Solution Mathematics 1. Point of intersection. Point of intersection ( 1 k, 1 k ) = 1 2. P [ ] = 28 2 ] = 70

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1 Solution Mathematics 1. Point of intersection y = kx, x = ky x = k (k x 4 ) x 3 = ( 1 3 k ) x = 1 k Point of intersection ( 1 k, 1 k ) Area = ( x k, kx ) dx = 1 ( 1 k = k 0 k x 3 kx3 ) k = 1 3k 1 3k = 1 k = 1 3. P [ ] = 46 C [ ] = 8 M [ 140 ] = 70 n (P C M) = n(p) + n(c) + n(m) n(p C) n(c M) n(m P) + n(p M C) = [ ] [ ] [140 6 ] + [ ] = = 10

b 3. (x 4 x )dx a From figure minimum area is (, ) 4. dy dx + 3(3 sec x)y = sec x This is linear differential equation Integration factor = e 3 sec x dx 3 tan x = e Hence y. e 3 tan x = e e3 tan x.

2 b 3. (x 4 x )dx a From figure minimum area is (, ) 4. dy dx + 3(3 sec x)y = sec x This is linear differential equation Integration factor = e 3 sec x dx 3 tan x = e Hence y. e 3 tan x = e e3 tan x.sec x dx ye 3 tan x = e3tanx 3 + c y = Ce 3 tan x Given y ( π 4 ) = = Ce C = e 3 Hence y ( π 4 ) = e3. e = e P( 7 or 8) = P(H) P(7 or 8) + P(T) P(7 or 8) = = = i=1 6. ( Now 0 C i 1 ) 3 0 C i + 0 C i 1 0 C i 1 0 C i + 0 C i 1 = 0 C i 1 1 C i = i 1 Let given sum be s, so S = 0 i=1 (i)3 = (1) 3 (0.1 ) = Given S = k k = 100 1

3 7. If the third term in the expansion of (1 + x log x ) 5 is 560 then the value of x is (A) 1 4 (B) (C) 1 8 (D) 4 8. Let 5, 5r, 5r be the sides of a triangle then value of r cannot be (A) 7 4 (B) 3 (C) 5 4 (D) Let ABC is a triangular plane such that AB = 7, AC = 6, BC = 5, and D is mid point of AC. A tower stand at mid point of AC subtending angle equal to 30 o at vertex B, then height of tower is (A) 7 3 (B) 7 3 (C) 7 (D) The equation of tangent to the hyperbola, 4x 5y = 0, which is parallel to x y =, is (A) x y 3 = 0 (B) x y 9 = 0 (C) x y + 1 = 0 (D) x y + 5 = 0

4 11. Let P be the nearest to ( 3, 0), then normal at P will pass through (3, 0), Let Co- ordinates of P be s ( t 4, t ) Hence equation of normal is y + tx = 1 + t3 4 This line passes through s ( 3, 0) 3t = t + t3 4 t = 0, {( ) is rejected} hence nearest point is (1, 1) distance = ( 3 1) + (1 0) = 5 1. μ = x+y 5 5 = 1 + x + y x + y = 13 (i) σ = Σ(x i μ) N 9. = x + y = 74 + x + y 171 = 74 + x + y 97 = x + y (ii) (x + y) = x + y + xy = xy xy = 36 T = So ratio is 4 9 or 9 4

5 13. Case I c 5 > 0 (i) f(0) > 0 c 4 > 0 (ii) f() < 0 4(c 5) 4c + c 4 < 0 c < 4 (iii) f(3) > 0 9(c 5) 6c + c 4 > 0 4c 49 > 0 c > 49 4 (iv) Here (i) (ii) (iii) (iv) c ( 49 4, 4) Case II c 5 < 0 (i) f(0) < 0 c < 4 (ii) f() > 0 c > 4 (iii) f(3) < 0 c < 49 4 (iv) (i) (ii) (iii) (iv) c φ Ans. c ( 49 4, 4) 14. x + y + z = 1 x + 3y + 5z = β 3x + 4y + αz = 9

