KVPY SA Stream Solution 2011

ANSWER KEY 

ANSWER KEY 

HINTS & SOLUTIONS (YEAR-2011)

HINTS & SOLUTIONS (YEAR-2011)

PART-A (1 Mark)

PART-A (1 Mark)

MATHEMATICS

MATHEMATICS

1. 1. P(x) =P(x) = )) c c a a )( )( b b a a (( )) c c x x )( )( b b x x ((

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+ + )) a a b b )( )( c c b b (( )) a a x x )( )( c c x x ((

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+ + )) b b c c )( )( a a c c (( )) b b x x )( )( a a x x ((

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L Leett ff((xx) ) = = PP((xx) ) – – 11 f(a) = 1 + 0 + 0 – 1 = 0 f(a) = 1 + 0 + 0 – 1 = 0 f(b) = 0 + 1 f(b) = 0 + 1 + 0 + 0 – 1 = 0– 1 = 0 f(c) = 0 + 0 + 1 – 1 = 0 f(c) = 0 + 0 + 1 – 1 = 0

f(x) is a polynomial of degree atmost 2, and also attains same value (i.e., 0) for 3 distinct values of x f(x) is a polynomial of degree atmost 2, and also attains same value (i.e., 0) for 3 distinct values of x (i.e. a,b,c).

(i.e. a,b,c).

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f(x) is f(x) is an identity with only value equal to zero.an identity with only value equal to zero.

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f(x) = 0f(x) = 0

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 x x

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RR

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P(x) = 1,P(x) = 1,

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 x x

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RR

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2. Using cauchy schwartz’s inequalityUsing cauchy schwartz’s inequality (a

(a22 _{+ b + b}22_{) (x) (x}22 _{+ y + y}22₎₎

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 _{(ax + by) (ax + by)}22

equality holds at equality holds at x x a a  =  = yybb

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ay – bx = 0ay – bx = 0 Aliter : Aliter : a a22 _{+ b + b}22_{==8811 ...((ii))} x x22 _{+ y + y}22 _{= =112211 ...((iiii))} a ax x + + bby y = = 9999 ...((iiiiii)) (a

(a22 _{+ b + b}22_{) (x) (x}22_{+ y+ y}22_{) ) = = 881 1 × × 112211 ...((iivv)) [[((ii) ) × × ((iiii))]]}

(ax + by)

(ax + by)22 _{= 99 = 99}22 _{...((vv)) [[ssqquuaarriinng g ((iiiiii))]]}

(iv) – (v) (iv) – (v) Q Quueess.. 11 22 33 44 55 66 77 88 99 1100 1111 1122 1133 1144 1155 1616 1177 1188 1199 2200  Ans.  Ans. AA CC AA BB BB DD AA AA AA CC BB CC BB BB CC AA AA CC DD AA Q Quueess.. 2211 2222 2233 2244 2255 2266 2277 2288 2299 3030 3311 3322 3333 3344 3355 3366 3377 3388 3399 4400  Ans.  Ans. BB CC CC DD CC BB BB DD AA CC DD CC DD DD AA DD AA AA BB DD Q Quueess.. 4411 4422 4433 4444 4455 4466 4477 4488 4499 5050 5511 5522 5533 5544 5555 5566 5577 5588 5599 6600  Ans.  Ans. DD DD BB BB CC BB DD CC BB CC DD AA CC CC AA BB AA BB BB DD Q Quueess.. 6611 6622 6633 6644 6655 6666 6677 6688 6699 7070 7711 7722 7733 7744 7755 7766 7777 7788 7799 8800  Ans.  Ans. CC CC BB BB AA BB BB BB CC AA AA CC BB AA AA BB DD CC AA CC

3. 3.  Then, Then, aa x x 1 1 x x

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 and and bb x x 1 1 x x 3 3 2 2

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2 2 2 2 a a x x 1 1 x x

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x x22 _+ + 2 2 x x 1 1 +2 = a +2 = a22 _...(i)...(i) 3 3 x x 1 1 x x

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 =  = aa33 x x33 _+ +

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x x 1 1 x x 3 3 x x 1 1 3 3  = a = a 3 3 _{...(ii)...(ii)}

add equation (1) and (2) add equation (1) and (2)

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x x 1 1 x x 3 3 2 2 x x 1 1 x x x x 1 1 x x22 ₃₃ 33 _{22  = a = a}22 _{+ a + a}33  b +  b +

