IITJEE - 2013
Vidyamandir Classes
Solutions to JEE Mains/Test Series - 9/IITJEE - 2013 [CHEMISTRY] 1.(C)
By conservation of moles PV P1V1 PV = + 2 2 RT RT RT P×4=2×1+3×2 P = 2 atm.
2.(C)
Statement-II violates the inert pair effect. Hence it is false.
3.(A)
KMnO4 will react with FeSO4 only. 100 1 × moles = 5 × 2 × 1000 moles = 1 χ FeSO4 = 1/ 3
4.(D) 5.(D)
Only Cl2 is oxidsing among the given options Cl
∴ 6.(D)
2 → KMnO K 2 MnO 4 4
∆S = n C v n
T2 V + n R n 2 T1 V1
= nC v n2 + n R n
1 2
= (C v − R) n 2 7.(D)
P-anthranilic acid has less acidic nature due to the +M effect of − NH 2 group.
2NH 3 (g) N 2 (g) + 3H 2 (g)
8.(B)
t=0
a
t = t eq
a 2
PNH 3
3a
3a 2 a P = P= . 3a 3
a
9.(C)
→ H 3 BO3 + 3HBF4 4BF3 + 3H 2O
10.(D)
[H + ] =
11.(B)
Ka1C1 + Ka 2C2 (Learn as a result)
=
1.8 × 10−5 × 0.01 + 6.3 × 10−5 × 0.01
=
8.1 × 10−7 = 9 × 10−4 .
Cr2 O27 − on reduction converts to Cr+3 which is green in colour.
VMC/JEE-2013/Solutions
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JEE Mains/Test Series-9/ACEG
Vidyamandir Classes n ( n + 2 ) = 1.73 = 3
12.(C)
n (n + 2) = 3 n = no. of unpaired e−s = 1 v = 4s2 3d3. Hence Vanadium must exist in the form V4+ compound is VCl4 ⊕
13.(C)
AgNO3 → AgCl ↓ + CH 2 = CH − C H 2 Allylic carbocation. CH 2 = CH − CH 2 − Cl
14.(A)
→ Mg ( NH 4 ) PO 4 + Na 2SO4 + H 2O . MgSO 4 + NH 4 OH + Na 2 HPO 4
15.(C)
P + I2 Mg / Ether HCHO H 2O → A → → C →D CH 3 − CH 2 − OH B
is CH 2 − CH 2 − I
(A)
is CH 3 − CH 2 − Mg − I
(B) ⊕
is CH 3 − CH 2 − CH 2 − O M gI (D)
(C)
is CH 3 − CH 2 − CH 2 − OH
16.(B)
17.(B)
O O || || CH 3 − C − CH 2 − C − CH 2 − CH 3 Most acidic hydrogen (Active methylene group)
18.(D) NF3 dipole > NH3 dipole.
19.(D) R.D.S. of cannizaro involves transfer of H − ion to the carbonyl group.
20.(A)
O O || || → R − C − Nu + Z − R − C − Z + Nu − Leaving group capacity of Cl − is maximum among the given options hence & fastest reaction when Z = Cl.
21.(C) Both (1) & (2) contain chiral carbon will undergo SN2 reaction with alc. KCN, hence inversion. 22.(B)
π = iCRT
0.75 = 2.47 × Moles = 23.(A)
E=
1 = 0.03 30
CZ2
εH ε
moles × 0.0821 × 300 2.5
Be3+
n2 =
2 Z2H n Be3+
n 2H
Z2 3+ Be
=
1 2
1
×
22
4
2
=
1 4
24.(D) Graphite: free e − are spread out between the structure & thus graphite is conducting. 25.(D) V1 = volume for complete neutralization of Na2CO3. V2 = Volume for complete neutralization of NaHCO3. V1 =x 2 V1 + V2 = x + y Hence V2 = y − x
VMC/JEE-2013/Solutions
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JEE Mains/Test Series-9/ACEG
Vidyamandir Classes 26.(D)
− → Cl(g) is exothermic, rest all endothermic. Cl(g) + e −
27.(D)
28.(C) Metal oxides an basic (or more ionic character ≈ more basic). TiO > VO > CrO > FeO ∴ 29.(C)
dx = k[A]2 dt dx log = log k + 2 log[A] dt y = 2x + log k On comparing with y = mx + c. Where m ≡ slope
c ≡ constant 30.(A)
4r = a 2 r=
2 a 4
Solutions to JEE Mains/Test Series - 9/IITJEE - 2013 [PHYSICS] 31.(C) A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically repeated in three dimensions on a lattice. The spacing between unit cell in various directions is called its lattice parameters or constants. Increasing these lattice constant will increase or widen the band-gap (Eg), which means more energy would be required by electrons to reach the conduction band from the valence band. Automatically Ec and Ev decreases. 32.(D)
K = K1 + K 2 +
T = 2π
33.(B)
v1 t
v2 t
34.(A)
=
K3 K 4 60 × 30 = 50 + 30 + = 100 N / m K3 + K4 90
m 0.01 π = 2π = Hz . 100 50 K
d1 − ρ d2 − ρ
⇒
0.18 20 − 2 = 10 − 2 v2
⇒
v 2 = 0.08 m / s t
t
C ||E × B & CB = E in magnitude.
