SOLUTIONS. 1. Since each compete two times with each other, hence boys Vs Boys total plays =

Transcription

1 Mathematics SOLUTIONS 1. Since each compete two times with each other, hence boys Vs Boys total plays =. m c Boys Vs Girls total plays = 4m As per given condition. m c = m m 5m 84 = 0 m 1m + 7m 84 = 0 (m 1)(m + 7) = 0 m = 1, 7x Hence (A) is the correct answer.. Since, n C 4, n C 5, n C 6 are in A. P. n C 5 = n C 4 + n C 6 n! 5! (n 5)! = n! 4! (n 4)! + n! 6! (n 6)! 30 + (n 5)(n 4) = 5! (n 5)! 6! (n 4)! 6 (n 4) = 30 + n 9n + 0 n 1n + 98 = 0 n = 7 or As we know (P ) = P, hence given expression ( P ) = P 4. The given integral can be written as I = 3. x 3 + x 5 ( + 3x + x 4 ) 4 dx Let + 3x + x 4 = t 6x 3 4x 5 dx = dt (3x 3 + x 5 )dx = dt

2 I = 1 dt t 4 = 1. t = t 3 + C = 1 6. ( + 3x + x 4 ) 3 + C = x 1 6(x 4 + 3x + 1) 3 + C 5. Given series can be written as ( 3 4 )3 + ( 6 4 )3 + ( 9 4 )3 + ( 1 4 )3 + ( 15 4 ) terms = ( ) = (15 ) = = Since, given 5K = 7 5 K = 7 e 6. Let I 1 = ( x e )x x log e dx 1 Let ( x e )x = t ln x dx = 1 t dt 1 I 1 = t 1 t dt 1 e = t 1 1 e = 1 1 e e and I = ( e x )x log x e dx 1 Let ( e x )x = t ln x dx = 1 t dt 1 I = t. ( 1 t ) dt e = e 1 Hence required integral is I 1 I = ( 1 1 ) (e 1) = 3 e 1 e e

3 7. lim n + n n +1 n r=1 n n n n n +(n) = n n + r = 1 1 n 1 + ( r n ) r=1 1 = 1 + x dx = [tan 1 x] 1 0 = tan 1 8. Given h(x) = f(f(x)) h (x) = f (f(x)). f (x) h (x) = f(f(x)). f(x) (as f (x) = f(x).(i) Now f (x) = f(x) ln f(x) = x + c f(x) = k. e x f(x) = e. ex as f(1) = From equation (i), h (x) = e e( ex ). ( e. ex ) h (1) = e. e. = 4e Hence (A) is correct answer. 9. Given limit can be written as π sin 1 x L = lim π + sin 1 x x 1 1 x π + sin 1 x L = lim x 1 π sin 1 x 1 x ( π + sin 1 x) = lim cos 1 x. x 1 1 x ( π + sin 1 x) cos Let K = lim 1 x put x = cos θ, we get x 1 1 x θ K = lim. θ 0. sin θ = L =. π = π Hence (A) is correct answer.

4 10. As we know, if the line y = mx + c is a tangent to x = 8y, then c = m,, So equation of tangent with slope is y = mx m.(i) Since tangent is inclined at as θ, with the positive x axis, have m = tan θ. Equation of tangent is y = tanθ. x tan θ y cot θ = x tanθ Hence (A) is correct answer. 11. General term is 60 C r ( r 5) ( 310) 1 r We will get rational term when r is multiple of 10 and 60 r is multiple of 5 r = 0, 10, 0, 30, 40,50, 60 Number of irrational term = Total term - Number of rotational term = 61 7 = Let rest two observations are x & y x + y = 0 x + y = 9.(i) Variance, 5. = x +y (4) x + y = 65 (ii) from (i) & (ii), xy = 8 (x y) = x + y xy = = 49 x y = Given (x+)(x )(x 3) = 1 x =,, 3 n(a) = 3 Again, 3 < x 1 < 9 1 < x < 5 x = 0, 1,, 3, 4 n(b) = 5 n (A B) = 3 5 = 15 number of subset of A B = det(a) = 1(1 + sin θ) sin θ( sin θ + sin θ) + 1 (sin θ + 1) = + sin θ θ [ 3π 4, 5π 4 ] sin θ [0, 1 ] det(a) [, 3]

