Questions with Solutions

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1 JEE-MAIN 07 Offline Paper Set-A Questions with Solutions PART A - PHYSICS ALL THE GRAPHS / DIAGRAMS GIVEN ARE SCHEMATIC AND NOT DRAWN TO SCALE.. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor by: () 9 () /9 () 8 () /8 Initial stress Where = mg A d l b h g = = dhg l b d = density, l= Length, b=width, h=height Final stress = d 9h g = 9 (Initial stress).. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time? () () () () For upward motion, velocity will be positive but decreases till maximum height and reaches zero. Then it will increase in negative direction.. A body of mass m = 0 kg is moving in a medium and experiences a frictional force F = kv. Its initial speed is v 0 = 0ms. If, after 0 s, its energy is (/8)mv 0, the value of k will be : () 0 kg m () 0 kg s () 0 kg m () 0 kg m s F = kv K.E. at 0 s dv m kv dt = v0 v = v 0 dv k v = m v 0 dt mv0 v0 = m+ 0kv k + = 0 m = m + 0 kv 0 v v m 0k = + v v m 0 mv0 v = m+ 0kv 0 0 m = 0 kv k = = 0 kg/s. A time dependent force F = 6t acts on a particle of mass kg. If the particle starts from rest, the work done by the force during the first sec. will be : Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

2 ().5 J () J () 9 J () 8 J F = 6t dv m 6t dt = v 0 dv = 6 tdt 0 v = 6 = w = K.E. = =.5 J. 5. The moment of inertia of a uniform cylinder of length I and radius R about its perpendicular bisector is I. What is the ratio l / R such that the moment of inertia is minimum? () Ans.: () () ML MR L R I= + = M + L R = dπ R L + ( ) dπr L L R = + dπr L L = + R 5 dπr L L =. R + R Let L x R = 5 dπr So, I =.x( x + ) As x increases, I increases for all x Lowest option is. O () () 6. A slender uniform rod of mass M and length I is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle 6 with the vertical is: () g sinθ () g sinθ () g cosθ () g cosθ l l l l Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

3 τ = I α 0 0 l ml mg sin θ=. α gsinθ α= l Z O θ mg X l sinθ 7. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius) : () () () () GMd for d R R g = GMd for d R R 8. A copper ball of mass 00 gm is at a temperature T. It is dropped in a copper calorimeter of mass 00 gm, filled with 70 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75 C. T is given by: Ans.: () (Given: room temperature = 0 C, specific heat of copper = 0. cal/gm C) () 800 C () 885 C () 50 C () 85 C Heat lost by copper ball = Heat gained by copper calorimetric + Heat gained by water (T 75) = (75 0) + 70 (75 0) φ(t 75) = 5φ+ 756φ T = = 885 C 9. An external pressure P is applied on a cube at 0 C: so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and a is coefficient of linear expansion. Suppose We Want to bring the cube to its original fize by heating. The temperature should be raised P () αk P k = v/v Or Pv V = k For expression, Pv α V T = k P T = α k Pv V = k () P α K () α PK () PKα Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

4 0. C p and C v are specific heats at constant pressure and constant volume respectively. It is observed that C p C v = a for hydrogen gas C p C v = b for nitrogen gas The correct relation between a and b is : () a= b () a = b () a = b () a = 8 b For molar heat capacities If C p and C C = R C p v C p are specific heats, then p R Cp = M Where M molar mass a MN 8 So, = = = b M A = b. H. The temperature of an open room of volume 0 m increases from 7 C to 7 C due to the sunshine. The atmospheric pressure in the room remains 05 Pa. If n f and n i are the number of molecules in the room before and after heating, then n f n i will be: ().6 0 ().8 0 () () PV PV n f n = i RT RT f = = 9 0 i = = 9 5 = A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like : () () () () At t = 0 i.e.at equilibrium position, K.E. is maximum. At At T t = i.e. at extreme position, K.E. zero T t =, K.E. is again maximum and so on. Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