As
C = 3 × 108 iˆ and E = 720 ˆj
35.(C)
300 − 0 10,000 = 9500 300 − v
36.(B)
Only electric force acts on the particle.
B = 2.4 × 10−6 kˆ
⇒ ⇒
300 − v = 285 ⇒ v = 15 m / s
37.(C) 38.(A) Lowest frequency = HCF of 420 Hz & 315 Hz = 105 Hz.
VMC/JEE-2013/Solutions
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JEE Mains/Test Series-9/ACEG
Vidyamandir Classes 39.(D)
φ = BAN sin ωt
ε =−
dφ = − BAN ω cos ωt dt
40.(B)
ε =−
dφ = − ( 20t − 50 ) = − 10v at t = 3s dt
41.(B)
D1 is not conducting but D2 is conducting 12 =2A i= ⇒ 4+2
42.(B)
Energy of proton = 8 × 7.06 − 7 × 5.6
⇒
ε m = BAN ω
= 17.28 MeV
d 2v dx
2
and
= 3x 2 − 1 < 0 for x = 0
d 2v
⇒
⇒
x3 − x = 0
⇒ x = 0 or x = ± 1
x = 0 is a point of maximum for V
= 3x 2 − 1 > 0 for x = ± 1
dx 2
dv =0 dx
⇒
43.(A) K.E. is maximum when P.E. is minimum
⇒
x = ± 1 are points of minima for V
1 1 −1 − = J at x = ± 1 4 2 4
⇒
minimum potential energy =
⇒
−1 9 maximum KE = 2 − = 4 4
9 1 2 = mvmax & m = 1 kg ⇒ 4 2
⇒
vmax =
3 m/s 2
44.(C) Let Q be total charge on the two spheres & let q1 & q2 be charges on A & B respectively in equilibrium. (rA = 1mm, rB = 2mm) q1 q2 Q 2Q q1 + q2 = Q and = . q1 = and q2 = ⇒ ⇒ 4π ∈0 rA 4π ∈0 rB 3 3
E1 / E2 = q1 / rA2 : q2 / rB2 = 1 / 3 :
45.(A)
µR < µB
⇒
= 2 :1
or for small angled prism D = A ( µ − 1)
(A)
46.(D) 48.(A)
( 2 )2
A + D sin 2 ∵ µ = sin A / 2
⇒ D1 < D2
Ans.
2/3
47.(C)
v2 1 ∝ n R R
T =
⇒
1+ 2π R ∝ R V
v∝
n −1 2
1 n −1) / 2 ( R
⇒
T ∝ R
n +1 2
H 49.(D)
50.(B)
T2
K
⇒
x f = 1/3
2K
H =
⇒
Ans.
(D)
⇒
Ans.
(B)
4x
| ∆P | = E / C − ( − E / C ) = 2 E / C
VMC/JEE-2013/Solutions
⇒
T1
4
KA (T2 − T1 ) T2 − T1 = x 4x 3x + 2 KA KA
JEE Mains/Test Series-9/ACEG
Vidyamandir Classes ⇒ the object is placed at the centre of the equivalent mirror.