6 1 1 1 D = = (3α 0) (α 15) + (4 9) 3 4 α D = 0 for α = 5 For this α equations first and third are 3x + 4y + 5z = 9 (i) x + y + z = 1 (ii) (1) 5 () x + 3y + 5z = 13 While remaining equation is x + 3y + 5z = β Hence for infinite solution α = 5, β = z 1 = 9 z 4z 1 z 1 = 9z z z 1 3z = 3z z 1 Given Z = z 1 + 3z z = 3z + 3z = real number 3z z 1 z 1 z 1 i j k 16. n = n 1 n = 3 1 = 7i 7j + 7k 1 3 Equation of plane is 7(x 4) 7(y + 1) + 7(z 3) = 0 7x 7y + 7z = 0 x + y = z 17. I ( ax 1+bx +cx 3 a+b+c, ay 1+by +cy 3 ) a+b+c 8(6) + 6(0) + 10(0) = (, 4 10(0) + 6(8) + 8(0) ) = (, ) 4

7 18. sin θ + cos 4 θ = cos θ + cos 4 θ = 3 4 Let cos θ = t t t = 0 (1 1 ) = 0 t = 1 cos θ = 1 cos θ 1 = 0 cos 4θ = 0 4θ = (n + 1) π θ = (n + 1) π 8 θ = π 8, 3π 8 [0, π ] Sum of values of θ is π 19. (sinn θ sin θ) 1 n cos θ dθ sin n+1 θ = (t n t) 1 ndt t n+1 t (1 1 = t n 1) t n+1 (1 1 = t n 1) t n 1 n 1 n dt (Put sin θ = t) dt Put 1 1 (n 1) tn 1 = z dt = dz t n I = 1 n 1 z 1 ndz = z 1 n +1 = n n+1 n 1 (1 1 sin n 1 θ ) n + c ( 1 + c = n(1 t1 n ) n + 1) (n 1) n 1 1 n +1 + c (1 x +sin 1 x ) sin([1 x] π 0. lim ) x x [1 x] (1 x + sin(x 1)) sin ( π = lim ) x 1 + (x 1)( 1) (x 1) + sin(x 1) = lim = = 0 x 1 + (x 1)

8 1. From the graph we can easily conclude that f(x) is non-derivable at a =, 1, 0, 1,. Normal to there curves are y = m(x c) bm bm 3 y = mx 4 am am 3 If they how a common normal (c + b)m + bm 3 = 4am + am 3 c + b 4a = (a b)m (m = 0 corresponds to axis) m = c a b > 0 3. Tangent at (1, 1) is x(1) + y( 1) + (x + 1) 3(y 1) 1 = 0 3x 4y = 7 Required circle is (x 1) + (y + 1) + λ(3x 4y 7) = 0 It pass through (4, 0) λ(1 7) = 0 λ = x + y 8x + 10y + 16 = 0 radius = 5 4. P(m) = m m + 41 P(3) = = 47 P(5) = = 61 Hence P(3) and P(5) are both prime

9 5. Two digit numbers of the form 7λ + are 16, 3,.,93 Two digit numbers of the form 7λ + 5 are 1, 19,..,96 Sum of all the above numbers equals to ( ) + (1 + 96) = = 6. f(x) = x 3 + ax + bx + c f (x) = 3x + ax + b f (x) = 6x + a f (x) = 6 a = f (1) = 3 + a + b a + b = 3 b = f = 1 + a a b = 1 c = f (3) c = 6 and a = 5, b = f(x) = x 3 5x + x + 6 f() = = Physics 1. ρ = M L 3 n 1 u 1 = n u 18[M 1 L 1 3 ] = n [M L 3 ] n = 18 [ M 1 M ] [ L L 1 ] 3 = 18 [ ] [ ]3 = = 40. Net torque= Torque on ring

R 0 dτ = μ τ = μfr F πr πx. xdx 3. Q A = K 1000 100 1 = 90 W m 4. ω = π n rad s dq = ρ. dx = ρ 0 l xdx di = dq.

10 R 0 dτ = μ τ = μfr F πr πx. xdx 3. Q A = K = 90 W m 4. ω = π n rad s dq = ρ. dx = ρ 0 l xdx di = dq. ω π dm = di A l dm = ω π ρ 0 l πx dx 0 M = π nρ 0l f 1 = ( ) f 340 f = ( ) f f = ( f ) = x = ε 1r = 1ε 13r l 13l ε = xl ε = 1ε 13l l l = 13l 4 7.