5 . An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 0 GHz. What is the frequency of the microwave measured by the observer? (speed of light= 0 8 ms ) () 0. GHz (). GHz () 7. GHz () 5. GHz For Doppler effect in light f = f 0 Vs C V s C Where, f = frequency of observation f 0 = frequency of source V s = speed of source C = speed of light So, f = 0 GHz. = 0 GHz. = 0 =7. GHz. C C C C = 0. An electric dipole has a fixed dipole moment p, which makes angle 8 with respect to x-axis. When subjected to an electric field E ˆ = Ei, it experiences a torque T ˆ =τk. When subjected to another electric field E = Eˆ j; it experiences a torque T = T. The angle θ is: T () 0 () 5 () 60 () 90 T = pesinθ = pe cosθ = pecosθ T = T pesinθ= pecosθ tanθ= θ= 60. θ 90 θ p E =E i p Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

6 5. A capacitance of µf is required in an electrical circuit across a potential difference of.0 kv. A large number of µf capacitors are available which can withstand a potential difference of not more than 00 V. The minimum number of capacitors required to achieve this is : () () 6 () () Desired configuration is So, minimum number of capacitor required = 8= 6. In the given circuit diagram when the current readies steady state in the circuit the charge on the capacitor of capacitance C will be: 7. r r r CE CE CE ( () CE () r + r ) () ( r+ r ) () ( r + r ) At steady state, current through capacitor will be zero. So, current through r = p.d. across capacitor, CE r Change on capacitor = CV = r + r V Er = Ir = r + r In the above circuit the current in each resistance is: () A () 0.5 A () 0.5 A () 0 A Taking potential of right side bottom end as zero and using concept of relative potential 6v v v v v v O Ω Ω Ω 6v v v O v v v p.d. across each resistor is zero, so current in each resistance is zero. Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

7 8. A magnetic needle of magnetic moment Am and moment of inertia kg m is performing simple harmonic oscillation in a magnetic field of 0.0 T. Time taken for 0 complete oscillations is: () 6.65 s () 8.89 s () 6.98 s () 8.76 s T = π I MB =. = 6.66 s When a current of 5 ma is passed through a galvanometer having a coil of resistance 5 to, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 0 V is : () Ω ().05 0 Ω ().55 0 Ω () Ω V 0 Rs = Rg = 5 = I 5 0 g = 985Ω = Ω 0. In a coil of resistance 00 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is: Ans.: () () 00 Wb () 5 Wb () 50 Wb () 75 Wb φ = q R = Area under i-t curve R = 50 wb. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If is the smallest possible wavelength of X-ray in the spectrum, the variation of log X min with log V is correctly represented in: () () () () Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

8 λ = min Lag hc ev hc λ min = lag lagv e log λ min So, graph is as shown lag λ min log V. A diverging lens with magnitude of focal length 5 cm is placed at a distance of 5 cm from a converging lens of magnitude of focal length 0 cm. A beam of parallel light falls on the diverging lens. The final image formed is: () real and at a distance of 0 cm from convergent lens () virtual and at a distance of 0 cm from convergent lens () real and at a distance of 0 cm from the divergent lens () real and at a distance of 6 cm from the convergent lens f = 5cm f= -5cm f=0cm 5cm For lens -; u =.f = 5 So, v = 5 For lens ; = - 0, f = 0 So, v = 0 cm. In a Young s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 50 cm away. A beam of light consisting of two wavelengths, 650 nm and 50 ran, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is : Ans.: () ().56 mm () 7.8 mm () 9.75 mm () 5.6 mm λ β= D d For light of 650 nm mm β = = 5 0 For light of 50 nm, mm β = = 5 0 Least distance from the common central maximum = L.C.M. of β and β = 7.8 mm Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

