Page # 1
CRASH COURSE BITSAT-2017MOCK TEST-4 (07.05.2017)
ANSWER KEY
PART-I_(PHYSICS)Q.1 B Q.2 D Q.3 B Q.4 C Q.5 B
Q.6 B Q.7 C Q.8 C Q.9 D Q.10 B
Q.11 B Q.12 D Q.13 A Q.14 A Q.15 D
Q.16 D Q.17 C Q.18 C Q.19 B Q.20 C
Q.21 C Q.22 C Q.23 B Q.24 D Q.25 C
Q.26 C Q.27 B Q.28 B Q.29 C Q.30 A
Q.31 B Q.32 C Q.33 B Q.34 C Q.35 C
Q.36 D Q.37 C Q.38 A Q.39 B Q.40 C
PART-II (CHEMISTRY)Q.41 C Q.42 C Q.43 A Q.44 B Q.45 B
Q.46 A Q.47 C Q.48 C Q.49 A Q.50 A
Q.51 A Q.52 C Q.53 A Q.54 A Q.55 A
Q.56 B Q.57 B Q.58 C Q.59 C Q.60 C
Q.61 C Q.62 C Q.63 B Q.64 D Q.65 C
Q.66 C Q.67 C Q.68 A Q.69 B Q.70 D
Q.71 A Q.72 B Q.73 C Q.74 D Q.75 D
Q.76 B Q.77 A Q.78 D Q.79 C Q.80 B
PART-III (A)ENGLISH PROFICIENCYQ.81 B Q.82 A Q.83 B Q.84 C Q.85 B
Q.86 C Q.87 D Q.88 C Q.89 D Q.90 C
Q.91 D Q.92 A Q.93 A Q.94 B Q.95 C
(B) LOGICAL REASONNGQ.96 C Q.97 C Q.98 A Q.99 D Q.100 C
Q.101 A Q.102 A Q.103 D Q.104 A Q.105 B
PART-IV (MATHEMATICS)Q.106 B Q.107 C Q.108 A Q.109 A Q.110 B
Q.111 B Q.112 B Q.113 D Q.114 C Q.115 B
Q.116 D Q.117 B Q.118 C Q.119 C Q.120 D
Q.121 D Q.122 A Q.123 A Q.124 B Q.125 C
Q.126 A Q.127 B Q.128 D Q.129 C Q.130 C
Q.131 C Q.132 A Q.133 C Q.134 B Q.135 D
Q.136 C Q.137 D Q.138 A Q.139 A Q.140 A
Q.141 B Q.142 C Q.143 D Q.144 C Q.145 A
Q.146 A Q.147 D Q.148 C Q.149 C Q.150 A
Page # 2
HINT & SOLUTIONPART-I_(PHYSICS)
Q.1 RX = RY = RZ = RR
R/2X, Y in parallel then Z in series.
Q.2 Resistance of circuit X = R/2
Resistance of circuit Y = 2R
Power in circuit X =)2/R(
E2
Power in circuit Y =R2
E2
Q.3 Gain in KE = work done = q times potential difference, potential difference is maintained constant.
Q.4 Gain in potential energy = charge x potential difference = 2 × (9–6) = 6J
Q.5 0r
r
r
r
4
r
r
4
q).Q(
r
r
4
q.Q 3
2
23
1
1
03
2
2
03
1
1
0
Q.6 First decrease current, increase voltage, then decrease voltage.
Q.7 I is not done because it will increase eddy current losses.
Q.8 a = g – kvacceleration decreases with increase in velocity or time, but variation with time exponential.
Q.9
6 rvFB
mg
mg = FB + 6rv
vT =9gr2 2
(object – liquid)
since r is not given : we cannot arrive at a value. But greatest value of VT will betime
depth =
4.4 × 10–5 m/s
Page # 3
Q.10 conservation of angular momentum at ends of major axis maxminminmax dvdv
Q.11 (149.5) × 1 × T = 0.5 × 540
T = 1.8°C (Temperature will decrease)
Q.12 Power radiated per unit area proportional to fourth power of temperature.