11 R = (X + Y) + XY = (X. + Y) XY = (X Y ). XY = (X Y )(X Y ) 8. D = h T R + h R R D = D = [ 8 + 8] D = [ 7 + ] D = [ 7 + ] D = [ ] = 64000m = 64km 65km 9. 1 f 1 = (μ 1 1) ( 1 R 1 ) 1 f = (μ 1) ( 1 R 1 ) f f 1 = μ 1 1 μ 1 = 1 μ 1 μ 1 μ 1 μ = λ (A 0 ) = V 10 = 150 V V = 150 = 80 kv Nearby value is 5 kev 11. e = Bvl = 0.1( 10 ) 6 = = 1mV

12 1. A 1 = π(r max ) 1 A π(r max ) = u 4 1 u4 = 1 16 [ max. Range = u g ] 13. q 1 a = q b = q 3 c = constant q 1 + q + q 3 = Q q 1 = q = q 3 = a (Q) a +b +c b (Q) a +b +c c (Q) a +b +c V = Kq 1 + Kq + Kq 3 a b c V = 1 4πε 0 [Q] ( a+b+c a +b +c ) 14. E = φ KE KE = E Q hv = φ + KE 15. F f r = ma (i) f r R = Iα = mr α (ii) For pure rolling a = αr (iii) From (i)(ii) and (iii)

13 mr α F = mαr F = 3 mr α α = F 3mR 16. V = fλ T μ = fλ = 100λ λ = 40 cm λ = 0cm 17. N mg = mg N = 3mg s = ut + 1 at g = (1) = g 4 W N = 3mg g 4 = 3mg 8 = 3mg 8

14 18. C = E 0A d C 1 = K 1E 0 A 3d C = K C 3 C 3 = K 3C 3 = K 1C 3 C eq = K eq = C 3 (K 1 + K + K 3 ) K eq = K 1 + K + K 3 3 = 36 3 = V e = v K. E = 1 m( v) = mv 0. r = r B r A = a i + a j 1.

15 Assume V F = as O as V E = V F = 0 Current in R B is zero V A V D = 10V I = 10 = 1 A from A to D 0. gh A = V (10 4 ) h = 9.8 (10 4 = 5.1 cm ) 3. η 1 = 1 T T 1 η = 1 T 3 T η 3 = 1 T 4 T 3 Given η 1 = η = η 3 T T 1 = T 3 T = T 4 T 3 T = T 1 T 3.(1) T = T 1 T 3.(1) (1) and () T 3 = (T 1 T 4 ) 1 3, T (T 1 T 4 )

16 V 1 + V = 0 K4Qa x KQa (x r) = 0 x = (x r) x = ( r 1 ) 5. Req = R 1R R 1 +R = = 4Ω 6. E = C E = B. C. = B V m = V m 7. d sinθ = nλ 8. λ = dsinθ n = n λ = = n n Putting n = 1, λ 1 = 500 nm n =, λ = 150 nm n = 3, λ 3 = 833 nm n = 4, λ 4 = 65 nm n = 5, λ 5 = 500 nm n = 6, λ 6 = 357 nm = 500 n (nm) Using relative velocity

17 a rel = 0 v rel = = v rel t t = = 1 sec v bullet = = 90 m s v particle = 10 1 = 10 m s S = = 95 m P i = P f = 0.05V V = 30 m s h = v g = 900 = 45 m 0 So from top of building 45 5 = 40 m 9. Green 5, Black 0 Red 10 Brown 10% tolerance I R = P I = = A I = 10 A = 0 ma

18 Chemistry 1. Acidic strength of phenol is dependent on the substituents attached to it. Presence of electron withdrawing groups increase the tendency to attract electrons (Acidity), thereby decreasing the pka value. NO is a strong withdrawing group, Cl is a weak electron withdrawing group as it has +M effects whereas OCH3 is an electron donating group. Hence, the correct order of pka values is: order of pka values: III < II < I < IV. A water supply with a BOD level of 3-5 ppm is considered moderately clean. In water with a BOD level of 6-9 ppm, the water is considered somewhat polluted because there is usually organic matter present and bacteria are decomposing this waste. Hence, B with a higher value of BOD level is more polluted. 3. Addition of alc. KOH to haloalkanes follows the Saytzeff rule which points towards greater stability. Saytzeff s rule implies that elimination reactions favour formation of alkene with greater number of substituents. Not only that, but it also gives the molecule an extended conjugate system which further enhances its stability.

