SOLUTION. Total number of ways of forming a term if two specific boys B 1, B 2 always join the team

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1 Mathematics 1. 7B 5G } (3B G ) SOLUTION Total number of ways of forming a team of 3 boys and girls = 7 C 3 5 C = = 350 Total number of ways of forming a term if two specific boys B 1, B always join the team = 5 C 1 5 C = 50 So, the total numbers of ways of forming a team is two specific boys never come together = = 300. x + x + = 0 < β α α, β = ± 4 8 = 1 ± i α, β = ( 1 ± i 1 ) = (cos 3π 4 ± isin 3π 4 ) α 15 + β 15 = [ (cos 3π 4 = 14 [(cos 45π 4 + isin 3π 4 )]15 + [ (cos 3π 4 isin 3π 4 )]15 45π 45π 45π + isin ) + (cos isin )] =. 15. cos 45π = cos 5π = = 56 π 0 3. I = cosx π 3 dx = cos 3 x 0 π = (1 sin x) cosx dx 0 π = [ cosxdx 0 = [sin sin3 x dx π sin xcosx dx π 3 ] 0 0 ] = [ ] = 4 3

2 4. I = x sin(x 1) sin(x 1) sin(x 1)+sin(x 1) dx Let x 1 = θ, x dx = 1 dθ I = 1 sinθ sinθ sinθ+sinθ dθ = 1 sinθ sinθcosθ sinθ+sinθcosθ dθ = 1 1 cosθ 1+cosθ dθ = 1 tan θ dθ = 1 ln (sec θ ) ( 1 ) + c = ln sec ( x 1 ) + c 5. Since a. c = 0 Angle θ between a c is π Now a c + b = 0 a c = b a c = b a c sinθ = b c = b a sin π = = 3 c = 3 5: x 1 a + bx: 1 < x < 3 6. f(x) = { b + 5x: 3 x < 5 30: x 5 For f(x) to be continuous at x = 5 lim x 5 f(x) = lim +f(x) = f(5) x 5 lim x 5 (b + 5x) = 30 b + 5 = 30 b = 5 For f(x) to be continuous at x = 1

3 lim x 1 + f(x) = lim + f(x) = f(1) x 1 lim x 1 (a + 5x) = 5 lim x 1 (a + 5x) = 5 Now at x = 3, LHL = lim x 3 (0 + 5x) = 15 RHL = lim x 3 (5 + 5x) = 18 f(x) is discontinuous for all a, b R 7. Let 1 st 5 students are x 1, x, x 3, x 4 and x 5 18 = E(x ) (E(x)) = x 1 +x +x 3 +x 4 +x 5 5 (150) x 1 + x + x 3 + x 4 + x 5 = new variance = x 1 +x +...+x (151) [Since new mean = = = ] 8. Given a, b, c are in G. P let common ratio of G. P be r, then b = ar, c = ar Given equation can be written as a + ar + ar = x. a. r x = r + 1 r + 1 As we know r > 0, then r + 1 r x 3 r < 0, then r + 1 x 1 r x (, 1] [3, ) = 3. ( 4 ) 100 = 8. (15 + 1) 100 = 8[ 100 C C C C 100 ]

4 = 8. [15λ + 1] = 8.15λ = 8λ = k = To solve this questions we need to apply the concept of rationalization two times y 4 Given limit is lim y 0 y y 4 = lim 1+ 1+y4 + y 0 y y 4 + = lim ( 1+y4 1) ( 1+y4 +1) y 0 y 4 ( 1+ 1+y 4 ( 1+y + ) 4 +1) = lim y4 y 0 y 4 ( 1+ 1+y 4 + )( 1+y 4 +1) = lim y 0 1 ( 1+ 1+y 4 + )( 1+y 4 +1) = There is an ambiguity is this question. Two possibilities are there Case1: When vertex is an RHS of y axis Clearly in this case vertex is (, 0). Since focus is S(4,0) a =. Hence equation of parabola is y = 4.(x ) y = 8(x ) Point (6, 8) does not lie on the parabola Case when vertex is on LHS of y axis