Q.13500
2001
Q.14 Due to input of energy, temperature and entropy rises
Q.15 M kg × 50 MJ/kg × 20% (efficiency) = 1000 MW
Q.16 Objects are at same temperature.
Q.17 Statistical law gives dependence on current condition not on history.
Q.18 40 = k × (2000)4
P = k × (2500)4
40
P =
4
4
5
P =4
4
5
× 40 = 97.6 W 98W
Q.19 Average velocity =time
ntdisplaceme =
t
R2 =
50
2502 = 10 m/s
Average speed =time
cetandis =
t
R =
50
25014.3 16 m/s
Q.20 Time of flight t remain some, R = vt, v 3v, R 3R
Q.21 S =
2
v0t v =
t
S2 =
11
282 = 5.1 m/s
Page # 4
Q.22 h = –12 sin 30° +2
1 (9.8) (5.6)2
= 147.6 m 148 m
Q.23 M = ib2
B M
Q.24 MiC2 + Ei = Mf C
2 + Ef
Ef > Ei fission reaction
So Mi > Mf
Q.25 Activity =NWhere = decay constantand N = number of nucleiGiven activity of S = Activity of T TNT = SNS
T
S
N
N =
S
T
Q.27 0.9 eV = (–1.6 eV) – (–2.5 eV)
Q.28 20Ca45 (21Sc45)* + – + (21Sc45)* 21Sc45 +
Q.29 A and E cancel, B and D gives resultant 1.14210 east, B + D + E gives 24.1
Q.30
R W
P
R + P = W
Q.31 µ 22 = µ 11 1
2
µ
µ =
2
1
=5.1
1 = 0.67
Q.32 =R
a =
2.0
5 = 25 rad/s2
Page # 5
Q.33 L = I
K =2
1I2 = 0.5
I
L2
Q.34 amax = 2xmax = 0
2
xT
2
= 2
2
T
4x0
Q.35 amax = 2A 40 = 2 × 0.02 = 520
f =2
=2
520 = 7.1 Hz
Q.36 Magnetic flux = Ad.B
; dot product give scalar result and energy is scalar..
Q.37 F = kx cm6
N20 = k
w =2
1k (x2
2 – x12) =
2
1 ×
6
20N/cm × (62 – 32) cm2 = 45 N-cm
Q.38 fwall = 50KHz
5–340
340
fecho = fwall
340
5340
= 50K
5–340
5340 = 51.5 KHz
Q.39 Superposition
Q.40 Apply dopplers formula twice, car as detector
f
f1 =oc
vc
car as source1
2
f
f =
vc
c
f2 = )vc(
)vc(f
~ f
c
v21 Binomial approx.
Page # 6
SOLUTION
PART-II (CHEMISTRY)
Q.41 Theory Based
Q.42 Theory Based
Q.43 2OW = 64 gm 2On = 2
2HW = 16 gm 2Hn = 8
2OP = atm105010
2
2HP = atm405010
8
Q.442
a
D C
BA O
Oa/2
a2
a/2
Q.45 2A(g) + B(g) C(g) H = – 300 Kcal/mol4 mole 3 mole
U = H – ng RT = – 300 +1000
50022 = – 298 Kcal/.mol
For 4 moles of A energy released will be = – 298 × 2 = – 596 Kcal/mol
Q.46 A(g) + 2B(g) l C(g)[A]eq = 3 M ; [B]eq = 4 M.
Let [C]eq = xM
Kc = 2eqeq
eq
]B[]A[
]C[ = 2)4(3
x =
48
x.
When the volume of vessel is doubled, reaction moves in the backward direction. Let the decrease inconcentration of C be yM. At new equilibriumA(g) + 2B(g) l C(g)
1.5 + y 2 + 2y y2
x
Given that 2 + 2y = 3 y =2
1 = 0.5
Page # 7
Kc = 2)3()5.05.1(
21-
x
=48
x
On solving, x = 4 M.