5 15. Let the line passing through ( 3, 4) intersect the coordinate axes at A & B. Clearly α+0 = 3, and β+ 0 = 4 α = 6 and β = 8 Hence equation of line is x 6 + y 8 = Since circle passing through origin intersect the coordinate axes at A & B, hence AB must be diameter and AB = r. Now, let foot of the perpendicular from origin upon AB be p(h, k). Equation AB is y k = h (x h) A +k k (h, 0) & B (0, h +k ) h k ( h + k h 0) + (0 h + k ) = r k (h + k ) ( 1 h + 1 k ) = 4r (h + k ) 3 = 4r h k Hence locus is (x + y ) 3 = 4r x y and (A) is correct option.

6 17. Given SBS = π OBS = π 4 b = ae.(i) Also, SBS = 8 1. ae. b = 8 aeb = 8.(ii) From (i) and (ii) b = 8 and a = 16 b = and a = 4 Length of the latus rectum b a =.8 4 = 4 Hence (C) is correct option. 18. Equation of line parallel to given line can be taken as y = x + c. Now this line touches the curve y = x 5x + 5, hence roots of the equation x 5x + 5 = x + c must be equal D = 0 4c = 9. Hence equation of tangent is 4y = 8x 9. Clearly it passes through ( 7, 1 4 ). Hence (C) is the correct answer. 19. a (b c ) = b (a. c )b (a. b )c = b b is not parallel to c a. c = 1 and a. b = 0 β = π 3 and α = π α β = π 6

7 0. cos (90 θ) = n 1.v n 1 θ sin θ = (i j kk ). (i + j k ) k k = k k = 5 + k k = ± NCC A, NSS B n(a) = 40, n(b) = 30, n(a B) = 0 n (A B) = n(a) + n(b) n(a B) = = 50 P (A B) = = 5 6 P (A B) = 1 P (A B) = = 1 6. At x = 1, f (x) = 0 f (x) = 3x 6(a )x + 3a 3 6 (a ) + 3a = 0 a = 5 f(x) 14 (x 1) = 0 x3 9x + 15x 7 (x 1) = 0 x 7 = 0 x = 7 3. dy dx + x y = x. It is in linear form I. F. = e x dx = e lnx = x y. x = x. x dx = x4 4 + c at x = 1, y =

8 1 = (1)4 4 + c c = 1 4 = 9 4 curve is y. x = x It passes through ( 3, 0). 4. AM GM sin 4 α + 4cos 4 β (4sin 4 α cos 4 β) 1 4 So AM = GM sin 4 α = 4cos 4 β = 1 sin 4 α = 1 α = π cos β = 1 β = π 4 Hence sinα sinβ = 1 1 = 5. Quadratic expression is positive, hence 1 + m > 0 and D < 0 m > 1 and 4(1 + 3m) 16 (1 + m)(1 + m) < 0 9m m 4(m + 3m + 1) < 0 m 6m 3 = 0 (m 3 + 3)(m 3 3) < 0 m (3 3, 3 + 3) m = 0, 1,, 3, 4, 5, 6 number of integral values = 7 6. tan 30 o = x 5 d

9 and tan 60 o = x+5 d tan 30o x 5 tan 60o = 1 = x 5 x+5 3 x+5 x + 5 = 3x 75 x = 100 x = 50m 7. C 1 (0, 0)& C (3, 4) r 1 = 9 & r = 4 C 1 C = 5 and r 1 r = 5 circles touch each other internally Z 1 Z min = 0 at the point of contact 8. P(success) = p(5 or 6) = 1 3 expectations equal to = 0 9. As we know, is four points P, A, B, C are ω planar, the vectors PA, PB, PC must be ω planar PA,. (PB PC ) = 0 λ 1 0 λ + 1 = 0 0 λ + 1 λ = ± 3

10 30. For non-trivial solution = 0 1 λ μ 1 = λ (1 λ)( λ + λ + 1) + 1( λ + 1) 1 ( 1 + λ) = 0 λ 3λ + 3λ 1 = 0 (λ 1) 3 = 0 λ 1 3 Singleton set Physics 1. + mg sin θ = μ mg cos θ mg sin θ + μ mg cos θ = 10 mg sin θ mg (sin θ μ cos θ) mg sin θ + μ cos θ = 10 = 1 5 (1 μ 3)5 = (1 + μ 3) 6 = 4 3 μ μ = 3. Let us take the equivalent circuit one by one.