5 Cleary a = 6 Equation of parabola y = 4(x + ). Since neither of the given points lies on this parabola This parabola is not considered here 1. Given complex number is z = 3+isinθ 1 isinθ z = (3+isinθ)(1+isinθ) (1+4sin θ) Since given complex number is purely imaginary Re(z) = 0 3 4sin θ = 0 sinθ = ± 3 θ = π 3, π 3, π 3 sum of all possible values of θ = π 3. cosθ sinθ 13. A = [ ] A sinθ = cosθ sinθ cosθ sinθ cosθ = cos θ + sin θ = 1 And adj(a) = [ cosθ sinθ sinθ cosθ ] A 1 = 1 cosθ adj(a) = [ sinθ A sinθ cosθ ] A = (A 1 ) = (A 1 )(A 1 ) cosθ sinθ cosθ sinθ = [ ] [ sinθ cosθ sinθ cosθ ] = [ cos θ sin θ sinθcosθ sinθcosθ cos θ sinθ ]

6 cosθ sinθ = [ sinθ cosθ ] It is visible that (A 1 ) n = [ cosnθ sinnθ sinnθ cosnθ ] A 50 = (A 1 ) 50 [ cos50θ sin50θ sin50θ cos50θ ] Now at θ = π 50π, sin50θ = sin = sin π = cos50θ = cos π 6 = 3 A 50 = [ ] 14. 1) (A B) (~A B) (B A) (~A B) B [A (~A B)] (using associate property) B (A B) A B ) = (A B) (~A B) (A ~A) B F B F 3) (A B) (~A B) (A ~A) B F B B 4) (A B) (~A B) [(A B) ~A] B (Using associate property) [B ~A] B ~A B 15. Given DE can be written as dy = x, which is a linear differential equation dx + y x If = e x dx = e ln x = x Solution is y. x = x. x dx

7 y. x = x4 4 + c Since given curve passes through (1, 1) c = 3 4 Hence y(x) = x x So y ( 1 ) = Using the concept of complementary events, P(x = 1) + P(x ) = 1 P(x = 0) = = = For given hyperbola the eccentricity is given as e = 1 + sin θ cos θ = 1 + tan θsec θ e = secθ length of latus rectum l = sin θ cosθ l = (e 1) ee = (e 1 e ) dl de = (1 + 1 e ) > 0 l is an increasing function l min = ( 1 ) = 3 range of latus rectum is (3, ) = tan θ secθ 18. Given l = 3 r + h = g Volume, V = 1 3 πr h V = 1 3 π(9 h )h = 1 3 π(9h h3 )

8 dv dh = 1 3 π(9 3h ) dv dh = 0 h = 3 d V dh = 1 3 π( 6h) at h = 3, cone has maximum volume V max = 1 π( ) = 3πcm x > 3 4 cos 1 3 = 4x sin 1 16x 9 4x Given cos 1 3x + cos 1 3 4x = π cos 1 3x + cos 1 3 4x = π + 16x 9 3x 4x ( 8 3 ) = 16x 9 16x = = x = x = Given condition is 3p + q + 4r = 0 3 p + 1 q + r = 0 (i) 4 And given family of line is px + qy + r = 0... (ii) From (i) we can say that equation (ii) always passes through ( 3 4, 1 ). Hence all lines are concurrent at ( 3 4, 1 ) Here, D = 3 3 a 1 = a 3 Hence at a = 3, D = 0. So system has no unique solution. At a = 4 and 3, D 0, So, system has unique solution. Hence net inconsistent.