Q.47 9]117[2
1pKpK
2
1pH
21 aa
Q.48 P4 can undergo oxidation as will as reduction
Q.49 2KI + HgI2 K2[HgI4] l 2K+ + [HgI4]2–
Thus, there is net decrease in number of ions present in solution and freezing point is raised
Q.50 Theory Based
Q.53 CH3COOAg is a salt of weak acid strong base. The solubility of any salt of weak acid strong base ishighest in acidic buffer, less in pure water and least in basic buffer.
CH3COOAg(s) l Ag+(aq) + CH3COO¯ (aq)CH3COO¯ (aq) + H+(aq) l CH3COOH(aq)
(from buffer)
Q.55 CH3CHO + CHOHC 2
CHOCHCHCH|
OH
23 OH2
CH3–CH=CH–CHO
CH3(CH=CH)3–CHO OH2
CHO|
CHCHCHCHCHCHCH|
OH
23 ]
Q.57Ph
Br
H
H
Ph
Br
dustZn
Ph Ph
H
Br
BrH
Ph Ph
H
Br
BrH
PhH
HPh
Page # 8
Q.58OH
42SOH.conc
Cl
1 Fraction
+
Cl
1 Fraction
h/Cl2
(B)
COOHCH)ii(
THF.BH)i(
3
3
(A)
+
Cl
(4) stereoisomer2 fraction
+
Cl(2) stereoisomer2 fraction
Cl
(4) stereoisomer2 fraction
Enantiomers give same fractions on fractional distillation.
Q.59 Rate of SN1 Stability of carbocation.
HN O
IV III I IIAromatic Aromatic Non
AromaticAnti aromatic
Q.60
OMe
OEt
HI (excess)DBC
DBC
OH
OH
+ MeI + EtI
Q.61
F
O N2 NO2
Cl
Cl
HO
OH
O N2 NO2
F
Q.62 CH2 = CH – CH2 – OH give resonance stabilized CH2 = CH – 2HC
carbocation therefore undergoes
SN1 mechanism and faster.
Page # 9
Q.63 CH3–CH2–COOH2Cl
Pred COOHCHCH|Cl
3
Example of HVZ reaction
Q.64 Reimer Tiemann reaction
Q.66 (I)
NH2
OCH3
(II)
NH2
CH3
(III)
NH2
CH3CH3H3C
(IV)
NH2
+ M +H +IRate tendency to donate lone pair from –NH2
Q.67 (A)
COOH
(C)
COOH
OCH3
(D)
COOH
CH3
Induced +M + M + H
Q.68
XeF ¯3
Xe
F
FF
90°
90°
SeF ¯4
Se
F
FF
F
XeF ¯5
XeF
FF
F
F
XeO ¯64
Xe¯ O
¯ O
O¯
O¯
O
O
XeF3¯ does not exist 2 lone pair can't exist at 90° or less.
Q.69 N
O
O
Cl
O O
sp2 / Bent sp2 / BentClO2 doesn't dimerised readily between odd electron delocalised in a vacant 3d orbital.
Page # 10
Q.70 [M(gly)(py)(OCN) (PPh3)N3]+2
(A) Calculation for oxidation state of central atomx – 1 + 0 – 1 + 0 – 1 = +2x = + 5 Oxidation of central atom
(B) [M(gly)(py)(OCN) (PPh3)N3]+2
[M(AB)cdef]2+ Coordination number = 6
(C) [M(AB)cdef] 12 Geometrical isomers are possible(D) [M(gly)(py)(OCN) (PPh3)N3]
+2
and –––––– Both complexes are linkage isomers of each other.[M(gly)(py)(NCO) (PPh3)N3]
+2]
Q.71 Complexes of M(AB)3 type shows both geometrical and optical isomerisms.
