March 2011 - Career Point

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P( A) = Number of cases not favourable to event A Total number of cases n(A) = n(S) (B) Odd in favour and odds against an event : As a result of an experiment if “a” of the outcomes are favourable to an event E and b of the outcomes are against it, then we say that odds are a to b in favour of E or odds are b to a against E. Thus odds in favour of an event E Number of favourablecases a = = Number of unfavourable cases b Similarly, odds against an event E Number of unfavourable cases b = = Number of favorablecases a Note : If odds in favour of an event are a : b, then the probability of the occurrence of that event is a and the probability of non-occurrence of a + b b that event is a + b . a a + b If odds against an event are a : b, then the probability of the occurrence of that event is b and the probability of non-occurrence of a + b a that event is a + b . (C) P(A) + P( A ) = 1 0 ≤ P(A) ≤ 1 P(φ) = 0 P(S) = 1 If S = {A 1 , A 2 , ..... A n }, then P(A 1 ) + P(A 2 ) + .... + P(A n ) = 1 If the probability of happening of an event in one trial be p, then the probability of successive happening of that event in r trials is p r . (D) If A and B are mutually exclusive events, then P(A ∪ B) = P(A) + P(B) or P(A + B) = P(A) + P(B) If A and B are any two events, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B) or P(A + B) = P(A) + P(B) – P(AB) If A and B are two independent events, then P(A ∩ B) = P(A) . P(B) or P(AB) = P(A) . P(B) If the probabilities of happening of n independent events be p 1 , p 2 , ...... , p n respectively, then (i) Probability of happening none of them = (1 – p 1 ) (1 – p 2 ) ........ (1 – p n ) (ii) Probability of happening at least one of them = 1 – (1 – p 1 ) (1 – p 2 ) ....... (1 – p n ) (iii) Probability of happening of first event and not happening of the remaining = p 1 (1 – p 2 ) (1 – p 3 ) ....... (1 – p n ) If A and B are any two events, then ⎛ B ⎞ P(A ∩ B) = P(A) . P ⎜ ⎟ or ⎝ A ⎠ ⎛ B ⎞ P(AB) = P(A) . P ⎜ ⎟ ⎝ A ⎠ ⎛ B ⎞ Where P ⎜ ⎟ is known as conditional probability ⎝ A ⎠ means probability of B when A has occurred. Difference between mutually exclusiveness and independence : Mutually exclusiveness is used when the events are taken from the same experiment and independence is used when the events are taken from the same experiments. (E) P(A A ) = 0 P(AB) + P( AB ) = 1 P( A B) = P(B) – P(AB) P(A B ) = P(A) – P(AB) P(A + B) = P(A B ) + P( A B) + P(AB) Some important remark about coins, dice and playing cards : Coins : A coin has a head side and a tail side. If an experiment consists of more than a coin, then coins are considered to be distinct if not otherwise stated. Dice : A die (cubical) has six faces marked 1, 2, 3, 4, 5, 6. We may have tetrahedral (having four faces 1, 2, 3, 4,) or pentagonal (having five faces 1, 2, 3, 4, 5) die. As in the case of coins, If we have more than one die, then all dice are considered to be distinct if not otherwise stated. Playing cards : A pack of playing cards usually has 52 cards. There are 4 suits (Spade, Heart, Diamond and Club) each having 13 cards. There are two colours red (Heart and Diamond) and black (Spade and Club) each having 26 cards. In thirteen cards of each suit, there are 3 face cards or coart card namely king, queen and jack. So there are in all 12 face cards (4 kings, 4 queens and 4 jacks). Also there are 16 honour cards, 4 of each suit namely ace, king, queen and jack. XtraEdge for IIT-JEE 50 MARCH 2011
STATEMENT OF OWNERSHIP Statement about ownership and other particulars about newspaper (XTRAEDGE FOR IIT JEE) required to be published in the first issue every year after the last day of February. FORM IV/SEE RULE 8 1 Place of Publication : 112A, Shakti Nagar, Kota, Rajasthan 2 Periodicity of its publication : Monthly 3 Printer’s Name : Mr. Nawal Kishore Maheshwari Nationality : Indian Address : 112A, Shakti Nagar, Kota, Rajasthan- 324009 4 Publisher’s Name : Mr. Pramod Maheshwari Nationality : Indian Address : 112A, Shakti Nagar, Kota, Rajasthan- 324009 5 Editor’s Name : Same as above Nationality : Same as above Address : Same as above 8 Names and addresses of individuals who own the newspaper and partners or shareholders holding More than one per cent of the total capital. : - I, Pramod Maheshwari, hereby declare that the particulars given above are true to the best of my knowledge and belief. Sd/- Pramod Maheshwari Signature of Publisher March 1, 2011 XtraEdge for IIT-JEE 51 MARCH 2011
Page 3 and 4: Volume - 6 Issue - 9 March, 2011 (M
Page 5 and 6: S Volume-6 Issue-9 March, 2011 (Mon
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Page 11 and 12: l 3 F × 2 = Iα F × 2 3l = (2ml 2
Page 13 and 14: P T = P e + P m = 22.69 + 43.48 = 6
Page 15 and 16: MATHEMATICS 11. Show that the value
Page 17 and 18: XtraEdge for IIT-JEE 17 MARCH 2011
Page 19 and 20: (A) Qr x +Q x (B) r 2Qr x S x (C) +
Page 21 and 22: PHYSICS Students'Forum Expert’s S
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Page 25 and 26: Perfect gas equation : From the kin
Page 27 and 28: The internal energy of n molecules
Page 29 and 30: * The binding energy per nucleon is
Page 31 and 32: KEY CONCEPT Organic Chemistry Funda
Page 33 and 34: KEY CONCEPT Inorganic Chemistry Fun
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Page 45 and 46: 6 6 ⎛ 55 ⎞ ⎛ 11 ⎞ 2 4 3 3 -
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Page 49: MATHS PROBABILITY Mathematics Funda
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Page 55 and 56: 25. The number of stereoisomeric pr
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Page 61 and 62: MOCK TEST FOR IIT-JEE PAPER - II Ti
Page 67 and 68: SECTION - IV Integer answer type an
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Page 79 and 80: 66. The solution to the differentia
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Page 85 and 86: 37. Two blocks are connected by a m
Page 87 and 88: 21. CCl 3 1eqv.of Br2 / ⎯⎯ ⎯
Page 89 and 90: ⎪⎧ 2 4 a a ⎪⎫ 11. 2 ⎨1 +
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Page 95 and 96: or circle 3 − 2i 2 1 z − = whic
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17. [1] ⇒ tan θ =2 dv → For mi
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SOLUTION FOR MOCK TEST PAPER - II A
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t eq . - 2p p 2p + p = 3 p = 1atm k
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c 1 = 1 & c 2 = 1 3 x ∴ required
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PHYSICS 1. [A] u 2 = 5gR ∴ v 2 =
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N cos θ = mg …..(1) [⊗ indicat
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1. [C] mvr = CHEMISTRY nh = 4 × 2
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2. [A] f(x) = 1 + x sin x [cos x] Q
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∴ Area of rectangle = 2te -t = f(
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39.[C] f(x) = cos -1 ⎛ | x | ⎞
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Only 2 is correct : Sentence 2 is a
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March 2011 - Career Point

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