9 . A particle A of mass m and initial velocity v collides with a particle B of mass (m/) which is at rest. The collision is head on, and elastic. The ratio of the de-broglie wavelengths λ A to λ B after the collision is: Ans.: () λ A () = λ B λ A () = λ Let v and v be velocities after collision m By momentum conservation, Or m B λ A () = λ m mv + v = mv Solving () & (), we get v v v m v + = v v Or v+ v = v..() & v Using equation of e, rest m v v = v..() = v B v = v = λ h p p = = λ p h P A B B B A A m v. = = v m. λ A () = λ B 5. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ / λ is given by : () r = / () r = / () r = ¾ () r = / E hc = λ Or hc λ= E E λ E So, r = = = = λ E E 6. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.. Then, t is given by : T log log. () t= () t = T () t = T log (.) () log. log T t = log(.) Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

10 Ans.: () λ A B t = 0 N 0 0 t x 0.x Now, x+ 0.x = N0 Or N 0 x =. Since, N= N.e λt N0 λt = Ne 0. t e λ =. ln. t = λ ln (.) t =.T ln lag (.) t = T lag 7. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be : () 5º () 90 º () 5 º () 80 º Based on theory 8. In amplitude modulation, sinusoidal carrier frequency used is denoted by ω and the signal frequency is denoted by ω m. The bandwidth ( ω m ) of the signal is such that ω m << ω c. Which of the following frequencies is not contained in the modulated wave? () ω m () ω c () ω m + ω c () ω c ω m Frequency contained in modulated wave (wc w m) w wc+ wm So, w is not contained in the modulated wave. m 9. Which of the following statements is false? () Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude. () In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed. () A rheostat can be used as a potential divider. () Kirchhoff's second law represents energy conservation. Ans.: () Based on theory. 0. The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D =.5 0 m; rise of water, h =.5 l0 m. Using g = 9.80 m/s rhg and the simplified relation T = 0 N/m, the possible error in surface tension is closest () 0.5% ().5% ().% () 0% Ans.: () Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

11 hg T = = T D h = + T D h T 0 0 = + T T 00 = 00 T = =.5%.. Given PART B - CHEMISTRY C(graphite) + O(g) CO (g) ; r H = 9.5 kj mol H(g) + O(g) HO( l) ; r H = 85.8 kj mol CO (g) + H O( l) CH (g) + O (g) ; r H = kj mol Based on the above thermochemical equations, the value of r H at 98K for the reaction C(graphite) + H (g) CH (g) () 7.8 kj mol ().0 kj mol () kj mol () +.0 kj mol. C+ O CO H= 9.5 kj mol - H + O HO H= 85.8 kj mol - CO+ HO CH + O H= 890. kj mol - C+ H CH M =? eq (I) + eq (II) + eq (III). gram of a carbonate (M CO ) on treatment with excess HCl produces mole of CO. The molar mass of M CO in g mol is : () 8.6 ().86 () 86 () 8. MCO + HCl MCl+ CO+ HO gm.086 mole means.58 gm.58 gm - gm gm CO - 8.gm.58 =. U is equal to : () Adiabatic work () Isothermal work () Isochoric work () Isobaric work. Adiabatic work For Adiabatic u= w Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

12 . The Tyndall effect is observed only when following conditions are satisfied : (a) The diameter of the dispersed particles is much smaller than the wavelength of the light used. (b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude. (d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. () (a) and (c) () (b)and(c) () (a)and(d) () (b) and (d) both b & d correct 5. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be : Ans.: () () a () For fcc d = a a () a () a 6. Given E =.6 V E o o, Cl /Cl Cr /Cr o + = 0.7 V, E Among the following, the strongest reducing agent is : o + =. V, + = CrO 7 /Cr MnO /Mn E.5V () Cr + () Cl () Cr () Mn + The strongest reducing agent is Cr 7. The freezing point of benzene decreases by 0.5 C when 0. g of acetic acid is added to 0 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (K f for benzene = 5. K kg mol ) () 7.6% () 9.6% () 6.6% () 80.% Ans.: () Tf = i Kfm = i i =.057 i for association = n.057 = =.95=9.5% 8. The radius of the second Bohr orbit for hydrogen atom is : Ans.: () (Planck's Const h = Js; mass of electron = kg; charge of electron e = C; permittivity of vacuum ε 0 = l0 kg m A ) () 0.59 Å (). Å ().65 Å ().76 Å nh r= π mze.59x.59() r = Aº = =.Aº z Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

