’) will move to its outer surface. = f
With these charges, write the net electric field at a point inside V 15 cos 45
the plate and equate it to zero. This relation can be simplified to
get the value of q’ and hence the conclusions. f ' f 1.5kHz = 3/2kHz
3. A uniform rod AB of mass m...................................... 9. A man of mass m starts moving..................................
Sol. (B,C,D) Point A is constrained to move along circular path. Ans. 3m/s
Initial velocity of point A is zero.
ac = 0 i.e. along the string (1) f
at 0 i.e. perpendicular to string (1) Sol. Magnification m m = –2
3f
4. A hydrogen atom of mass.............................. f
Sol. (A,D) From conservation of momentum 2
hf h Taking the direction in the right hand side Vim m 2 Vom
mv
c
Assuming all the energy is taken by the photon. Vim
velocity of image w.r.t. mirror
3 3 E0 h
hf E E V Vom
velocity of object w.r.t mirror
4 0 4mc m
5. Charge +q is uniformly distributed................................ Vim 4u
Sol. (A,C) Energy conservation
mu u
1 2 1 2 kq 2 Vm
× 2 + kq = 0 + 0 + q × velocity of platform (minor)
mu + mu + q × m 2m 3
2 2 3R 6R
Vim Vi Vm
kq
× 2 + kq
2 5kq2
u= Vi
velocity of image w.r.t ground
R 2R 3mR
6. Column-I gives certain situations................................... u 13
Vi = 4u Vi u
Ans. (A) – s, t ; (B) – p, s ; (C) – p, r ; (D) – q, r 3 3
Q V Vi = 3m/s
Sol. (A) E or E
A 0 d 10. A glass rod of rectangular cross-section................................
Ans. 2
for Q = const.
Sol. A ray entering through surface A and travelling along the inner
E = const.
side of the rod will be reflected by the outer side with the smallest
If V = const. and d increases. E decreases (s, t)
angle , at which the reflected ray is tangent to the inner side.
Q2 1
(B) U or U CV 2
2C 2 1
If the plates are pulled apart, C decreases.
critical sin 1
n
Q = const.
U increases 1
for V = const. sin n
U decreases. (p, s)
(C) On insertion of a dielectric, the capacitance of a capacitor R
always increases. (p, r) but sin R d
Q2 1
(D) U or U CV 2
2C 2 R 1
for Q = const.
U = decreases
R d 3/ 2
for V = const
U increases. (q, r)
Rd 3
7. A closed organ pipe P1 .................................. R 2
Ans. (A) – r ; (B) – q ; (C) – s ; (D) – t
Sol. Let f1 and f2 be the fundamental frequencies of P1 and P2.
RESONANCE SOL010513 - 7
d 1 R mu 2 3 3
2 N= mg
R 2 d R 2
R FMin 2N cos 300 mg
2
d min
3mu 2 9
= mg mg FMin 1 newton
11. Figure shows the variation of................................ R 2
Ans. 2
Sol. For rectangular hyperbola 15. To reduce the light reflected by....................................
XY = constant Ans. 3
U = constant
n CVT = constant Sol. 2d 2n 1 (n = 1, 2, 3,.....)