51.(A) u = v
1 =−2 Fm
1.5 1 1 1 − 1 − + /2 1 ∞ − 30 − 30
⇒
Fm = − 10 cm .
R = | 2 Fm | = 20 cm 6 = 4A ( 2||6 ) + 1.5 ||3
52.(C)
i=
53.(A)
X 20 4X x ; = = Y Y 80 100 − x
54.(B)
x = 50 cm
Let : length of the wire ; R : radius in first case; r : radius in second case. ⇒ 2π R = & n(2π r ) =
B= B' = 55.(B)
⇒
T = 2π
µ0i 2R
µ0 π i
=
n µ0i n 2 µ0 π i = = n2 B 2r
I MB
where
I: moment of Inertia =
m 2 12
M : pole strength ×
I' =
m ( / s) 12
2
= I / 9 ; M ' = ( pole strength remain same ) × 3 × / 3 M
I /9 T = = 2 / 3 sec M .B 3 56.(D) As P.d across L & C are out of phase.
⇒
T ' = 2π
57.(B)
induced current = −
58.(C)
ω=
59.(B)
ε =
1 = LC
( change influx ) × no of
1 L' (2C )
(
turns
total resistance × time
⇒
(W − W1 ) =− 2 × n ÷ t R + 4 R ) (
= −
n (W2 − W1 ) 5RT
L' = L / 2
)
1 1 2 ∆ω2 = × 0.2 × 10−4 × 5 × (1) = 5 × 10−5V 2 2 = 50 µ v
60.(A)
⇒
VMC/JEE-2013/Solutions
truth table of OR gate.
5
JEE Mains/Test Series-9/ACEG
Vidyamandir Classes
Solutions to JEE Mains/Test Series - 9/IITJEE - 2013 [MATHEMATICS] 61.(D) For the graph to touch x-axis, the polynomial y = x 2 − 2 px + p + 1 would have to be a perfect square. Therefore,
(
)
p 2 = p + 1 , or p 2 − p − 1 = 0 , p = 1 ± 5 / 2 .
62.(C)
Z1 = ( 8 sin θ + 7 cos θ ) + i ( sin θ + 4 cos θ ) Z 2 = ( sin θ + 4 cos θ ) + i ( 8 sin θ + 7 cos θ ) Z1 = x + iy where x = ( 8 sin θ + 7 cos θ ) and y = ( sin θ + 4 cos θ ) Z 2 = y + ix
Hence,
(
Z1 .Z 2 = ( xy − xy ) + i x 2 + y 2
)
= a + ib ⇒ a = 0 ; b = x 2 + y 2
x 2 + y 2 = ( 8 sin θ + 7 cos θ ) + ( sin θ + 4 cos θ ) 2
Now,
2
= 65 sin 2 θ + 65 cos 2 θ + 120 sin θ cos θ
= 65 + 60 sin 2θ
( a + b )max = 125
Hence, 63.(C)
(
)
( )
| 2 a × b | + 3 a .b
= 2|a||b|sin θ + 3 |a||b|cos θ , where θ is angle between a and b = 12 sin θ + 18 cos θ ≤ 122 + 182 = 6 13 64.(B)
Line ax + y = 0 and x + by = 0 intersect at O(0, 0).
ax + y = 0
If AB subtends right angle at O(0, 0), then ax + y = 0 and x + by = 0 are perpendicular.
⇒
( −a ) −
1 = −1 b
⇒
a+b = 0
65.(D) The first two columns of first determinant are same as first two rows of second. Hence, transpose the second. Add the two determinants and use C1 → C1 + C3 . It gives D = 0.