11 C eq = C 7 3 C+ 7 = C = C C = 7 6 C = 7 11 μf. 3. y = 5 sin 3πt + 3 cos 3π t = 5 + ( 3) sin (3πt + Q) = 8 sin ( πt 3 Time period = 3 s + Q) Amplitude = 8 m 4. From momentum conservation mu + 0 = mv 1 + mv..(1) Using k energy conservation 1 mv = 1 mv mv.() Hence it retain 36%. Hence v 1 = 0.6v. On solving we get M = 4m.

12 5. When electron pass through the Hg vapor it loses some of its energy. The loss in KE of electron = = ( )eV = 4.9 ev. energy of radiation emitted = 4.9 ev. wavelength of radiation, λ = A 50 nm. 6. When we put our system in water, the focal length of lens is increased. Now the object is inside F. Hence image will be magnified 7. It is problem of kirchhof s current law. ln R 1, current is 0.4 A. ln R, current is 0.8 A O. 4A = 0.4A ln R 3, current is ( )A = 1.1 A. 8. When cylinder is rotated, water will rise at the edge and will dip at the centre. The difference in pressure at A and O is P A P 0 = gh. Due to ω 0 Due to rotation, pressure difference P A P 0 =. a cm R =. (ω. R ) R gh = ω R = 0.0 m = cm h = ω R g = (π ) (0.05) 10

13 9. Inductive reactance x L = ωl = 100x 3 10 Ω = 10 3 Ω Capacitive reactance x c = 1 ω c = Ω = 0 v 3 Ω For i 1, phase difference between i 1 & voltage, tanφ 1 = x L R = = 3 1 = 60 o, current lagging For i, phase difference, tanφ = x C φ 1 = 60, current leading. difference in phase = 90 o. = 0 v 3 R 0 = v I x = I cm + mx I x = 5 mr + mx Parabola opening upward 11. dv dt = k v = kt 4 3 πr3 = kt r = ( 3 kt 1 3 4π ) P ex 4T r = 4T ( 3KE 4π ) 1 3 log P ex = log C 1 log t 3 y = c mx = 4T (4π)1 3 (3 k) 1 3 t 1 3 = ct 1 3

14 1. K = 1 mv = 1 m ( GM) = GMm m R R R K A K B = m A m B R B R A = m m R R = V = (50 + R) = 50 + R 50 5 = 50 + R R = = 600 Ω 14. hv = W + V 0 e hv = W + V 0e on solving we get, W = 3 h v hv 0 = 3 h v v 0 = 3 v 15. Range = hr To double the range h have to be made 4 times = (μ f 1 1) ( 1 1 ) 1 = (μ 1 R ω f 1) ( 1 1 ) R ω When joined together 1 = = μ 1 + μ 1 = μ μ 1 f eq f 1 f R R R = f eq = R μ μ 1

15 17. εind = Bvl sin 45 0 = = V 18. When switched on, V CE = 0 V CC R C i C = 0 i c = V CC 5 = R C = A i c = βi B i B = i C β = = A =.5 μa 00 using KVL at input side, V BB i B R B V BE = 0 V BB = V BE + i B R B = = = 3.5 V 19. λ 1 4 λ 4 v = 11 cm. +e = 11 cm + e (i) 51 4 v = 7 cm. +e = 7 cm + e (ii) 56 4 Equation (ii)-(i) v = 0.16 v = = m s = 38 m s.