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3 3 x x 1 1 x x  _{+ 2 + 3a = a + 2 + 3a = a}22_{+ a+ a}33 2 2 3 3 x x 1 1 x x

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_{= a= a}33 _{+ a + a}22 _{– 3a – b – 2 – 3a – b – 2} 4. 4. |a – b| = 2, |b – c| = 3, |c – d| = 4|a – b| = 2, |b – c| = 3, |c – d| = 4 a – b = ± 2 a – b = ± 2 b – c = ± 3 b – c = ± 3 c – d = ± 4 c – d = ± 4

possible values of (a – d) are ± 9, ± 5, ± 3 possible values of (a – d) are ± 9, ± 5, ± 3 |a – d| = 9, 5, 3, 1

|a – d| = 9, 5, 3, 1

Sum of all possible values are Sum of all possible values are 1818 5.

5. Given 0 < r < 4Given 0 < r < 4

in all the obtain in all the obtain (Base) (Base)xx _= = 5 5 9 9

the option having least base will give the largest x. the option having least base will give the largest x. So, in option B base

So, in option B base

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17 17 r  r  1

1  _{is minimum for 0 < r < 4. is minimum for 0 < r < 4.} Aliter : Aliter : (1 + (1 +

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xx ₌₌ 5 5 4 4

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x log (1 +x log (1 +

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) = log (1.8)) = log (1.8)

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For x

For x to be to be maximummaximum log (1 +

log (1 +

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) should be minimum) should be minimum

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 = =

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r r  oorr 22rr oorr    17 17 r  r  or  or  r  r  1 1 in 0 < r < 4 in 0 < r < 4 17 17 r  r   is minimum.  is minimum. 6.

6.  Angle  Angle bisector bisector theoremtheorem

6 6 x x  =  = z z y y x x 66 z z y y 10cm 10cm22

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xz = 6yxz = 6y

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20 = 6y20 = 6y

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y =y = 3 3 10 10 cm cm ALITER : ALITER : Let

Let

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 ACB = ACB =

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BC = 6 cos

BC = 6 cos

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, AB = 6 sin, AB = 6 sin

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BD : CD BD : CD = AB : = AB : ACAC  = 6 sin  = 6 sin

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 : 6 : 6  = sin  = sin

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 : 1 : 1 BD = BD =

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sinsin

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sin sin 1 1 cos cos 6 6 ...(i) ...(i) CD =

CD = ₁₁66coscos_sinsin

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11

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...(ii) ...(ii) BD = CD sin BD = CD sin

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Now, area of 

Now, area of 

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 ADC = 10  ADC = 10 cmcm22

2 2 1 1 × 6(CD) sin × 6(CD) sin

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 =  = 1010 CD sin CD sin

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 = 10/3 cm = 10/3 cm 7. 7.

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Slant height = 13Slant height = 13

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byby   _{= r  = r} 

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2 2

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.5 = 13.5 = 13

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13 13 10 10

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8. 8. x x 12 12  =  = 8 8 x x 20 20

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xx22 _{– 20x + 96 = 0 – 20x + 96 = 0} x = 8, 12 x = 8, 12 ALITER : ALITER : B(0,0) B(0,0) C(20,8) C(20,8) E E((xx,,00)) DD((2200,,00))  A(0,12)  A(0,12)  AC =  AC = ₂₀₂₀22

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₄₄22 _{== 44 55}22

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_{11 == 44 2626}

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(12(1222 _{+ x + x}22_{) + (8) + (8}22 _{+ (20 – x) + (20 – x)}22_{) ) = = 2020}22 _{+ 4 + 4}22 2x 2x22 _{– 40x + 400 + (12 – 40x + 400 + (12}22 _{– 20 – 20}22_{) + (8) + (8}22 _{– 4 – 4}22_{) = 0) = 0} 2x 2x22 _{– 40x + 144 + 12.4 = 0 – 40x + 144 + 12.4 = 0} x x22 _{– 20x + 72 + 24 = 0 – 20x + 72 + 24 = 0} x x22 _{– 20x + 96 = 0 – 20x + 96 = 0} x = 12, 8 x = 12, 8 9. 9.  AP = 2 sin  AP = 2 sin 60° =60° = ₃₃