66.(D)
dr dh dr dt = c and h = ar + b ; also dt = 3 dt (given)
VMC/JEE-2013/Solutions
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JEE Mains/Test Series-9/ACEG
Vidyamandir Classes ∴
a
dr dr =3 ⇒ a=3 dt dt
Hence, h = 3r + b When
r =1 ; h = 6 ⇒ 6 = 3+b ⇒ b = 3
∴
h = 3 ( r + 1)
(
V = π r 2 h = 3π r 2 ( r + 1) = 3π r 3 + r 2
(
)
)
dV dr = 3π 3r 2 + 2r dt dt where
dV r =6; = 1 cc / sec dt dr dr 1 = 3π (108 + 12 ) ⇒ 360π = 1 dt dt
Therefore,
dV Again when r = 36 , =n dt
)
(
2 dr n = 3π 3 ( 36 ) + 2.36 dt
n = 3π . 36 (110 ) . (1 / 360 π ) n = 33 67.(A)
sin 3 x a L = lim 3 + 2 + b x→0 x x
= lim
sin 3 x + ax + bx3
x3
x→0
3 = lim
x→0
sin 3 x + a + bx 2 3x x2
For existence of limit 3 + a = 0
⇒
a = −3
∴
L = lim
sin 3 x − 3 x + bx 3
x→0
=−
x
27 +b = 0 6
3
= 27 lim
t →0
sin t − t t3
(Apply LH rule to get lim
+ b = 0 ( 3x = t )
t →0
sin t − t t
3
=
−1 6
⇒
b=
9 2
68.(C) Given integral 1
=
dx
∫ ( x + cos α )2 + (1 − cos 2 α ) 0
1
=
dx
∫ ( x + cos α )2 + sin2 α 0
= =
1 x + cos α tan −1 sin α sin α 1 sin α
1 0
cos α −1 1 + cos α − tan −1 tan sin α sin α
VMC/JEE-2013/Solutions
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JEE Mains/Test Series-9/ACEG
Vidyamandir Classes
69.(B)
α −1 −1 tan cot 2 − tan ( cot α )
=
1 sin α
=
1 −1 π α π −1 tan tan − − tan tan − α sin α 2 2 2
=
1 sin α
π α π α 2 − 2 − 2 − α = 2 sin α
3 log 2 x − 1 − log 2 x + 2 > 0 ( x > 0 ) 2 3 1 ( log 2 x − 1) + > 0 2 2
⇒
log 2 x − 1 −
Let
log 2 x − 1 = t ≥ 0
⇒
log 2 x ≥ 1 ⇒ x ≥ 2
∴
3 1 t − t2 + > 0 2 2
⇒
2t − 3t 2 + 1 > 0 ⇒ 3t 2 − 2t − 1 < 0
⇒
−
. . . .(i)
1 < t
Solutions to JEE Mains/Test Series - 9/IITJEE - 2013 [CHEMISTRY] 1.(C)
By conservation of moles PV P1V1 PV = + 2 2 RT RT RT P×4=2×1+3×2 P = 2 atm.
2.(C)
Statement-II violates the inert pair effect. Hence it is false.
3.(A)
KMnO4 will react with FeSO4 only. 100 1 × moles = 5 × 2 × 1000 moles = 1 χ FeSO4 = 1/ 3
4.(D) 5.(D)
Only Cl2 is oxidsing among the given options Cl
∴ 6.(D)
2 → KMnO K 2 MnO 4 4
∆S = n C v n
T2 V + n R n 2 T1 V1
= nC v n2 + n R n
1 2
= (C v − R) n 2 7.(D)
P-anthranilic acid has less acidic nature due to the +M effect of − NH 2 group.
2NH 3 (g) N 2 (g) + 3H 2 (g)
8.(B)
t=0
a
t = t eq
a 2
PNH 3
3a
3a 2 a P = P= . 3a 3
a
9.(C)
→ H 3 BO3 + 3HBF4 4BF3 + 3H 2O
10.(D)
[H + ] =
11.(B)
Ka1C1 + Ka 2C2 (Learn as a result)
=
1.8 × 10−5 × 0.01 + 6.3 × 10−5 × 0.01
=
8.1 × 10−7 = 9 × 10−4 .
Cr2 O27 − on reduction converts to Cr+3 which is green in colour.
