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

Qq

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

Page # 23

Q.149 log7log50x5x2

= log7 1

log5(x2 + 5 + x)1/2 = 1 = log5 5 (x2 + 5+ x)1/2 = 5

(x2 + x + 5) = 25 x2 + x – 20 = 0 (x – 4)(x + 5) = 0 x = 4, –5 x = 4.

Q.150 Given system of equation can be written as

1862

zyx

113213111

As,

113213111 0. So unique solution exists.