9 At a = 3, D = 0 and nd & 3 rd equation are parallel. Hence no solution are possible. Hence, system is inconsistence for a = 3. 3(cosθ sinθ) 4 + 6(cosθ + sinθ) + 4sin 6 θ = 3((cosθ sinθ) ) + 6(cosθ + sinθ) + 4sin 6 θ = 3(cos θ + sin θ sinθcosθ) + 6(cos θ + sin θ + sinθcosθ) + 4sin 6 θ = 3(1 sinθcosθ) + 6(1 + sinθcosθ) + 4sin 6 θ = 3(1 4sinθcosθ + 4sin θcos θ) + 6(1 + sinθcosθ) + 4sin 6 θ = 9 + 1sin θcos θ + 4sin 6 θ = 9 + 1(1 cos θ)cos θ + 4(1 cos θ) 3 = 9 + 1cos θ 1cos 4 θ + 4(1 3cos θ + 3cos θ cos θ) = 13 4cos 6 θ 3. AB = (a + b) (b a) = ab BC = (a + c) (c a) = ac AC = (b + c) (c b) = bc AC = AB + BC bc = ab + ac 1 a = 1 b = 1 c 4. Given f (J(f 1 (x))) = f 3 (x) f (J ( 1 x )) = 1 1 x 1 J ( 1 x ) = 1 1 x J ( 1 x ) = x = x 1 x

10 J(x) = 1 x 1 1 x = 1 x 1 = 1 1 x J(x) = f 3 (x) 5. A variable point on line x + y z = 0 = x + y 3z + 5 is ( t + 5, t 5, t) DR of variable point & ( 4, 1, 3) is ( t + g, t 6, t 3) Since line is parallel to x + y + z = 3 t + g + t 6 + t 3 = 0 t = 0 DR of line s (9, 6, 3) or ( 3,, 1) Equation of line is x+4 3 = y 1 = z Intersection points of given curves are (±, 6) for 1 st curve dy dx = x m 1 = 4 or ( 4) For nd curve, dy dx = x m = 4 or (4) tanθ = 4 ( 4) 4 4 or 1+4( 4) 1+( 4)4 = 8 15 (In both cases) 7. Let required plane is (using concept of family of plane) x + y + z 1 + t(x + 3y z 4) = 0 (1 + t)x + (1 + 3t)y + (1 t)z 1 4t = 0 Again DR, of y axis is (0, 1, 0) Since plane is parallel to y axis Hence (1 + t) 0 + (1 + 3t) 1 + (1 t) 0 = 0 t = 1 3 required plane is 1 x + 4 z + 1 = x + 4z + 1 = 0 Which satisfy (3, 1, 1)

11 8. Let tangent to parabola y = 4x is y = mx + 1 m... (i) If equation (i) is tangent to give circle whose centre is (3, 0) and radius is 3 then length of perpendicular from centre of circle to equation (i) is equal to radius of circle. Hence, 3m+ 1 m m +1 = 3 3m + 1 = 3m 1 + m 9m 4 + 6m + 1 = 9m + 9m 4 m = ± 1 3 common tangents are y = x or y = x dy dx = x slope of tangent at (, 3) m = 4 equation of tangent is y 3 = 4(x ) point T (0, 5) area of APTM = 1 8 = 8 Area of curve PAMP = = 3 [(y + 1)3 ] 1 3 = y + 1dy Required shaded area = = 8 3

12 Physics 1. Block will be at rest it mgsinθ + 3N = P + f = P + μmgcosθ ( To get P minimum friction must be minimum) P min = = = 3N. For path ABC, ΔQ = ΔU + ΔW 60 = ΔU + 30 ΔU = 30J For path ADC ΔQ = ΔU + ΔW As initial and find points are same, change in internal energy is same in both the cases ΔQ = = 40J 3. From given data 1 v 1 u = 1 f 1 1 = 1 10 ( 10) f f = 5cm

13 When glass plate is placed Shift s = t (1 1 μ ) = 1.5 (1 3 ) = 0.5 cm Shift will be in the direction of incoming rays u = = v 1 u = 1 f 1 = v 5 ( 9.5) v = Shift = = = Angular momentum L = mr w Area of sector, da = 1 r dθ Areas velocity da = 1 dθ r dt dt (i) Keplers laws 5. v = yi + xj v x = y = dx dt v y = x = dy dt dx dy = y x x dx = y dy x = y + C or y = x + constant 6. When block attains maximum speed F = Kx Or x = F K From work energy theorem