2
5 M 2n 1
n R T constant d
2 V 4
5 VM For light of wavelength
P constant P = constant (isobaric process) d Layer
2 V 1 700nm
5 700 2100 3500 Glass
U BA 7 2 5J nRT d1 , ,
2 4 4 4
2J nRT W 2J for light of wavelength 2 420nm
12. The potential energy function ................................... 420 1260 2100
Ans. 2 d2 , ,
Sol. The force acting on the particle can be given by 4 4 4
U
F 6 î 4 ĵ 2100
r 2100
d min 7
Thus acceleration of this particle would be ax = 6ms-2 and ay = 4 4
4ms-2 4
when the particle crosses the 'x' axis
d min 300nm 3 107 m
1
8 4t2 t2 = 4 t = 2s
2
13. Two uniform rods tied together....................................... PART-II (Chemistry)
Ans. 3 16. (AC)
Sol. the FBD of any one rod is
T
2 R 320
Sol. V1 =
T N ...(i) 4
mg = N ...(ii) N 2 R 640
Taking torque about point 'P' mg 37
0
V2 =
8
L N P V = 0
mg cos 37 0 = TL sin 37º so w = 0
2 H = nCp T
mg 4 2mg 7
=2× × 2 ×320 = 4480 Cal
T= 2
2 3 3 (B) E = q + w
E = q = 3200 Cal
2mg
mg 23 (D) E = nCV dT
3 5
=2× × 2 × 320 = 3200 Cal
14. A hollow cylinder of mass m = 1kg .................................. 2
Ans. 1
17. (ABD)
mV 2 3
Sol. mg N V O
R 2 N N
%s %s
P P
1 1 3 %s %s
mg mg
mV 2 mu 2 mgR O O O O
2 2 2
O O
Sol. O P P O P P
2 2 O O
mV mu
3mg O O O O
R R
P P
mu 2 3
3mg = mg N O
R 2
RESONANCE SOL010513 - 8
P–O–P bond length in P4O6 is more than P4O10
P4O6 due to the absence of bond, p – d bond is not present paramagnetic
CH2 =
diamagnetic
(ix)
PC PD 10
so, = = 5.
2 2
(ii)
RESONANCE SOL010513 - 9
28. (3) PART-III (Mathematics)
31. (ABC)
Sol. Plane P1 = 2(x – 1) – 2(y + 1) + 2(z – 1) = 0 x – y + z = 3
OH Plane P2 = 2(x – 1) – 3(y – 2) – 2(z – 1) = 0
Sol.
2x – 3y – 2z + 6 = 0
so P1 + P2 = 0 (x – y + z – 3) + (2x – 3y – 2z + 6) = 0
(1 – 1 – 3) + (2 + 2 + 6) = 0
3
=
+ 10
10(x – y + z – 3) + 3(2x – 3y – 2z + 6) = 0
16x – 19y + 4z – 12 = 0
put z = 0 in P1 and P2
x – y = 3, 2x – 3y = – 6
By solving we get x = 15, y = 12, z = 0
32. (AD)
+ r
Sol. (A) If R is its radius then, sin (/n) =
R–r
(y = 3)
29. (4)
I / OH
2 Na metal R = r(1 + cosec (/n))
Sol. C9H12O4 CHI3 + y 2H2 + salt
H 2
DU = 4 (No. of active H = 4)
sin cos
2 2
Since 1 + cosec
sin
33. (AD)
Sol. f(x) = max {|2 – x|, 2 – x3}, then
2 I / OH f(x) is continuous x R
H and f(x) is non-differentiable at x = ±1, 0, 2
f(x) = max {|2 – x|, 2 – x3},
COOH
| Na metal
HOOC — C — COOH + 4CHI3 2H2 + C(COONa)4
|
COOH
Ans. 4 moles of CHI3 formed.
30. (6)
1 1
f(x) = –
cos x sin x
sin x – cos x
=
sin x cos x
f(x) change sign –t0 + at x =
4
so f(x) takes minimum value at x =
Ans. 5 molecules of MeMgBr. 4
RESONANCE SOL010513 - 10
/4 /2 1 i 3
(C) z = ±
f(x) = sec d cos ec d 2
0 /4
2
Principal argument of z is or –
= nsec tan /4
+ ncos ec – cot /2 3 3
0 /4
1 i 3
= n 2 1 – n (1 0) n1 – n ( 2 – 1) (D) z = ±
2
2 1
n 2 2
= 2 – 1 = n
2 1 = 2n ( 2 1) Principal argument of z is –
3
or
3
.