13 9. Two reactions R and R have identical pre-exponential factors. Activation energy of R, exceeds that of R by 0 kj mol. If k and k are rate constants for reactions R and R respectively at 00 K, then ln(k /k ) is equal to : Ans.: () (R = 8.Jmol l K l ) () 6 () () 8 () logk = loga 0+ E log k = log A RT Ea it is given E = 0 + E (0 0 ) RT 0 E log k = log A RT RT... (I) E log k = log A... (II) RT E 0 E logk logk = loga loga+ + RT RT RT K K RT log = = = 0. pk a of a weak acid (HA) and pk b, of a weak base (BOH) are. and., respectively. The ph of their salt (AB) solution is : () 7.0 ().0 () 7. () 6.9 For weak acid & weak base ph = pkw+ pka pkb = () + (.) (.) ph = 6.9. Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect, is : Ans.: () both form nitrides () nitrates of both Li and Mg yield NO and O on heating () both form basic carbonates () both form soluble bicarbonates () both form basic carbonates Li & mg not form basic carbonates. Which of the following species is not paramagnetic? () O () B () NO () CO CO Total No. of el. in CO are & all are paired. So it is diamagnetic.. Which of the following reactions is an example of a redox reaction? () XeF 6 + H O XeOF + HF () XeF 6 + H O XeO F + HF () XeF + O F XeF 6 + O () XeF + PF 5 [XeF] + PF6. XeF + O F XeF 6 + O Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

14 . A water sample has ppm level concentration of following anions F = 0; SO = 00; NO =50, The anion/anions that make/makes the water sample unsuitable for drinking is/are : () only F () only SO () only NO () both SO and NO only F.5 ppm of F - causes tooth decay and more concentration causes poison effect. 5. The group having isoelectronic species is: () O, F, Na, Mg + () O, F, Na +, Mg + () O, F, Na +, Mg + () O, F, Na, Mg + O, F, Na +, Mg + + O,F,Na &mg + all have 0 electronic 6. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are: () Cl and ClO () Cl and ClO Cl and ClO NaOH+ Cl NaCl+ NaClO+ HO hence Cl & ClO. () ClO and ClO () ClO and ClO. 7. In the following reactions, respectively acting as a/an : Ans.: (a) ZnO + Na O Na ZnO (b) ZnO + CO ZnCO () acid and acid () acid and base () base and acid () base and base () acid and base ZnO is amphoteric oxide. 8. Sodium salt of an organic acid 'X' produces effervescence with conc. H SO. 'X' reacts with the acidified aqueous CaCl solution to give a white precipitate which decolourises acidic solution of KMnO. 'X' is: () CH COONa () Na C O () C 6 H 5 COONa () HCOONa Ans.: () Na C O COONa COONa HSO CO + NaSO + H O CaCl COO Ca COO COOH COOH KMnO H + CO + H O 9. The most abundant elements, by mass in the body of a healthy human adult are : Ans.: Oxygen (6.%); Carbon (.9%), Hydrogen (0.0%); and Nitrogen (.6%). The weight which a 75 kg person would gain if all H atoms are replaced by H atoms is: () 7.5 kg () 0 kg () 5 kg () 7.5 kg () 7.5 kg for H 0% Total mass of body 75 kg H = 7.5 kg. H = 7.5 kg H = 5 kg. Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