Q.73 In extraction of Cu from copper pyrite(A) 2CuFeS2 or Cu2S · Fe2S3 + 4O2 Cu2S + 2FeO + 3SO2 (Roasting process)(B) Cu2O + FeS Cu2S + FeO (Roasting process)(C) 2CuO 2Cu + O2(D) FeO + SiO2 FeSiO3 (In Bessemer's converter)
(slag) ]
Q.74 Rubby copper : Cu2OMalachhite green : Cu(OH)2. CuCO3Azurite : Cu(OH)2. 2CuCO3Siderite : FeCO3 ]
Q.75 Due to small size of He, it escapes from interstitial spaces / voids of molecular lattice of quinols ]
Q.76 Na2CO3 precipitates out Ca++ and Mg++ from hard water as insoluble carbonates Calgon; (NaPO3)6 ; exchanges Ca++ and Mg++ from hard water Permutit : common type of zeolite, it exchanges Ca++ and Mg++ from hard water with its Na+
and also does not allow to pass large sized molecules from it.
Q.77 Li < Be > B Due to the stable fully filled s-orbital configuration the IE of Be is greats than Li & B]
Q.78)A(
232 )NO(Hg HCldil/SH2 SNa2
Page # 11
Q.79 NaNO3 C500
below
NaNO2 + 2
1O2
Cu(NO3)2 CuO + 2NO2 +2
1O2]
Hg (NO3)2 Hg + 2NO2 + O2
AgNO3 Ag + NO2+2
1O2
Q.80 Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3(g)(A)
3CuO + 2NH3 3Cu + N2 + 3H2O (A) (B) (C)
(NH4)2Cr2O7 N2 + Cr2O3 + 4H2O
NH4Cl NH3 + HCl
NH4NO3 N2O + 2H2O
(NH4)2SO4 2NH3 + H2SO4
Page # 12
SOLUTIONS
PART-III (A) ENGLISH PROFICIENCY
(B) LOGICAL REASONING
Q.96 In this simple alternating subtraction and addition series; 1 is subtracted, then 2 is added, and so on.
Q.97 The first letters are in alphabetical order with a letter skipped in between each segment: C, E, G, I, K.
The second and third letters are repeated; they are also in order with a skipped letter: M, 0, Q, S, U.
Q.98 In this series, the third letter is repeated as the first letter of the next segment The middle letter, A, remains
static The third letters are in alphabetical order, beginning with R.
Q.99 In this series, the letters remain the same: DEE The subscript numbers follow this
series: 1,1,1 ; 1,1,2; 1,2,2 ; 2,2,2; 2,2,3
Q.100 There are two alphabetical series here. The first series is with the first letters only: STUVW. The second
series involves the remaining letters: CD, EF, GH, IF, KL.
Q.101 The middle letters are static, so concentrate on the first and third letters. The series involves an alphabetical
order with a reversal of the letters. The first letters are in alphabetical order: F, G, H, I, J. The second and
fourth segments are reversals of the first and third segments. The missing segment begins with a new
letter.
Q.102 This series consists of a simple alphabetical order with the first two letters of all segments: B, C, D, E, F,
G, H, I, J, K. The third letter of each segment is a repetition of the first letter.
Q.103 There are three series to look for here. The first letters are alphabetical in reverse: 2, Y, X, W, V. The
second letters are in alphabetical order, beginning with A. The number series is as follows: 5,4, 6, 3,7. ]
This set contains sequence questions that use a series of nonverbal, nonnumber symbols. Look carefully
at the sequence of symbols to find the pattern.
Q.104 Look at each segment. You wilI notice that in each, the figure on the right and the figure on the left are the
same; the figure in between is different. To continue this pattern in the last segment, the diamond on the
left will be repeated on the right. Choice a is the only possible answer.
Q.105 Each arrow in this continuing series moves a few degrees in a dockwise direction. Think of these arrows
as the big hand on a dock. The first arrow is at noon. The last arrow before the blank would be 1240.
Choice b, the correct answer, is at 1245.