14 1 (K F ) + F ( F ) = 1 MV K max F K = 1 MV max V max = F KM 7. Magnetic field due to AD and BC are zero at O. Magnetic field due to AB B AB = μ 0 10 π 4π (outwards) Simplifying field due to DC B DC = μ 0 10 π 4π (inwards) 10π Net field B = μ o 4π 4 10 (1 1 ) 3 5 = T (outwards) 8. Electric field at a distance h from centre on axis E = 1 Qh 4πε 0 (R +h ) 3

15 To get maximum E, de dh = 0 0 = 1 3 Q(R +h ) Qh 3 (R +h 1 ) h 4πε 0 (R +h ) 3 h = ± R 9. Activity A = λn 10 = λ A N A 0 = λ B N B 1 = λ A λ B λ A λ B = B = 0.4 sin(50πt), A = m ε = dφ dt dq = ε R dt = 1 R dφ t = 10 ms Q = 1 R = 140μc t=10ms t= sin(50πt)dt 11. i = neav d 1.5 = v d v α = 0.0 mm s 1. T 1 T = 10 C We define thermal resistance R = L KA Given circuit can be reduced to

16 Equivalent resistance R eq = 8R 5 Thermal current i = T 1 T 8R 5 T A T B = T 1 T ( 8R 5 ) ( 3R 5 ) = = mv = (m + m)v (m + m)v = (m + m + M)v f final velocity v f = Initial energy = 1 mv m m+m v Final energy = 1 mv (m + M) ( m+m ) 1 Given that 5 6 mv + 1 mv mv = 1 6m = m + M M m = 4 m v (m+m) (m + M) ( mv m+m ) 14. (V RMS ) He (V RMS ) Ar = M Ar M He = 40 4 = λ 1 = 340nm, λ = 540 nm, V 1 V = hc λ 1 = φ + 1 m(v)

17 hc λ = φ + 1 mv hc hc 4 = φ 4φ λ 1 λ hc ( 4 540nm 1 340nm ) = 3φ φ = 1.85eV 16. Apply KCL i + i + i 1 = 0 V + V 10 + V 0 = 0 4 V + V 10 + (V 0) = 0 5V 50 = 0 V = 10V i 1 = 10 = 5A 17. Net force on charge at O is zero KQq ( d ) + KQ d = 0 q = Q 4

18 18. As we know E = C B E = V m B = E C = B = T 19. U = P. B F = U db = P r r dr F = i (πa ) d dr (μ 0i 1 πr ) F = i 1i π 0 a F ( a r ) As r = d F a d 1 r 0. I max = ( I 1 + I ) I min = ( I 1 I ) I max = ( I 1+ I ) I min ( I 1 I ) 16 = ( I 1+ I ) 1 ( I 1 I ) 4 1 = I 1+ I I 1 I I 1 = 5 I 9 1. V = A. e O = da A + dl l da A = dl l R = l A If dr is change in resistance due to change in l and A, then dr R =. dl A = dl e da A

19 % charge in R = 0.5% = 1%. At equilibrium, the centre of mass will be below the point of suspension. Taking torque about point O, mg l sinθ + mg (lsinθ l cosθ) = 0 3l sinθ = l cosθ tanθ = 1 3 θ = tan Since the length of the rod does not change, the increment due to increase in temperature will be balanced by the compression due to applied force. Y = F A Δl l Also Δl l F AY Y = Δl l = αδt = α ΔT F A.αΔT = F A.Y 4. Consider a small element at a distance x. It is a two capacities in series. dc = dc 1+dc dc 1 +dc

20 dc = dc = C = ε 0.adx xtanθ k +d xtanθ 1 ε 0 K.adx xtanθ(k 11)+d.k 0 a ε 0 k.adx xtanθ(k 1)+d.k = ε 0k.a atanθ(k 1) +d.k ln tanθ(k 1) d.k = ε 0ka ln k d(k 1) 5. ρ = E J = E neυ d = 1 neμ μ = mobility ρ = 1 = 1 Ω m = 0.4Ω m. 6. For standing car For accelerating car υ = (Mg) +(Ma) μ 60.5 = (Mg) +(Ma) 60 Mg ( ) = (Mg) +(Ma) Mg 1 + a = g 1 = + 1 a g a = g 4 10 = g 30