15 50. On treatment of 00 ml of 0. M solution of CoCl.6H O with excess AgNO ;. 0 ions are precipitated. The complex is : Ans.: () () [Co(H O) 6 ]Cl () [Co(H O) 5 Cl]Cl.H O () [Co(H O) Cl ]C.H O () [Co(H O) Cl ].H O mole 000 M = vol.in ml mole 000. = 00 mole =.0 mole of CoCl 6 H O for AgNO ions - mole. 0 ions -.0 mole -.0 mole mole =.0 = mole of Cl will ppt.0 =.0 mole 5. Which of the following compounds will form significant amount of meta product during mono-nitration reaction? () () () () NH group is O & P directing but in this reaction most of the aniline convert into Anilinium ion (NH + ) which is meta directing and about 7% m - Nitro compound is formed. 5. Which of the following, upon treatment with tert-buona followed by addition of bromine water, fails to decolourize the colour of bromine? () () () () Here is no β 'H' No double bond formation in presence of base Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

16 5. The formation of which of the following polymers involves hydrolysis reaction? () Nylon 6, 6 () Terylene () Nylon 6 () Bakelite CH O C N HOH HO O C (CH ) 5 NH CH CH CH CH 5. Which of the following molecules is least resonance stabilized? Ans.: () () () () () Aromaticity is not present in structure 55. The increasing order of the reactivity of the following halides for the S N l reaction is: () (I) < (III) < (II) () (II) < (III) < (I) () (III) < (II) < (I) () (II) < (I) < (III) 56. The major product obtained in the following reaction is : () (+) C 6 H 5 CH(O t Bu)CH C 6 H 5 () ( ) C 6 H 5 CH(O t Bu)CH C 6 H 5 () (±) C 6 H 5 CH(O t Bu)CH C 6 H 5 () (+) C 6 H 5 CH=CHC 6 H 5 Br CH 6 5 C CH CH 6 5 OH H CH 6 5 C OHC H 6 5 CHCH 6 5 = CH CH 6 5 H t. BuOK H 57. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution? () () Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

17 () () 58. -Methyl-pent--ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is : () Two () Four () Six () Zero CH CH C Or CH H C C CH CH CH CH CH HBr 5 HO CH H Br CH C C * * H CH 5 two chiral carbon atoms are present in the product hence n = 59. The correct sequence of reagents for the following conversion will be : () CH MgBr, [Ag(NH ) ] + OH, H + /CH OH () [Ag(NH ) ] + OH, CH MgBr, H + /CH OH () [Ag(NH ) ] + OH, H + /CH OH, CH MgBr () CH MgBr, H + /CH OH, [Ag(NH ) ] + OH. O O [ Ag(NH) ] + OH + H + /CH OH CHO O HO CH COOH CHHgBr COOCH HO CH CH Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

18 60. The major product obtained in the following reaction is : () () () () O O OH DIBAL H CHO COOH CHO DlBAL reduces the ester into Aldehydes and acids into aldehyde CH Al CH HC H CH PART C - MATHEMATICS 6. The function f:r, defined as x f(x) = + x is: () injective but not surjective. () surjective but not injective. () neither injective nor surjective. () invertible. Ans.: () f:r, x f(x) = x R (, ) + x - ' (+ x). x.x [x+ ).(x ) f(x) = = (+ x) (+ x) (0, 0) So, f(x) is subjective but not injective - (-, ) Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

19 6. If, for a positive integer n, the quadratic equation, x(x + l) + (x + l)(x + ) ( x+ n (x ) + n) = 0n has two consecutive integral solutions, then n is equal to : () 9 () 0 () () nx + [(0+ ) + (+ ) + (+ )...(n ) + n)x sum of roots + [.+. (n )n) = 0 n [+ +...n ] + n α+β= n αβ=. n(n+ ) [ α β ) = (n ) n + n = n n n = n n(n )(n+ ) = 0n ( α β ) = ( α+β) α+β= n 0 n = n n + = n = n = 6. Let ω be a complex number such that ω + l = z where z=. If ω ω = k, then k is equal to: ω 7 ω () z () () () z ω+ = Z Z = = ω ω = k ω ω ω ω ω ( ω ω ) + ( ω +ω + ) = k ω ω ω+ω + +ω +ω = k ω+ ω = k w ω= k ( Z ) = k k = 7. Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