VMC/JEE-2013/Solutions
1
JEE Mains/Test Series-9/ACEG
Vidyamandir Classes n ( n + 2 ) = 1.73 = 3
12.(C)
n (n + 2) = 3 n = no. of unpaired e−s = 1 v = 4s2 3d3. Hence Vanadium must exist in the form V4+ compound is VCl4 ⊕
13.(C)
AgNO3 → AgCl ↓ + CH 2 = CH − C H 2 Allylic carbocation. CH 2 = CH − CH 2 − Cl
14.(A)
→ Mg ( NH 4 ) PO 4 + Na 2SO4 + H 2O . MgSO 4 + NH 4 OH + Na 2 HPO 4
15.(C)
P + I2 Mg / Ether HCHO H 2O → A → → C →D CH 3 − CH 2 − OH B
is CH 2 − CH 2 − I
(A)
is CH 3 − CH 2 − Mg − I
(B) ⊕
is CH 3 − CH 2 − CH 2 − O M gI (D)
(C)
is CH 3 − CH 2 − CH 2 − OH
16.(B)
17.(B)
O O || || CH 3 − C − CH 2 − C − CH 2 − CH 3 Most acidic hydrogen (Active methylene group)
18.(D) NF3 dipole > NH3 dipole.
19.(D) R.D.S. of cannizaro involves transfer of H − ion to the carbonyl group.
20.(A)
O O || || → R − C − Nu + Z − R − C − Z + Nu − Leaving group capacity of Cl − is maximum among the given options hence & fastest reaction when Z = Cl.
21.(C) Both (1) & (2) contain chiral carbon will undergo SN2 reaction with alc. KCN, hence inversion. 22.(B)
π = iCRT
0.75 = 2.47 × Moles = 23.(A)
E=
1 = 0.03 30
CZ2
εH ε
moles × 0.0821 × 300 2.5
Be3+
n2 =
2 Z2H n Be3+
n 2H
Z2 3+ Be
=
1 2
1
×
22
4
2
=
1 4
24.(D) Graphite: free e − are spread out between the structure & thus graphite is conducting. 25.(D) V1 = volume for complete neutralization of Na2CO3. V2 = Volume for complete neutralization of NaHCO3. V1 =x 2 V1 + V2 = x + y Hence V2 = y − x
VMC/JEE-2013/Solutions
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JEE Mains/Test Series-9/ACEG
Vidyamandir Classes 26.(D)
− → Cl(g) is exothermic, rest all endothermic. Cl(g) + e −
27.(D)
28.(C) Metal oxides an basic (or more ionic character ≈ more basic). TiO > VO > CrO > FeO ∴ 29.(C)
dx = k[A]2 dt dx log = log k + 2 log[A] dt y = 2x + log k On comparing with y = mx + c. Where m ≡ slope
c ≡ constant 30.(A)
4r = a 2 r=
2 a 4
Solutions to JEE Mains/Test Series - 9/IITJEE - 2013 [PHYSICS] 31.(C) A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically repeated in three dimensions on a lattice. The spacing between unit cell in various directions is called its lattice parameters or constants. Increasing these lattice constant will increase or widen the band-gap (Eg), which means more energy would be required by electrons to reach the conduction band from the valence band. Automatically Ec and Ev decreases. 32.(D)
K = K1 + K 2 +
T = 2π
33.(B)
v1 t
v2 t
34.(A)
=
K3 K 4 60 × 30 = 50 + 30 + = 100 N / m K3 + K4 90
m 0.01 π = 2π = Hz . 100 50 K
d1 − ρ d2 − ρ
⇒
0.18 20 − 2 = 10 − 2 v2
⇒
v 2 = 0.08 m / s t
t
C ||E × B & CB = E in magnitude.
As
C = 3 × 108 iˆ and E = 720 ˆj
35.(C)
300 − 0 10,000 = 9500 300 − v
36.(B)
Only electric force acts on the particle.
B = 2.4 × 10−6 kˆ
⇒ ⇒
300 − v = 285 ⇒ v = 15 m / s
37.(C) 38.(A) Lowest frequency = HCF of 420 Hz & 315 Hz = 105 Hz.