19 4. KE max = bυ hυ o. Photoelectric effect. Intercept must be ve 5. The most inorganized system has a b c and α β γ = 90 o is called the triclinic system. 6. Steric no. = 6, hybridisation = sp 3 d Geometry = octahedral No. of lone pairs on Xe = 1 7. Possible Structural isomers No of geometrical isomer by each [MCl(NH 3 )(NO )(SCN)] 3 [MCl(NH 3 )(ONO)(SCN)] 3 [MCl(NH 3 )(NO )(NCS)] 3 [MCl(NH 3 )(ONO)(NCS)] 3 Total number of isomers will be Hall Heroult s process: At cathode: Al e Al At cathode: O + C CO + e Overall reaction: Al O 3 + 3C 4Al + 3CO

20 9. H O acts as an oxidizing agent, as well as a reducing agent in both acidic as well as basic mediums. 10. According to MOT; electronic configuration of N + is: σ1s σ 1S σs σ S πp x πp y σp z 1 B. O. = 1 (9 4) =.5 i.e., π bonds and 0.5 σ bonds. 11. Chlorotris(triphenylphosphine)rhodium(I), is known as Wilkinson s catalyst. It is used as a homogeneous hydrogenation catalyst. It is a square planar 16-electron complex. The oxidation state of Rhodium in it is K P = K C (RT) Δng K P K C = (RT) Δng Equations 1: Δng = 0 K P K C = 1 Equations : Δng = 1 K P K C = RT Equations 3: Δng = K P K C = (RT) 13. Since, P 0 A = Pa; P 0 B = Pa and X A = 0.4 According to Raoult s Law: P T = P 0 A x A + P 0 B x B P T = ( )(0.4) + ( )(0.6) P T = Pa Method 1 Now, P 0 A x A = P T y A

21 y A = P A 0 x A P T = y B = 1 y A Method We have the total pressure and the individual partial vapour pressures of the two liquids, P T = Pa, P A = 8 10 Pa and P B = 7 10 Pa. Each gas exerts a partial pressure proportional to its mole fraction. Hence, mole fractions of A and B in vapour phases are 0.8 and 0.7 respectively. 14. (A) Since Combustion of Coal is a fast and spontaneous process, no catalyst is required. (B) Catalyst used is Ni(s) (C) Catalyst used is Fe O 3 (s), K O (S) and Al O 3 (s). (D) Catalyst used is Pt (s) 15. W CaSO4 = M CaSO4 = (136) = 0.95g [OH ] =? Remaining is moles The solution will have OH - ions coming from both Ca(OH) and NaOH. The concentration of NaOH is 0.007, which is negligible as compared to 0.1 moles of Ca(OH). [OH ] = M

22 16. Isotopes of Hydrogen: Among all three isotopes only Tritium is radioactive. 17. Dichloromethane and water are immiscible in nature. Also, Dichloromethane has higher density than water so forms bottom layer (Layer II) in the separating funnel. 18. Higher is the reduction potential, higher will be its tendency to get oxidized and thus higher will be its reducing property. Consequently, its reduction ability will be low. 19. EN of Al = 1.61 EN of Be = 1.57 EN of B =.04 EN of C =.55 Due to diagonal relationship between Be and Al, they exhibit similar properties and thus, have similar electronegativity values NaBH 4 doesn t reduce esters and alkenes.. The speed of hydrolysis will depend on how likely a molecule is to accepting OH - ions. More acidic a molecule, higher will be the speed of reaction. NO is a strong EWG (Electron withdrawing group), Cl is a weak EWG and OCH3 is an EDG. Hence, the increasing order of rate of hydrolysis will be: 3 < < 4 < 1

23 3. Cyclic anhydrides in 7 membered ring are less stable. 4. Allylic radial is more stable due to resonance. Rate of reaction Stability of free radical 5. Ea RT 6. Arhaneous Solution: K = Ae K decreases exponentially with B a The second graph should be an asymptotic curve tangetial to positive x axis. 1 st graph is correct and second is incorrect. 7. Beryllium is used in making windows in X-ray tubes. 8. G = H T S. (Gibbs Helmholtz Equation) For a reaction to be spontaneous, G < 0 So, H T S < 0 H < T S H S < T i.e. Minimum temp. should be H S = 5K

then the ordered pair (a, b) for which the value of

then the ordered pair (a, b) for which the value of QUESTION PAPER Mathematics 1. If the area bounded by the curves y = kx 2 and x = ky 2 is 1, then value of k is (k > 0) (A) 2 3 (B) 1 3 (C) 3 (D) 3 2 2. If there are 140 students numbered from 1 to 140;