20 6. If A =, then adj (A + A) is equal to () 8 7 () 6 7 () 8 5 () 6 9 A =,A = A = A + A = adj ( A + A) 5 6 = = If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = l x + ay + z = ax + by + z = 0 has no solution, then S is : () an infinite set () a finite set containing two or more elements () a singleton () an empty set = a = 0 a b (a b) ( a) +.(b+ a) = 0 a= = = = 0 But a = and b = Then x + y + = there is no solution. x+ y+ = A man X has 7 friends, of them are ladies and are men. His wife Y also has 7 friends, of them are ladies and are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting ladies and men, so that friends of each of X and Y are in this party, is : () 68 () 69 () 8 () 85 X L M L M Y Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

21 c c c c = c c c c = + c c c c = + c c c c = 0 0 Ans The value of ( C 0 C ) + ( C 0 C ) + ( C 0 C ) + ( C 0 C ) +. + ( C 0 0 C 0 ) is: () 0 () 0 9 () 0 0 () ( C 0 ) ( 0 ) ( 0 C + C C +... C0 C0 ) ( 0 C C + C C0 ) C C +... C C... C 0 0 ( 0 ) For any three positive real numbers a, b and c, 9(5a + b ) + 5(c - ac) = 5b(a + c). Then : () b, c and a are in AP () a, b and c are in AP () a, b and c are in GP () b, c and a are in GP. ( ) ( ) ( ) 5a + b + 5c (5a)(5c) (5a)(b) b 5c = 0 (5a b) + (b 5c) + (5c 5a) = 0 c b = 5 a + b = c b, c, a an in A.P. c a = 69. Let a, b, c R. If f(x) = a + bx + c is such that a + b + c= and f(x + y) = f(x) + f(y) + xy, x, y R, then f(n) equal to : () 65 () 90 () 55 () 0 f(x) = ax + bx+ c, a+ b+ c = and f(x+ y) = f(x) + f(y) + x.y x, y R 0 n= a(x+ y) + b(x+ y) = ax + bx+ c+ ay + by+ c Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

22 Comparing coefficient of x 5 f(x) = + 0 n= n= x f(n) =? 0 n 5 +.n x,x and find 5 a = b =, C = ( ) + ( ) cotx cosx lim is equal to: π x ( π x) () /6 () /8 () / () / cotx cosx lim π ( x) π x R.H.L. approach π cot + h cos + h π lim π π + h h 0 tanh+ sinh lim 8h h 0 (tanh sinh) lim + 8h h 0 ( cosh sinh lim 8 h h 0 lim h 0 h h h 8 h sin sin cos h sin = h lim h If for x 0, the derivative of 6x x tan 9x is () x x 9x 6x x y = tan 9x x () 9x x 0, () + 9x x.g(x), then g(x) is: 9 () + 9x Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

23 .(x ) = tan = tan.x x As.x 0, 8 dy = x x dx + 9x 9 =. + 9x g(x) 9 = + 9x. x 7. The normal to the curve y(x )(x ) = x + 6 at the point where the curve intersects the y-axis passes through the point: (), (), (), Normal to the curve at point P intersect the y-axis x = 0 y(0 )(0 ) = 0+ 6 y = point (0, ) dy dx dy dx dy dx P(0,) P(0,) p = (x 5x+ 6) + y(x 5) = (6) 5) = Equation of normal at point (0,) (y ) = (x 0) dy dx (y ) = x Ans. () p Satisfy the point in equation of normal x = and =. y = (), 7. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is : Ans.: () () 0 () 5 () 0 ().5 Total length = 0 r + r + rθ= 0 θ= 0 r r Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