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d =d = ₂₂

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₃₃ 10. 10. 51251233 _{– (253 – (253}33 _{+ 259 + 259}33₎₎ = 512 = 51233 _{– [(512) (253 – [(512) (253}22_{+ 259+ 259}22 _{– 253.259)] – 253.259)]} = 512 (512 = 512 (51222 _{– ((512) – ((512)}22 _{– 3(253)(259)) – 3(253)(259))} = 512 (3.253 – 259) = 512 (3.253 – 259) = 2 = 299_{. 3. 253 . 7.37. 3. 253 . 7.37} = 2 = 299_{. 3. (11) . (23). 7.37. 3. (11) . (23). 7.37}

So, number of distinct prime divisors are 6. So, number of distinct prime divisors are 6. 1

11.1. TTop layer has (13 × op layer has (13 × 13) balls13) balls

Simillary one layer below top layer will have (14 × 14) balls and we have 18 lesens to t

Simillary one layer below top layer will have (14 × 14) balls and we have 18 lesens to t otal number of ballotal number of ball N = (13) N = (13)22 _{+ (14) + (14)}22 _{+ ...+ (30) + ...+ (30)}22 N = N = 6 6 61 61 31 31 30 30

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 –  – 6 6 25 25 13 13 12 12

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N = 8805 N = 8805

12.

12. Let distance is 6dLet distance is 6d M Muudd :: TTaarr :: SSttrreeaamm D Diissttaannccee dd :: 33dd :: 22dd S Sppeeeedd 33VV :: 55VV :: 44VV time time V V 3 3 d d :: V V 5 5 d d 3 3 :: V V 4 4 d d 2 2

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1100 :: 1188 :: 1155 ((NNootte e : : oorrddeer r iis s cchhaannggeed d iin n qquueessttiioonnss)) 13.

13. 3 step3 step

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14. The clock well show 1 in an hour for 19 timThe clock well show 1 in an hour for 19 tim e for 11 hours it will show the incorrect time for (19 × 11) time.e for 11 hours it will show the incorrect time for (19 × 11) time. The last 12

The last 12thth _{hour will always show the incorrect time so total incorrect tim hour will always show the incorrect time so total incorrect tim ee}

(19 × 11 + 60) min = 269 min (19 × 11 + 60) min = 269 min

there are 24 hours in a day to = 269 × 2 = 538 min there are 24 hours in a day to = 269 × 2 = 538 min 538 min = 538 min = 30 30 269 269 = = 9 h9 houoursrs

the fraction day when the clock shows correct tim the fraction day when the clock shows correct tim e is =e is =

24 24 9 9 1 1

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= = 8 8 5 5 8 8 3 3 1 1

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15. 15. Ri

Righght t WrronW ong g UnUnaatttetempmptetedd 1 155 1155 00 14 14 44 13 13 88 1 122 1122 1 111 1166 1 100 2200 6 cases only 6 cases only

PHYSICS

PHYSICS

16.

16. By mechanical energy conservationBy mechanical energy conservation KE KE_ii +  + UU_ii =  = KEKE_f f  +  + UU_f f  O + U O + U_ii = 0 + U = 0 + U_f f  U U_ii = U = U_f f  h h_ii = h = h_f f 

So D will lie on line AB So D will lie on line AB 17.

17. Since toy is not accelerating so net external forSince toy is not accelerating so net external for ce on toy is zero. So (A)ce on toy is zero. So (A)

18. 18. |S |S₁₁| + |S| + |S₂₂| = H| = H 2 2 gt gt 2 2 1 1  + ut –  + ut – gtgt22 2 2 1 1  = H  = H u ut t = = H H .. ...((II)) H = H = g g 2 2 u u22 ....(II) ....(II) t = t = ₂₂uu_gg S S₂₂ = ut – = ut – gtgt22 2 2 1 1  =  = 33₈₈uu_gg g g 4 4 u u g g 2 2 1 1 g g 2 2 u u 4 4 2 2 2 2 2 2   = = HH 4 4 3 3 19. 19.

By concept of centre of mass By concept of centre of mass 36x = 9(20–x) 36x = 9(20–x) 36x = 180 – 9x 36x = 180 – 9x 45x = 180 45x = 180 x = 4m x = 4m 20.

20.  All three wi All three will be ll be in thermal equilibrium in thermal equilibrium with air with air of room. so of room. so temperature of the temperature of the three will three will be samebe same 21.