VMC/JEE-2013/Solutions
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JEE Mains/Test Series-9/ACEG
Vidyamandir Classes 39.(D)
φ = BAN sin ωt
ε =−
dφ = − BAN ω cos ωt dt
40.(B)
ε =−
dφ = − ( 20t − 50 ) = − 10v at t = 3s dt
41.(B)
D1 is not conducting but D2 is conducting 12 =2A i= ⇒ 4+2
42.(B)
Energy of proton = 8 × 7.06 − 7 × 5.6
⇒
ε m = BAN ω
= 17.28 MeV
d 2v dx
2
and
= 3x 2 − 1 < 0 for x = 0
d 2v
⇒
⇒
x3 − x = 0
⇒ x = 0 or x = ± 1
x = 0 is a point of maximum for V
= 3x 2 − 1 > 0 for x = ± 1
dx 2
dv =0 dx
⇒
43.(A) K.E. is maximum when P.E. is minimum
⇒
x = ± 1 are points of minima for V
1 1 −1 − = J at x = ± 1 4 2 4
⇒
minimum potential energy =
⇒
−1 9 maximum KE = 2 − = 4 4
9 1 2 = mvmax & m = 1 kg ⇒ 4 2
⇒
vmax =
3 m/s 2
44.(C) Let Q be total charge on the two spheres & let q1 & q2 be charges on A & B respectively in equilibrium. (rA = 1mm, rB = 2mm) q1 q2 Q 2Q q1 + q2 = Q and = . q1 = and q2 = ⇒ ⇒ 4π ∈0 rA 4π ∈0 rB 3 3
E1 / E2 = q1 / rA2 : q2 / rB2 = 1 / 3 :
45.(A)
µR < µB
⇒
= 2 :1
or for small angled prism D = A ( µ − 1)
(A)
46.(D) 48.(A)
( 2 )2
A + D sin 2 ∵ µ = sin A / 2
⇒ D1 < D2
Ans.
2/3
47.(C)
v2 1 ∝ n R R
T =
⇒
1+ 2π R ∝ R V
v∝
n −1 2
1 n −1) / 2 ( R
⇒
T ∝ R
n +1 2
H 49.(D)
50.(B)
T2
K
⇒
x f = 1/3
2K
H =
⇒
Ans.
(D)
⇒
Ans.
(B)
4x
| ∆P | = E / C − ( − E / C ) = 2 E / C
VMC/JEE-2013/Solutions
⇒
T1
4
KA (T2 − T1 ) T2 − T1 = x 4x 3x + 2 KA KA
JEE Mains/Test Series-9/ACEG
Vidyamandir Classes ⇒ the object is placed at the centre of the equivalent mirror.
51.(A) u = v
1 =−2 Fm
1.5 1 1 1 − 1 − + /2 1 ∞ − 30 − 30
⇒
Fm = − 10 cm .
R = | 2 Fm | = 20 cm 6 = 4A ( 2||6 ) + 1.5 ||3
52.(C)
i=
53.(A)
X 20 4X x ; = = Y Y 80 100 − x
54.(B)
x = 50 cm
Let : length of the wire ; R : radius in first case; r : radius in second case. ⇒ 2π R = & n(2π r ) =
B= B' = 55.(B)
⇒
T = 2π
µ0i 2R
µ0 π i
=
n µ0i n 2 µ0 π i = = n2 B 2r
I MB
where
I: moment of Inertia =
m 2 12
M : pole strength ×
I' =
m ( / s) 12
2
= I / 9 ; M ' = ( pole strength remain same ) × 3 × / 3 M
I /9 T = = 2 / 3 sec M .B 3 56.(D) As P.d across L & C are out of phase.
⇒
T ' = 2π
57.(B)
induced current = −
58.(C)
ω=
59.(B)
ε =
1 = LC
( change influx ) × no of
1 L' (2C )
(
turns
total resistance × time
⇒
(W − W1 ) =− 2 × n ÷ t R + 4 R ) (
= −
n (W2 − W1 ) 5RT
L' = L / 2
)
1 1 2 ∆ω2 = × 0.2 × 10−4 × 5 × (1) = 5 × 10−5V 2 2 = 50 µ v
60.(A)
⇒
VMC/JEE-2013/Solutions
truth table of OR gate.
5
JEE Mains/Test Series-9/ACEG
Vidyamandir Classes
Solutions to JEE Mains/Test Series - 9/IITJEE - 2013 [MATHEMATICS] 61.(D) For the graph to touch x-axis, the polynomial y = x 2 − 2 px + p + 1 would have to be a perfect square. Therefore,
(
)
p 2 = p + 1 , or p 2 − p − 1 = 0 , p = 1 ± 5 / 2 .