24 Area = = 0r r 0 r r θ= r r d = 0 0 r = 0 dr r = 5 = 50 5 = 5. r θ 7. Let In = n n I = tan xdx, (n > ). If I + I 6 = a tan 5 x + bx 5 + C, where C is a constant of integration, then the ordered pair (a, b) is equal to : (),0 5 I + I 6 = tan xdx,(n> ) (), 5 6 tan xdx+ tan x.dx = + ( + ) = Put I + I 6 = tan xdx tan x dx sec x.tan xdx tanx= tsec x.dx= dt 5 t t dt = + c 5 5 tan x I + I 6 = + c 5 a =,b = 0. 5 (),0 5 (), The integral π dx is equal to cosx π + () () () () I = π dx.() + cosx π π π x + x I = I = π π π π dx + cos( π x) dx.() cosx Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

25 π dx dx I = + + cosx cosx I π π = π sin x dx π I x cosec x dx = I= π [ ] cotx π π π I= cot cot I =+. π 76. The area (in sq. units) of the region {(x, y) : x 0, x + y, x y and y + x } is: Area = () / () 7/ () 5/ () 59/ x (+ x)dx + ( x)dx dx x+y= (,0) (,) (,) y=+öx x= y Area = 5 (,0) (,0) π + sinx + (y+ )cosx = 0 and y(0) =, then y is equal to: dx 77. If ( ) dy () / () / () / () / dy (+ sinx) + (y+ )cosx = 0 dx dy (+ sinx) = (y+ )cosx dx dy cosx = dx (y+ ) + sinx log(y+ ) = log[ + sinx) + c Find c y(0) = n = y log(+ ) = log() + c log() = c π Again y = find"y" log(y+ ) = log() + log y+ = y =. Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

26 78. Let k be an integer such that the triangle with vertices (k, k), (5, k) and ( k, ) has area 8 sq. units. Then the orthocenter of this triangle is at the point: (), k k 5 k = 8 k 5k + k 6= 0 k = and = k 5 K is integer k = Orthocentre, (), (-5,)B (), H k, A (,-6) D C (-,) (), 79. The radius of a circle, having minimum area, which touches the curve y = x and the lines, y = x is : Ans.: () () ( ) () ( ) () ( + ) () ( + ) Given curves are y = x and y = x There are two circle satisfy the ratio of circle is r Length = (0, - r) Circle touch the line y = x in first quadrant ( r) = r r = ( ). 80. The eccentricity of an ellipse whose centre is at the origin is (/). If one of its directrices is x =, then the equation of the normal to it at () x y = l () x + y = 7 () x + y = () y x = Directrix of ellipse a = x = a = e e ae = a b = b x a y + = b dy = dx m = P y = x y x x y dy + = 0 dx e = a = = b dy x = dx y b = Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

27 Equation of normal at, y = (x ) x y = 8. A hyperbola passes through the point P(, )has foci at (±, 0). Then the tangent to this hyperbola at P also passes through, the point: () (, ) () (, ) () (, ) () (, ) Equation of hyperbola x a y = b = a b = a a 8 a a = a a a 9a + 8 = 0 a 8a a + 8 = 0 Equation of tangent at P (, ) xx a yy = b x y = a b And has foci ( ±,0) Find a & b b e = + a (ae) = a + b = a + b () Put a = 8 = 8+ b b a = = b b = x y = Equation of tangent y x = At point (, ) satisfying the equation of tangent Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