21. Pressure of gas is same Pressure of gas is same everywhere in the vessel.everywhere in the vessel. 22.

22. TTo travel from P to Q in minimum o travel from P to Q in minimum time, she should travel on path PCQ.time, she should travel on path PCQ. 23. 23. i = 45°i = 45°

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 C C C = 45° for minimum C = 45° for minimum

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 sin 45 = 1 sin 45 = 1

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== ₂₂  = 1.42 = 1.42

24. 24.

In this case only half part of lens

In this case only half part of lens is used to form is used to form the image so intensity will reducethe image so intensity will reduce

25. 25. R =R =  A  A    R = R = _  A  A R = R = V V 2 2  

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R = R = V V )) 2 2 ((  22

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 = 4R  = 4R 26. 26. E

E₁₁ = Electric field due to +Q = Electric field due to +Q E

E₂₂ = Electric field due to –2Q = Electric field due to –2Q There resultant is 0 at this point There resultant is 0 at this point

27. 27.

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 = = 100100 750 750 tt // mgh mgh _ = = ₁₀₀₁₀₀ 750 750 60 60 6 6 10 10 300 300    = = 100100 750 750 300 300_ = = 40%40% 28. 28.

29. 29. pbpb314314₈₂₈₂

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 2e 2e –1 –1 _+ + 44 2 2 He He  + + XX₈₂₈₂210210

So 82 proton and 128 Neutron So 82 proton and 128 Neutron 30. 30. PV = NKTPV = NKT 10 1055 _{× 100 = N × 1.38 × 10 × 100 = N × 1.38 × 10} –23 –23 _{× 273 × 273} N N

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3 3 × × 10102727

CHEMISTRY

CHEMISTRY

31.

31. Since pressures of the gases are same Since pressures of the gases are same in both the containers. So, the final pressure will not changein both the containers. So, the final pressure will not change 32.

32. Reativity towards Friedel-Crafts alkylation is prReativity towards Friedel-Crafts alkylation is pr oportional to electron density in the benzene ring.oportional to electron density in the benzene ring.

33. 33. n = 2n = 2 l = 0, 1 l = 0, 1 m m = 0, = 0,

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1 1 1 1 0 0 34.

34.  Average  Average Kinetic Kinetic Energy Energy depends depends only only on teon temperaturemperature (KE)

(KE)_avgavg = = 2 2 kT kT 3 3  per molecule  per molecule 35.

35. Ideal gasIdeal gas

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H = 0H = 0

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S > S > 00 (randomness increases) (randomness increases) 36. 36. (NH(NH₄₄))₂₂ Cr  Cr ₂₂OO₇₇

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 Cr  Cr ₂₂OO₃₃ + N + N₂₂ +  + 4H4H₂₂OO 37.

37.  Acc Accordording ing to to the the gragraph ph solusolu bilibility ty 40° 40° is is apprappr ox. ox. 200 200 g g per per 100 100 mlml. . For For 50 50 ml, ml, amoamount unt is is 100 100 gg approx.

approx.

38.

38.  Aldehyde, ketones  Aldehyde, ketones with acetywith acetyl group CHl group CH₃₃ –  – – show – show Iodoform test.Iodoform test.

39. 39. 16 16 1 1 2 2 1 1 n

n

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n, number of half-lives = n, number of half-lives = 4 = 2 hrs. hal4 = 2 hrs. half-life = 30 min.f-life = 30 min.

40.

40. ZnSZnS

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ZnSOZnSO₄₄ during roasting, sulphide ore is during roasting, sulphide ore is converted into sulphate.converted into sulphate.

41. 41. 42. 42. ClCl₂₂ + 2KBr  + 2KBr  → → 2KCl + Br 2KCl + Br  2 2 reddish brown reddish brown

43.

43. (i) and (iv) are hetro aromatic (i) and (iv) are hetro aromatic and the resonance form of azulene (iii) is aromand the resonance form of azulene (iii) is arom atic.atic.

(ii) Is nonaromatic. (ii) Is nonaromatic.

44.