62.(C)
Z1 = ( 8 sin θ + 7 cos θ ) + i ( sin θ + 4 cos θ ) Z 2 = ( sin θ + 4 cos θ ) + i ( 8 sin θ + 7 cos θ ) Z1 = x + iy where x = ( 8 sin θ + 7 cos θ ) and y = ( sin θ + 4 cos θ ) Z 2 = y + ix
Hence,
(
Z1 .Z 2 = ( xy − xy ) + i x 2 + y 2
)
= a + ib ⇒ a = 0 ; b = x 2 + y 2
x 2 + y 2 = ( 8 sin θ + 7 cos θ ) + ( sin θ + 4 cos θ ) 2
Now,
2
= 65 sin 2 θ + 65 cos 2 θ + 120 sin θ cos θ
= 65 + 60 sin 2θ
( a + b )max = 125
Hence, 63.(C)
(
)
( )
| 2 a × b | + 3 a .b
= 2|a||b|sin θ + 3 |a||b|cos θ , where θ is angle between a and b = 12 sin θ + 18 cos θ ≤ 122 + 182 = 6 13 64.(B)
Line ax + y = 0 and x + by = 0 intersect at O(0, 0).
ax + y = 0
If AB subtends right angle at O(0, 0), then ax + y = 0 and x + by = 0 are perpendicular.
⇒
( −a ) −
1 = −1 b
⇒
a+b = 0
65.(D) The first two columns of first determinant are same as first two rows of second. Hence, transpose the second. Add the two determinants and use C1 → C1 + C3 . It gives D = 0.
66.(D)
dr dh dr dt = c and h = ar + b ; also dt = 3 dt (given)
VMC/JEE-2013/Solutions
6
JEE Mains/Test Series-9/ACEG
Vidyamandir Classes ∴
a
dr dr =3 ⇒ a=3 dt dt
Hence, h = 3r + b When
r =1 ; h = 6 ⇒ 6 = 3+b ⇒ b = 3
∴
h = 3 ( r + 1)
(
V = π r 2 h = 3π r 2 ( r + 1) = 3π r 3 + r 2
(
)
)
dV dr = 3π 3r 2 + 2r dt dt where
dV r =6; = 1 cc / sec dt dr dr 1 = 3π (108 + 12 ) ⇒ 360π = 1 dt dt
Therefore,
dV Again when r = 36 , =n dt
)
(
2 dr n = 3π 3 ( 36 ) + 2.36 dt
n = 3π . 36 (110 ) . (1 / 360 π ) n = 33 67.(A)
sin 3 x a L = lim 3 + 2 + b x→0 x x
= lim
sin 3 x + ax + bx3
x3
x→0
3 = lim
x→0
sin 3 x + a + bx 2 3x x2
For existence of limit 3 + a = 0
⇒
a = −3
∴
L = lim
sin 3 x − 3 x + bx 3
x→0
=−
x
27 +b = 0 6
3
= 27 lim
t →0
sin t − t t3
(Apply LH rule to get lim
+ b = 0 ( 3x = t )
t →0
sin t − t t
3
=
−1 6
⇒
b=
9 2
68.(C) Given integral 1
=
dx
∫ ( x + cos α )2 + (1 − cos 2 α ) 0
1
=
dx
∫ ( x + cos α )2 + sin2 α 0
= =
1 x + cos α tan −1 sin α sin α 1 sin α
1 0
cos α −1 1 + cos α − tan −1 tan sin α sin α
VMC/JEE-2013/Solutions
7
JEE Mains/Test Series-9/ACEG
Vidyamandir Classes
69.(B)
α −1 −1 tan cot 2 − tan ( cot α )
=
1 sin α
=
1 −1 π α π −1 tan tan − − tan tan − α sin α 2 2 2
=
1 sin α
π α π α 2 − 2 − 2 − α = 2 sin α
3 log 2 x − 1 − log 2 x + 2 > 0 ( x > 0 ) 2 3 1 ( log 2 x − 1) + > 0 2 2
⇒
log 2 x − 1 −
Let
log 2 x − 1 = t ≥ 0
⇒
log 2 x ≥ 1 ⇒ x ≥ 2
∴
3 1 t − t2 + > 0 2 2
⇒
2t − 3t 2 + 1 > 0 ⇒ 3t 2 − 2t − 1 < 0
⇒
−
. . . .(i)
1 < t