16 0. Since dq dt t=4s = 0 Therefore current = 0 1. [ L ] = CVR [L R ] = R CRV [L] [ 1 ] R RC [R] V = [T] [ 1 ] T [R] V = [ R ] = V [1] = 1 A 1. y = 5 & m(3πt) cos 3πt ω = 3π T = π ω = 3 sec 3. t = π ω V A = ωr 1 ( i ) V B = ωr ( i ) 4. Applying conservation of energy mgr = 1 mv 1 mv 1

17 V = gr + V 1 = V = 15 m s L 0 = mv B r = L 0 = 6 kgm s 5. E = σ = Q = 100 V m ε 0 Aε 0 Q = 100 A ε 0 Q = C = C 6. P 1 + mg A = P nrt + mg = nrt l a A A l A m = nrt g ( 1 l 1 l 1 ) = nrt g (l 1 l l 1 l ) 7. Change in Z = (6 ) 4 = 8 Change in A = 6 4 = TH 6α 4β 08 8 P B

18 8. I = B 0 μ 0 C B 0 = μ 0I C B rms = μ 01 = 10 8 C 4π T 9. δ = ρ 0vg L Ay δ = (ρ 0 ρ L )vg L Ay δ = 3 mm δ^ δ = ρ 0 ρ L ρ 0 = 8 8 Chemistry 1. Li Al H 4 is a strong reducing agent as it can reduce carboxylic acids, aldehydes, ketones and alcohols. But it cannot reduce double bonds. Also Na OET acts as a solvent in the above reaction and does not take part chemically. Hence, the reaction will be. H O H O Volume strength = M = = Nylon 66 is made up by hexamethylene diamine (H N (CH ) 6 NH ) and Adipic acid (HCOO (CH ) 4 COOH). NH (CH ) 6 NH + HCOO(CH ) 4 COOH

19 4. C atom has the tendency to link with one another through covalent bonds to form chavis and rings (catenation) this is because C C bonds are very strong. Down the group the size increases tendency to show catenation decreases. This is because of bond enthalpy. The order of catenation is C > Si > Ge Pb does not show catenation

20 7. 8. In Fischer projechai formula β H present at leaving roup. sayzeff product will from. (more stable alkene) 9. In the mentioned, H will get added to the carbon having slight positive charge. Presence of Cl increases the positive charge whereas an electron donating group like NH decreases the tendency. Hence, the correct order will be: D > C > B > A 10. Photochemical smog, often refused to as summer smog consists of nitrogen oxides, volatile organic compounds, tropospheric ozone, etc., It is formed when there components react in the presence of sunlight. CF Cl, a chlorofluorocarbon, is not present in photochemical smog, but is a major cause of ozone-layer depletion.

21 11. C + O CO ΔH = x We get above equation add n of and 3 C + 1 O CO Δx = y +CO + 1 O CO Δx = z C + O CO 3 rd ΔH = x x = y + z 1. Mn + will have configuration = 3d 5 4s 0 μ = 5.93 = n(n + ) or, n = 5 as 35 = 5.93 If has 5 unpaired electron which are not getting paired up. This implies that the ligand paired up. This implies that the ligand is a weak field ligand. Among the given options, weak ligand is en (Ethylenediamine) 13. Ions will be Cl, CO Δy = ky m i (i) n = because benzoic acid is deamarised β = 80% = 0.8 mwy of C 6 H 5 COOH = 11 here i = ( 1 1) β + 1 n i = ( 1 1) = 0.6 Put the value in equation (i) = 0.6 wt of benzoic acid wt of benzoic acid = 4.4gm

22 15. πr = nλ πa 0 n Z = nλ n πa 0 Z = n1.5πa 0 n Z = 1.5 = n 1 = 8 40 = 0. n = = 1 Mole fraction of NaOH = = Molarity = = = (S + 0.1) S Or S = M 18. C Form most stable pπ pπ bonds 19. n 1 T 1 = n T n 300 = (n n 5 ) T 300 = 3 5 T T = 500K 1. Calcination is required for hydroxide carbonate and hydrated oxide ores 3. Theory based 5. Fact 6. P and S will do acid base reaction with Grignard reagent rates.