21 Chemistry 1. k b values are a measure of basicity. Higher the k b value, greater will be basicity. Also basicity implies the tendency to donate electrons Basic character α 1 % S character Nsp > Nsp 3 (% s character=more%s-character more will be electro negativity ) Than Nsp has more % s character than Nsp 3 is less basic as compare to Nsp 3 So R > p. In Q lone pair involved in Aromaticity it so difficult to donate as it is very less basic Thus, the correct order will be: R > P > Q. The first step is anti addition of Br across the double bond. In the second step, OEt nu ) gets added by SN 1 mechanism. The Br in dashed position acts as the having group as this will add to the formation of a stable benzylic carbocation, where OEt gets added. 3. Acidic character α stability of conjugate base (anion) As stability of an α M and I Cl and -F have ( I > +M effect) in which F is more electron withdrawing due to greater electronegativity. NO and CN have both I and M effect with NO being more electron withdrawing. Order of -I-effect NO >CN>-F>-Cl Hence, order of acidity (k a ) will be O NCH COOH > NC CH COOH > FCH COOH > Cl CH COOH (R) (S) (P) (Q) 4. The concentration of the ions for drinking water are Fe 0. ppm Mn 0.05 ppm Cu 3.0 ppm Zn 5.0 ppm Hence, only Mn + concentration is greater than allowed limit.

22 5. Acidic character α stability of conjugate base (anion) As stability of α M and I Electron withdrawing groups increase acidic strength. Among the given compounds, the electron withdrawing strength of substituents is CN > Cl > Br > I Hence, CH (CN) 3 will be most acidic 6. Chloroxylenol is a phenol derivative and hence can be identified with FeCl 3 test. The primary amine of sulphapyridine can be identified with earbylamine test. The aliphatic alkyne of Nore thindrone can be identified by the Baeyer s reagent. 7. The final product is an isocyanide as the AgCN bond has covalent nature, due to which C is unavailable for bond. Hence, N bonds with the benzylic carbon leading to formation of isocyanide. 8. In the first step, R CN gets reduced to RCH = NH which then gets hydrolyzed to RCH = O

23 9. Hence, the correct order is Arg > Lys > Gly > Asp

24 1. Moles of Na + = Weight of Na+ ions Molecular weight = 9 3 = 4 4, mole of Na + present in 1 kg of water Hence molarity Na + ions will be Ba (NO 3 ) does not have any water of crystallization. This happens because the large size of Ba leads to poor polarizing power and hence, if unit able to attract water molecules effectively. 14. According to the Freundlich adsorption isotherm, x m = Kp1 n Taking log on bot sides. log ( x ) = logk + 1 log P m n Slope = 1 n Slope from graph = y y 1 x x 1 = 4 = 1 or, 1 n =1 x m = Kp(1 ) or x m p(1 ) 15. Iron and copper are both present in Copper pyrite CuFeS Malachite CuCO 3. Cu(OH) Azurite CuCO 3. Cu(OH)

25 16. Millimoles of H SO 4 = = Millimoles of NH 4 OH = = 6 Hence, the solution will act as a basic buffer. poh = pk b + log [salt] [Base] [NH + ] = [(NH 4 ) SO 4 ] = 4 poh = pk b + log [NH 4 + ] [NH 3 ] = log 4 = = ph = 14 poh~9 17. According to molecular orbital theory Li + has electrons σ1s σ 1s σs 1 Li has 7 electrons σ1s σ 1s σs σ s 1 bond order = electron in bonding molecular orbital Electron in Antibonding molecular orbital Bond order of Li + = 3 = 1 Bond order of Li = 4 3 = 1 As both have similar bond order, both Li + and Li will be stable. Note: Li will be relatively unstable to Li + as it has more electrons in anti bonding orbitals, so have more energy and having more repulsion so Li is less stable as Li Henry s constant increases with increase in temperature while it decrease with increase in solubility. Hence value of K H increases as solubility of gas increases is a wrong statement. 19. Let Assume x & y are the order of reaction w,r,t of A & B respectively From equation (1) & () From equation (1) & (3) and put the value y = 0