28 8. The distance of the point (,, 7) from the plane passing through the point (,, ), having normal perpendicular to both the lines x y + z = = and x y + z + = = 7 () 0 8 () a(x ) + b(y+ ) + c(z+ ) = 0 a b + c = 0 a b c = 0 a b c = = + 5 b + a b c = = =λ 7 7 a = 7λ b = 7λ & 5 8 7( ) + 7(+ ) + ( 7+ ) PQ = PQ = 0 8 Q () 0 7 P (,,-7) () 0 7 7(x- ) + 7(y+ ) + (z+ ) = 0 c = λ 8. If the image of the point P(l,, ) in the plane, x + y z + = 0 measured parallel to the line x = y = z is 5 Q then PQ is equal to : () () () 6 5 () 5 Equation of PQ line x y + z = = = 5 R point line in the plane x y z = = and b = i + j + 5k 5 ( λ+ ) + (λ ) (5λ+ ) + = 0 λ= R(,,8) distance between P & R is ( ) + () + (5) = PQ = PR PQ = 8. Let a = i + j k and b = i+. j Let c be a vector such that c a = and a b be 0. Then a.c is equal to : a b c = and the angle between c, ( ) () () 5 () /8 () 5/8 a = i + j k and b =+ i j c a =, (a b) c) = a b. c.sin θ.n = x+y-z+=0 P (,-,) Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0 Q

29 i j k a b = 0 a b = i j + k a b = 9 = a b. c.sin θ.n = a b. c.sin0.n = c = c = c a = c a = 9 (c a).(c a). = 9 c.c a.c c.a+ a.a= 9 c a.c + a = 9 + 9= 9+ a.c a.c = 85. A box contains 5 green and 0 yellow balls. If 0 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is : () 6 () () 6/5 () /5 5 G 0 Y 0 balls one by one i i Variance = x.p ( xipi) 5 5 = = = For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = (/) and P(All the three events occur simultaneously) = (/6) Then the probability that at least one of the events occurs, is : () 7/6 () 7/6 () /6 () 7/ Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

30 P(exactly one of A or B occurs) = P(A) + P(B) P(A B) = P(exactly one of B or C occurs) = P(B) + P(C) P(B C) = P(exactly one of C or A occurs) = P(C) + P(A) P(C A) = Add these all P(A) P(A B) = (P(A)) P(A B) = 8 Now P(A B C) = (Given) 6 P(A B C) = P(A) (P(A B) + P(A B C) 7 = + = If two different numbers are taken from the set {0,,,,..., 0); then the probability that their sum as well as absolute difference are both multiple of, is: () /55 () /5 () 7/55 () 6/55 A= { 0,,,,...0} n(s) = C when's'denote sample space F = {(0,),(0,8),(,6),(,0),(,8),(6,0)} 6 P(E) = If 5(tan x - cos x) = cos x + 9, then the value of cos x is : () / () /9 () (7/9) () (/5) 5(tan x cos x) = cos x+ 5 5(tan x cos + x) = cos x + 5(sec x cos x) = cos x + = (cos x ) + 5sec x 5cos = cos x+ 5sec x 9cos = 5. + cos x 9 = + cos x 0 9(+ cos x) = (+ cos x) 0 9(+ cos x+ cos x) = + cos x 9cos x 8cos x= + cos x 9cos x+ cos x= = 0 Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0

31 9cos x+ cosx+ 9cos x+ (cos x+ )(cosx+ ) = 0 cos x = 7 = =. 9 9 cosx = cos x 89. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = AB. If BPC = β, then tan β is equal to: Ans.: () () / () /9 () /9 () 6/7 PCA x tanα= = x PBA tan( α ) + tan( β) tan( α+β ) = tan α.tanβ tanα+ tanβ + tanβ = tanβ.tanβ.tanβ 9 = tanβ= The following statement (p q) [(~ p q) q] is: () equivalent to ~p q () equivalent to p ~q () a fallacy () a tautology. hence Tautology Add. R-, Opp. Railway Track, Zone-II, M.P.Nagar, Bhopal (M.P.) Pin-60 Phone No.: , Mob.No.: , 0