44. Simple nomenclature of alkane.Simple nomenclature of alkane. 45.

45. The major product formed in the reaction :The major product formed in the reaction :

 Product is :  Product is : ((AA))ii ((BB)) iiii ((CC))iiiiii ((DD))iivv S Sooll. . ((CC))

The given reaction is S

The given reaction is S_NN2 which occurs at sp2 which occurs at sp33 _{carbon with good leaving group. carbon with good leaving group.}

PART-II (2 Marks)

PART-II (2 Marks)

MATHEMATICS

MATHEMATICS

61. 61. f(x) = axf(x) = ax22 _{+  + bbx x + + cc,, ff((11) ) = = 00} g giivveenn,, f(f(11) ) = = 00 a + b + c = 0 a + b + c = 0 a anndd 440 0 < < ff((66) ) < < 5500 40 < 36a + 6b + c < 50 40 < 36a + 6b + c < 50 40 < 35a + 5b < 50 40 < 35a + 5b < 50 8 < 7a + b < 10 8 < 7a + b < 10 7 7a a + + b b = = iinntteeggeer r = = 99 ...((11)) a anndd 660 0 < < ff((77) ) < < 7700 60 < 49a + 7b + c < 70 60 < 49a + 7b + c < 70 60 < 48a + 6b < 70 60 < 48a + 6b < 70 10 < 8a + b < 11.6 10 < 8a + b < 11.6 8 8a a + + b b = = 1111 ...((22)) solving equation (1) and (2)

solving equation (1) and (2) a = 2, b = –5, so c = 3 a = 2, b = –5, so c = 3 f(x) = 2x

62.

62. We can write the expression asWe can write the expression as

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 2011 2011 2 2 r  r  22 2 2 1 1 r  r  1 1 r  r  = =

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_{ }

_{ }

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1 1 r  r  1 1 r  r  2 2 1 1 = =

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11

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_r r 

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11₁₁

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_r r 1

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1₁₁

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Putting r = 2, 3, ..., 2011 Putting r = 2, 3, ..., 2011 = 2010 + 1 + = 2010 + 1 + 2011 2011 1 1 2012 2012 1 1 2 2 1 1

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= 2011 + = 2011 + ₂₂11

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_2011201111

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_2012201211

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this lies between (2011, 2011 this lies between (2011, 2011

2 2 1 1 )) 63.

63. Let initially 2 bases have radii 5 & Let initially 2 bases have radii 5 & r. and finally bases have radii (1.21 × 5) & r. and finally bases have radii (1.21 × 5) & r.r.

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Ratios of volumes =Ratios of volumes = 11..2121 V V V V 1 1 2 2

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V V₂₂ = = 3 3 h h

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 ((6.05)  ((6.05)22 _{+ 6.05 r + r  + 6.05 r + r} 22₎₎ V V₁₁ = = 3 3 h h

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(5 (522 _{+ 5r + r  + 5r + r} 22₎₎ 21 21 .. 1 1 V V V V 1 1 2 2

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_11..2121 r  r  r  r  5 5 5 5 r  r  r  r  05 05 .. 6 6 )) 05 05 .. 6 6 (( 2 2 2 2 2 2 2 2

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36. 36.6026025 + 65 + 6.0.055 r +r + r r 22 _{= 30.25 + 6.05r + 1.21 r  = 30.25 + 6.05r + 1.21 r} 22 .21r  .21r 22 _{= 6.3525 = 6.3525} r  r 22₌₌ 21 21 .. 3525 3525 .. 6 6 r = r = 2 2 11 11  cm. = 55 mm  cm. = 55 mm 64.

64. Let speed of B = V km/hr.Let speed of B = V km/hr. Let speed of A = 3V km/hr. Let speed of A = 3V km/hr. Given 4r = 2 × 60 km/hr 

Given 4r = 2 × 60 km/hr 

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 V = 30 km/hr  V = 30 km/hr 

Distance covered by then after 10 min. = 2 × 10 = 20 k Distance covered by then after 10 min. = 2 × 10 = 20 k mm So, remaining distance = (30 – 20) km = 10 km.

So, remaining distance = (30 – 20) km = 10 km. Time table b

Time table by B to coy B to cover 10 kmver 10 km == 30 30 10 10  = 20 min  = 20 min 65. 65.  A A BB CC 1 10 0 hhrr 220 0 hhrr 330 0 hhr  r   Exactly one pair of taps is open during

Exactly one pair of taps is open during each hour and every pair of taps is each hour and every pair of taps is open at least for open at least for one hour.one hour. So

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20 20 1 1 10 10 1 1  +  +

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₂₀₂₀11

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₃₀₃₀11

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 + +

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₃₀₃₀11

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₁₀₁₀11

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 = = 60 60 22 22 First

First then then second second then then Third Third thenthen

In three hours the tank will be f In three hours the tank will be f illedilled

th th 60 60 22 22

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 part  part

Now, for minimum tim

Now, for minimum tim e, the rest tank me, the rest tank m ust be filled with A and B taps.ust be filled with A and B taps.