Page # 13
SOLUTIONSPART-IV (MATHEMATICS)
Q.106 Given that
y = tan–1
1xx
12 + tan–1
3x3x
12 + tan–1
7x5x
12 + tan–1
13x7x
12 +
.... + upto n terms
General term
Tr = tan–1)rx)(1rx(1
1
Tr = tan–1)rx)(1rx(1
)1rx()rx(
Tr = tan–1 (x + r) – tan–1(x + r – 1)T1 = tan–1(x + 1) – tan–1(x + 1)
T2 = tan–1(x + 2) – tan–1 (x + 1)
Tn = tan–1(x + n) – tan–1 (x + n–1) y = tan–1 (x + n) – tan–1 x
y' = 2)nx(1
1
– 2x1
1
y'(0) = 2
2
2 n1
n1
n1
1
Q.1071x
nx....xxLim
n2
1x
(n N)
0
0 from
=2
)1n(n
1
nx....x21Lim
1x
Q.108 We have,
10150 – 9950
= (100 + 1)50 – (100 – 1)50
10050
50
100
11
– 10050
50
100
11
= 10050
....100
1·C
100
1·C2
3
350
150
= 50
5050
100
99101 = 1 + 2
...100
1C
3
350
Page # 14
50
5050
100
99101 = 1 + (positive quantity) > 1
10150 – 9950 > 10050 10150 > 10050 + 9950
Q.109 7log
15
16 + 5 log
24
25 + 3 log
80
81 = log
357
80
81·
24
25·
15
16 = log 2
Q.110 Now, x2 – 6x + 7 = (x – 3)2 – 2It is obvious that minimum value is –2 and maximum value is .
Q.111 e)x1(Lim)0(f)x(fLim x/1
0x0x
Q.112 Let a and b be two numbers, then
AM =2
ba 27 =
2
ba a + b = 54
and HM =ba
ab2
12 =
54
ab2 ab = 324
GM = ab = 324 = 18.
Q.1130x
Lim 2x
1)xcos(sin =
0xLim 2
2
x
2xsin
sin2
= 20x
Lim
22
22
x2
xsin
2xsin
·2
xsinsin
= –2(1)24
1 = –
2
1
Q.114 Given equation can be rewritten as
(x – 1)2 + (y – 3)2 =2
13
17y12x5
SP = PM
Here, focus is (1, 3), directrix 5x – 12y + 17 = 0
The distance of the focus from the directrix =14425
17365
=13
14 = 2a
Latus rectum = 2 ×13
14 =
13
28
Page # 15
Q.115 Since, A + B + C = But 2B = A + C
3B = B = /3
b
ca =
Bsin
CsinAsin
=
3sin
2CA
cos2
CAsin2
=
3sin
2CA
cos3
sin2
=
2CA
cos2 ]
Q.116 tan ( + ) =
tan.tan1
tantan =
31
·21
131
21
= 1 =4
Q.117 Length of ladder, AC =º60sin
36 = 12m
Q.118 Let d be the length of line, then projection on x-axis = dl = 3, projection of y-axis = dm = 4
and projection on z-axis = dn = 5
Now, d2(l2 + m2 + n2) = 50 d2 = 50 d = 25
Q.119 (1 + 3x + 3x2 +x3)6 = a0 + a1x + ... + a18x18 (19 terms)
Q.120 Since, f : R R and g : R R, given by f(x) = 2x – 3 and g(x) = x3 + 5 respectively, are
bijection, therefore f–1 and g–1 exists.
We have,
f(x) = 2x – 3
f(x) = y 2x – 3 = y x =2
3y f–1(y) =
2
3y
Page # 16
Thus, f–1 is given by
f–1(x) =3
3x , n R
Similarly, g–1(x) = (x – 5)1/3, x RNow,
(fog)–1 (x) = (g–1 o f–1)(x) = g–1(f–1(x)) = g–1
2
3x =
3/13/1
2
7x5
2
3x
Q.121 Let z = x + iy, then z2 + z = 0
(x2 – y2 + 2ixy) + (x –iy) = 0 x2 – y2 + x =0 and 2xy – y =0Now, 2xy – y = 0 y(2x – 1) = 0 y = 0 or x = 1/2
If y = 0, then
x2 – y2 + x = 0 x2 + x = 0 x = 0 or x = –1If x = 1/2, then x2 – y2 + x = 0
y = ±2
3
Thus, the given equations has four solutions.