26 the rate law will be R = K[A] 1 [B] 0 Hence, the reaction is of first order. K = R = = [A] 0.1 t1 for first order reaction is given by t1 = K = = In a group decreased down and in a group increase. (i) Electronegativity: Top to bottom decrease (radii α n so generally top to bottom increases) (ii) Electronegativity: - Top to bottom decreases Electron affinity: - generally top to bottom decreases not increases (iii) radii: - rαn top to bottom generally increases Electronegativity: - top to bottom increases (iv) Electron gain enthalpy: - top to bottom decreases Electronegativity: - Top to bottom decreases 1. W = nrt ln v f v i W = nrt lnv f nrt v i Intercept will be ve. (A) True. ( 0 ) A < ( 0 ) B, both complex have CN = 6; B has strong field ligand and A has weak field ligand so ( 0 ) A < ( 0 ) B (B) True. By the crystal field splitting parameter can be measured by wavelength of complementary colours for A and B respectively (C) True. In both complexes, Cr has 3 unpaired electrons and hence, will be paramagnetic [Cr(H O)]Cl 3

27 Cr +3 = 1S S p 6 3S 3p 6 3d 3 Cr +3 m in strong field ligand or weak field ligand t g 111, eg L = ligand = H O NH 3 Hybridization is d Sp 3 and Having 3 top lame pairs (D) False. The crystal field splitting cannot be measured by wavelength of yellow and violet colours for A & B respectively. 3. Silicones with medium length chains behave as viscous oils, jellies and greases. They are chemically inert, i.e., resistant to oxidation, thermal decomposition or to attach by organic reagents. Since silicones are surrounded by non-polar alkyl groups, they are hydrophobic by nature. Being biocompatible, they are used in surgical and cosmetic implants. Hence, all options are correct. 4. Change of one mole of e = 1faraday Pb(s) + SO 4 PbSO 4 + e mole of e involved the above reaction = faraday mole of electron = faraday current pass than PbSO 4 deposited = 303 gm mol 0.05 F faraday current pass than PbSO 4 deposited = = Piezoelectric material are those that produce an electric current when they are placed under mechanical stress. It was first discovered in 1880 by two French Physicists, brother Pierre and Paul- Jacques curie, in crystals of quartz, tourmaline and Rochelle salt.

28 6. Isotopes have same atomic number and different atomic mass ie. Same number of proton & different number of Neutron relative atomic mass Protein 1H 1 = H Deuterium 1H = D Trillian 1H 3 = T According to Rydberg s equation V = R H Z ( 1 n 1 1 n ) = R H 1 ( 1 n f 1 8 ) V = R H n R H f 64 Hence, slope for V & = 1 n f is + R H 8. The sum of the first three ionization enthalpies of at considerably decreases and is therefore after to form at +3 ions. However down the group due to poor shielding, effective nuclear charge holds ns electron tightly (responsible for in emf pair effect) & thereby, restricting results of this only P-orbital electrons may be involved in banding so till show + 1 & + 3 oxidation states. 9. Maximum number of unpaired electrons for transition metals = 5 or n = 5 Spin only magnetic moment = n(n + )BM = 5(5 + ) = 35 = 5.9BM 30. According to ideal gas equation pv = nrt = (0.5 + x)r 1000 = 0.5R + xr x = 0.5R R = 4 R R

ChemGGuru. JEE Main Online Exam 2019 [Memory Based Paper] Questions & Answer. Morning 9 th January 2019 CHEMISTRY

ChemGGuru. JEE Main Online Exam 2019 [Memory Based Paper] Questions & Answer. Morning 9 th January 2019 CHEMISTRY CAREER PINT ChemGGuru JEE Main nline Exam 019 [Memory Based Paper] Questions & Answer Morning 9 th January 019 CHEMISTRY Q.1 What is the correct order of acidic-strength? N CH CH ( I) NC CH( CH II) F CH