= x – · 2/3 [4x – 1]
dx 3 3 x 3x 2 / 3 Using IV property
x
hence f is for x >
1 ( 4 – x )x
4
I= (4 – x ) e
4– x
dx ........(ii)
Adding we get
x
( 4 – x )x
2I = 4e dx
1 4– x
and f for x <
4 x
x( 4 – x )
4 1/ 3 1 – 3
x – x
2 2I = 4(2) (Given e dx 2 )
f(x) = (non existent at x = 0 , vertical tangent) 4– x
3 3 I=4
4 1 1 2 1
n o w f(x) = 2 / 3 + · · 5/3 102
9 x 3 3 x –1
2 1 2 2x 1
(B) k = [ cot
0
x ] dx
= 2 x
9x 2 / 3 9x 2 / 3 x
RESONANCE SOL010513 - 11
Now from (1)
3 3
2 2 cos x
= (0 – 0) – sin dx
= 2 | b |2
2 (1) ( | b | ) – =8
2 2
2 2 | b |2
= 2
cos 3 – cos 2 = – 1 – 1 =8
2 2
4 2 | b |= 4
=– 2 I= – |2 | = 2
2
10 2 42. (0 )
(D) sgn (sin x ) dx = 5 sgn (sin x ) dx
xf ( x )
0 0 Sol. lim f(x) = xlim = lim (f(x) + xf(x)) = a
x x x
1 2 lim xf(x) = 0
5 1dx (–1) dx
1 2 x
=
= 5x 0 5– x 1 = 5 – 5 = 0
0 1 x 2f ' (x)
Now lim xf(x) = lim = lim (2xf(x) + x2f(x)) = 0
38. (2) x x x x
Sol. The number of integers in [– 5, 30] is 36 lim x2 f(x) = 0.
y is positive for all x. x
Discriminant < 0
4(a + 4)2 – 4(64 – 5a) < 0 43. (9 )
a2 + 13a – 48 < 0
(a + 16) (a – 3) < 0 dx x 2007 1 – x 2007
a (– 16, 3) Sol. = x( x 2007
dx
But a [– 5, 30] 1) x( x 2007 1)
possible integral values satisfying are – 5, – 4, – 3, – 2, – 1,
0, 1, 2, = 8 1 2006
– x
dx
8
2 = x 1 x 2007
Probability = = 2
36 9
1
39. (3 ) = nx – ln (1 + x2007)
2007
Sol. sin2x + sin22x – sin23x – sin24x = 0
or (sin2x – sin23x) + (sin22x – sin24x) = 0 x 2007
sin 2x [sin4x + sin6x] = 0
nx 2007 – n(1 x 2007 ) 1
C
= n 2007
2 sin2x sin 5x cos x = 0 2007 2007 1 x
sin2x = 0or sin5x = 0or cosx = 0
p + q + r = 6021 = 9
n n
x= or x= or x = (2n + 1) 44. (1 )
2 5 2
1 10 102 103 10 2n
Sol. 2n 2n 2n
n – n . C1 + n . C2 – n . C3 + ...... +
solutions in the interval 0,
2
81 81 81 81 81n
1
2 = [1 – 10. 2nC1 + 102 . 2nC2 – 103. 2nC3 + .............+ 102n ]
are x = , and 81n
2 5 5
Hence 3 solutions. 1
= [2nC0 – 10. 2nC1 + 102 . 2nC2 – 103. 2nC3 + .............+ 102n. 2nC2n]
40. (4)
81n
Sol. 77 ends with 01
74k + r ends with same two digit number as does 7r
1 1 1
2n 2n
= n [1 – 10] n (–9) = . 81n = 1
therefore the given number ends with the same two digits as 73. 81 81 81n
45. (2 )
41. (4) Sol. Equation of directrix is x = – 1 + but equation of directrix is x =
1 (given)
Sol. | a | = 1 , a.(b (a b)) = 8
= 2.
point on intersection of 2y = – 4x + 11 and x = 1 is (1, 7 / 2)
a. (b. b) a – (a. b ) b 8 tangent at Q passes through (1, 7 / 2) and is perpendicular to
2y = – 4x + 11
| a |2 | b |2 – (a. b)2 = 8 .........(1) equation of the tangent is x – 2y + 6 = 0
x2 = 6
since a. b = | a || b | cos The point (6,6) lies on the parabola.
(6 – )2 = 4 (6 – 2) = 16
= 2,10
1
= (1) | b |
2
2 | b |2
(a. b) =
2
RESONANCE SOL010513 - 12