Solution - Career Point

or 0.1 v α + 1 = 2 + 2 × 10 –2 × v(0.1vα + 1) 0.1vα×(100 + v) (1.00 + v) α –2K a == × 2 × 100.1v(1 − α)( 1−α)(100 + v)α∴ × 2 × 10 – 2= 10 –2 or( 1−α)or 2α = 1 – α or α = 1/3∴ 0.1 v × 31 + 1 = 2 + 2 × 10–2 × vα1− αv v 20vor – = 1 or = 130 50 30×50150or v = = 75 ml218.[D] For isohydric solution, Ka 1 C 1 = Ka 2 C 2= 21Column Matching19. [A] → r,s,t; [B] → p,r,s; [C] → s; [D] → q,tNaNO 2 / HClFor CH 3 CH 2 CH 2 NH 2 ⎯⎯⎯⎯→CH 3CH 3CH 3+CH3CH2CH2N2Diazonium ion issimply int ermediate;not the productCH 3 –CH 2 –CH 2 –OH+ O HN – CH 3 on heating doesn't give alkene.20. [A] → q,r; [B] → p,r; [C] → r,s; [D] → p,t−OH / Br 2C 6 H 5 CH 2 –CHO⎯⎯⎯→ C 6 H 5 – CH –CH = O|BrCH 3 CHOOH−⎯⎯⎯⎯→0 − 25º CCH 3 –OH|CH( ± )( ± )–CH 2 –CH = OCH 3 –CH 2 CH = O⎯ OH25ºC→ −CH 3 CH 2 –CH= C − CHO|CH 3Conc. OHCH 3 – CH − CH = O⎯⎯⎯⎯|CH 3OCH 3 – CH − C || –O – + CH 3 – CH − CH 2 OH||CH 3CH 3→ −MATHEMATICS1.[A] Let us first count the number of elements in F.Total number of functions from A to B is 3 4 = 81.The number of functions which do not containx(y) [z] in its range is 2 4 .∴ the number of functions which contain exactlytwo elements in the range is 3 . 2 4 = 48.The number of functions which contain exactlyone element in its range is 3.Thus, the number of onto functions from A to B is81 – 48 + 3 = 36[using principle of inclusion exclusion]n (F) = 36.Let f ∈ F. We now count the number of ways inwhich f –1 (x) consists of single element.We can choose preimage of x in 4 ways. Theremaining 3 elements can be mapped onto {y, z}is 2 3 – 2 = 6 ways.∴ f –1 (x) will consists of exactly one element in4 × 6 = 24 ways.Thus, the probability of the required event is24/36 = 2/32.[A]3.[D]4.[D]Let E 1 denote the event that the letter came fromTATANAGAR and E 2 the event that the lettercame from CALCUTTA. Let A denote the eventthat the two consecutive alphabets visible on theenvelope are TA. We have P(E 1 ) = 1/2, P(E 2 ) =1/2, P(A / E 1 ) = 2/8, P (A / E 2 ) = 1/7. Therefore,by Bayes' theorem we haveP(E 2)P(A / E2)P(E 2 / A) =P(E1)P(A / E1)+ P(E2) P(A / E2)4= 11Required probability = 1 – P (all the letters areput in correct envelops)The number of the ways of putting the letters inthe envelops = 4 P 4 = 4!The number of ways of putting letters in correctenvelops = 11 23∴ Required probability = 1 – = 24 24We have⎡5x0AB =⎢⎢0 1⎢⎣0 10x − 2⇒ x = 1/50 ⎤ ⎡10⎥⎥=⎢⎢05x⎥⎦⎢⎣00100⎤0⎥⎥1⎥⎦5.[B] Greatest term in the expansion of (x + y) n isk th ⎡(n+ 1)y ⎤term where k = ⎢ ⎥⎣ x + y ⎦XtraEdge for IIT-JEE 89APRIL 2010