Q.122 The determinant of the coefficient matrix of given system of equation is
955
312
321
= 1 (9 – 15) – 2(18 – 5) + 3(10 – 5) = 3 0
Hence, the system of given equation has unique solution.
Q.123 Given, series is
....3·7
1
3·5
1
3·3
1
3
1753 ...(i)
Putting,3
1 = x in Eq.(ii), we get
....3·7
1
3·5
1
3·3
1
3
1753
Page # 17
=
...
7
x
5
x
3
xx
753
=2
1loge
x1
x1
=
2
1loge
31
1
31
1
3
1x =
2
1loge2
Q.124 n+1Cn–2 – n+1Cn–1 100
n+1C3 – n+1C2 100
2
n)1n(
6
)1n(n)1n(
100
(n + 1)n (n – 1) – 3n(n +1) 600
(n + 1)n (n – 4) 600
The values of n satisfying this inequality are 2, 3, 4, 5, 6, 7, 8, 9.
Q.125 From given relation, we have
x = cos ± i sin Take x = cos + i sin and xn = cos n + i sin nand 1/xn = cos n – i sin n xn + 1/xn = 2 cos n x2n – 2xn cos n + 1 = 0
Q.126 Since, the probability lies between 0 and 1
13
p310
, 1
4
p10
, 1
2
p210
0 1 + 3p 3, 0 1 – p 4, 0 1 – 2p 2
3
1 p p
2, –3 p 1,
2
1 p
2
1... (i)
Again, the events are mutually exclusive
0 2
p21
4
p1
3
p31
1 0 13 – 3p 12
3
1p
5
13... (ii)
Page # 18
From Eqs. (i) and (ii), we get
max
3
1,
2
1,3
3
1p min
3
13,
2
1,1,
3
2
3
1p
2
1
Q.127 Put x2 + x = y, so that equation (1) becomes (y – 2) (y – 3) = 12 y2 – 5y – 6 = 0 (y – 6) (y + 1) = 0 y = 6, –1When y = 6, we get x2 + x – 6 = 0 (x + 3) (x – 2) = 0 or x = –3, 2When y = –1, we get x2 + x + 1 = 0
x = w, w2 and their sum is –1.
Q.128 AM > GM
22
22
tan9.cot42
cot4tan9
9tan2 + 4cot2 > 12.
Minimum value is 12
Q.129 Let r and R be the inradius and circumradius of incircle and circumcircle of the given regular
polygon of side n then we know that
r + R =2
a cot
n
+
2
a cosec
n
=
2
a
n/sin
n/cos1=
2
a
n2·cos
n2sin2
n2cos2 2
=2
acot
n2
Q.130 We have, 3f(x) – f
x
1 = logex
4
3f(x) – f
x
1 = 4 logex ... (i)
3f
x
1 – f(x) = 4loge
x
1
x
1byxreplacing
3f
x
1 –f(x) = – 4logex ... (ii)
Solving Eqs. (i) and (ii), we get
8f(x) = 8logexf(x) = logex f(ex) = x
Page # 19
Q.131 | z 1 + z2 | = | z1 | + | z2 | amp(z1) = amp(z2)
So, amp
2
1
z
z= amp(z1) – amp (z2) = 0
Q.132 The combined equation of the lines joining the origin to the points of intersection of
x cos + y sin = p and x2 + y2 – a2 = 0 is a homogeneous equation of second degree given
by
x2 + y2 – a2
2
p
sinycosx
= 0
x2(p2 – a2 cos2 ) + y2 (p2 – a2 sin2) – 2xya2 sin cos = 0
The lines given by this equation are at right angle, if coefficient of
x2 + coefficient of y2 = 0
p2 – a2 cos2 + p2 – a2 sin2 = 0
2p2 = a2
Q.133 Image (x,y) of the point (x1, y1) about the line ax + by + c = 0
221111
ba
)cbyx(2
b
yy
a
xx
10
)20(2
3
8y
1
3x
x = –1, y = –4
Q.134 f(x) = [x] cos
2
1x2
Only possible points of discontinuity can be integers as [x] is discontinuous at integers.