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60 60 9 9 20 20 1 1 10 10 1 1

So, the rest So, the rest

th th 60 60 38 38

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 Part of tank will take 5 hour more  Part of tank will take 5 hour more So, the tank will be filled in 8

So, the tank will be filled in 8thth _hour. hour.

PHYSICS

PHYSICS

66. 66. kx kx₁₁ + +

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₁₁vg vg ==



vgvg v = v = _ggkxkx _gg 1 1 1 1    kx kx₂₂ + +

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₂₂vg vg ==



vgvg v = v = _ggkxkx _gg 2 2 2 2    g g )) (( kx kx g g )) (( kx kx 2 2 2 2 1 1 1 1       

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xx₁₁ – –

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_xx₁₁ = =

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xx₂₂ – –

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_xx₂₂



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xx₁₁ – x – x₂₂) =) =

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_xx₁₁ – –

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_xx₂₂



 = = 2 2 1 1 2 2 1 1 1 1 2 2 x x x x x x x x     67. 67. 22f f  22ii 8 8 4 4 mv mv 2 2 1 1 mv mv 2 2 1 1 dx dx .. F F  



2 2 1 1  × 3 × 8 –  × 3 × 8 – 2 2 1 1  × 1.5 × 4 =  × 1.5 × 4 = 2 2 1 1  × 0.5 (v  × 0.5 (v_f f 22 _{– 3.16) – 3.16)}22 24 – 6 = 0.5 (v 24 – 6 = 0.5 (v_f f 22 _{– 3.16 – 3.16}22₎₎

68. 68. v v_ii =  = vv_f f  2 2 5 5 ×1 × V + ×1 × V + 2 2 5 5  × 0.5 × 4v =  × 0.5 × 4v = 2 2 5 5  P 5V  P 5V 3V = 5PV 3V = 5PV P = P = 5 5 3 3  = 0.6  = 0.6 69. 69. sin(90 –

sin(90 –

 

 

) =) =

 

 

sinr sinr  cos cos



 = = 4 4 x x h h 2 2 // x x 3 3 4 4 2 2 2 2_  cos cos



== 4 4 1 1 x x h h 1 1 3 3 2 2 2 2 2 2  == 16 16 4 4 16 16 49 49 1 1 3 3 2 2  = = 53 53 4 4 3 3 2 2 = = 53 53 3 3 8 8 70. 70. P P_ii = i = i22_RR P = P = 16 16 100 100 1 1 4 4 10 10 22

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P P_f f  = i = i22_RR 100 100 100 100 10 10

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22

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CHEMISTRY

CHEMISTRY

71.

71. 100 mL 0.1 M CH100 mL 0.1 M CH₃₃COOH 50 mL 0.4 M COOH 50 mL 0.4 M CHCH₃₃COONaCOONa

 

ChCh COOHCOOH



COO COO CH CH 3 3 3 3   = 1  = 1 pH = pKa + log 1 pH = pKa + log 1 = pKa = 4.76 = pKa = 4.76 72.

72. 9-structural isomers are possible.9-structural isomers are possible. CH CH₃₃CHCH₂₂CC

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CCHH,, CCHH₃₃ –C –C



C–CHC–CH₃₃ CH CH₂₂=CH–CH=CH=CH–CH=CH₂₂,, CHCH₂₂=C=CH–CH=C=CH–CH₃₃ 73. 73. 74.

74. CoCo+3+3 _{: 3d: 3d}66_{4545°° ststrorong ng fifieleld ld ligiganand (d (NHNH} 3 3)) Ni Ni+2+2 _{: 3d: 3d}88_{45° 45° SFL SFL (NH(NH} 3 3)) Cr  Cr +3+3 _{: 3d: 3d}33_{45° 45° WFL (HWFL (H} 2 2O)O) Fe Fe+2+2 _{: 3d: 3d}66 _{45 45° ° WFL (HWFL (H} 2 2O)O) Co

Co+3+3 _{will bwill be diamagnetic e diamagnetic (i)(i)}

75.

75. ReReacactition on ququototririenent t Q Q ==

]] [[ ]] H H [[ ]] H H [[ 2 2 2 2 2 2



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 =  = 2 2 .. 0 0 1 1 .. 0 0 )) 4 4 .. 0 0 (( 22



= 8= 8 Q < Q < KK

reaction proceeds in the forward direction. reaction proceeds in the forward direction.