Let x = a (a integer)
f(a–) = (a – 1) cos
2
1a2 = 0
2
ofmultipleoddis2
)1a2(
f(a) = a cos
2
1a2 = 0
f(a+) = a cos
2
1a2 = 0
Page # 20
Q.135 Given, 0x7dx
dy4
dx
dy
On squaring, we getdx
dy = 16
2
dx
dy
+ 49x2 + 56xdx
dy
Here, order is 1 and degree is 2.
Q.136 1000
0
])x[x( dxe
ex–[x] is periodic with period 1.
1000
0
])x[x( dxe = 1000 1
0
xdxe = 1000(e – 1)
Q.137 0a0bba00ba
a(a2) – b(–b2) = 0 a3 = –b3 1b
a3
Q.138 f(x) – f(1) = f(n – 1) – f(0)f(n) = f(n – 1) + f(1) as f(0) = f(1)
f(n) = f(n – 2) + f(1) + f(1) f(n) = f(n – 3) + f(1) + f(1) + f(1) f(n) = f(0) + nf(1) = nf(1)
Let f(x – 1) + f(x+ 3) = 2f(x + 1)f(x + 1) – f(x) = f(x) – f(x – 1)
Q.139 Clearly, from diagram such line is 2x = 3
Page # 21
Q.140 l o g 1/2 1|z|2
4|z|2|z|2
2
< 0 = log1/2 1 1|z|2
4|z|2|z|2
2
< 1
or | z |2 +| z | + 4 > 2 | z |2 + 1 or, | z |2 – 2 | z |3 < 0
or (| z | + 1) (|z | – 3) < 0 | z | < 3 Ans.
Q.141 Since x2 + 1 = 0, gives x2 = –1 x = ±i
x is not real but x is real (given)
No value of x is possible.
Q.142 Given that,yx
100
b
10
a
10
ba
= 1000.
Let a = 0 and b = 1
1000100
1
10
1yx
10–x = 10–2y = 103 x = –3, y = –2
3
Now, 3
1
3
2
3
1
y
1
x
1
Q.143 f (x) = x )1(f)16(f
)25(f
=
116
25
=
5
5= 1
Q.144 g(x) = 22 )x('f)x(f
Differentiating w.r.t. x g'(x) = 2f(x) f '(x) + 2f '(x) f "(x)
= 2f '(x) [f(x) + f "(x)] = 2f '(x) . 0 = 0
(By hypothesis f (x) + f "(x) = 0)
g(x) = constant Thus g(8) = g(3) = 3
Q.145 Given curve x2 = 3 – 2y
Differentiate w.r.t. x, 2x = 0 – 2dx
dy
dx
dy= –x
Slope of the tangent of the curve = –xFrom the given line, slope = –1, x = 1 and from equation (i), y = 1.
Co-ordinate of the point is (1, 1).
Page # 22
Q.146 I =
02 dx
x1
1
x
1xlog
Put x = tan dx = sec2 d
I =
dsec
sec)cotlog(tan 2
22/
0
I =
2/
0
d)cotlog(tan
I =
2/
0
2
dtan
)tan1(log I = 2
2/
0
2/
0
dtanlogdseclog
I = 2
2/
0
dseclog
2/
0
0tanlog
I = –2
2/
0
dcoslog I = –2×2
log 2
2log2
coslog2/
0
I = log 2
Q.147 = mC2 =2
)1m(m
C2 = m(m–1)/2C2 =
1
2
)1m(m
2
)1m(m·
2
1
=8
1m(m – 1) (m – 2) (m + 1) =
8
1(m + 1) m(m – 1) (m – 2) = 3 . m+1C4
Q.148 The equation of tangent at (2, 3) to the given parabola is x = 2y – 4
Required area = 3
0
2 dy}4y21)2y{(
=
3
0
23
y5y3
)2y(
=
3
1– 9 + 15 +
3
8= 9 sq. unit.
(2, 3)
(–4, 0)(y – 2) = (x – 1)2
x
y
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