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I feel there can be only one case. When the polarity of the solvent increases sn2 is not favoured and the rate of the reaction decreases. In case of sn1, polarity of the solvent favours the transition carbocation formed, hence increases the rate of the rxn. In the first two examples, sn2 will be possible. But in the last 2 e.gs sn1 will be favoured as in the intermediate stage, three degree (tertiary carbocation) will be formed which is very stable ( by 9 hyperconjugated structures and +I effect of the three methyl groups which stabilises the carbocation)
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Hiiii, I'm sure all of you must have arrived at the step where, 23xx2 = 564 ............. eqn (1) Let me do the rest for you. 2log5x will be defined when x>0 Now, from eqn. (1), it is clear that the R.H.S is an odd integer. As such, x cannot be irrational & also x cannot be a postive integer otherwise then the L.H.S will be even which is a contradiction. Hence, there is no solution. Plz tell me iff I'm wrong anywhere!!
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Hiiii, f ' (x) = [ {ln(e+x)}/(pie+x) - {ln(pie+x)}/(e+x) ] / [ln(e+x)]^2 pie means  Now, for x>=0, {ln(e+x)}/(pie+x) - {ln(pie+x)}/(e+x) < 0 i.e f(x) is a decreasing function, Now, the domain of the function f(x) is (-e, infinity) For, x>=1-e, f ' (x) < 0 i.e f(x) is decreasing function. Again, for -e<x<1-e, f ' (x) < 0 i.e f(x) is decreasing function i.e f(x) is a decreasing function throughout its domain i.e (-e, infinity) Plz tell me if I'm wrong anywhere. 
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I am very confident about my method & I understand what Magico is trying to say. But here this method is vey well applicable, as a,b,c,d denote the no. of stations. Satisfy yourself by taking small examples and try to find if there is any contradiction. I am sure that you will find none!
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The answer is 744. This question was discussed earlier. So plz refer to the previous problems that has been discussed in the Algebra section.
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Sir/Madam, I did not take the test conducted by GOIIT on 16th & 17th December as I did not know about this website then. Can you plz tell me how can I get the questions & solutions of the above test. Also, plz tell me whether such tests will be conducted soon before the IIT exam to test our preparations?? Plz reply at your earliest convinience
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The above one is not a correct answer!
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Hii, The thermo-emf, V = at^2 + bt a,b depend on the material. t = temp. diff between the hot & the cold junction. So, the thermoemf does not vary linearly with temp. diff. From, the given data we can find a relation between a & b but can't find its individual values. So, I don't think this is solvable. Plzzzzzzzz tell me if I am wrong ??
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Hiii, The solution is easy. One set of persons ( whether they be 2,3or 4 people) can leave the floor in 10 ways. One of the two remaining sets can leave in 10-1 or 9 ways. The remaining people can leave the lift in 8 ways. Total no. of ways = 10 x 9 x 8 = 720
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Let there be 'a' stations before the 1st stoppage, 'b' stations between the 1st & 2nd stoppage, 'c' stations between 2nd & 3rd stoppage & d stations after the 3rd stoppage. Now, by the conditions of the given problem: a & d are non-negative. b & c are positive. Let b = e + 1 c = f + 1 By the problem, a + b + c + d = 100-3=97 i.e a+d+e+f=95 Hence, the total no. of ways = total no. of non-negative integral solns of the above eqn. = 98C3
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First, let us put one ball in each of the 3 boxes. So, now no box remains empty & we have to just distribute 27 balls in 3 identical boxes. The no. of ways by which this can be done = (The no. of non-negative integral solutions of the eqn. a + b + c = 27 ) = [(3+ 27-1)C27]/3 = 29C27 = 406 Here, I have made an assumption that the case when x no. of balls are in the 1st box, y no. of ball are in the 2nd box and the remaining balls in the 3rd box & the case when y no. of balls are in the 1st box, x no. of ball are in the 2nd box and the remaining balls in the 3rd box are distict cases and both of them are taken into account. Strictly speaking, these two are not distinct cases as the boxes are identical. So, actual no. of cases = 406/3 But since this is not an integer, I have made the above assumption.
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Yes, KAB. I didn't notice that the denominator of both the expressions (a/c & b/c) were same. Your method is perfectly correct. I owe an apalogy to you for commenting that ur method was wrong. Sorry !!!!!!!! When a>b>c>0, (a-c)/(a+c) > (b-c)/(b+c). You are absolutely correct !!!!!!  Take one salute from me !!
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Thanks, Malay ur answer of 220 is correct. So, take a salute from me !!!
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There are 14 intermediate stations between C & B. A train starts from C towards B and stops at exactly 3 of the intermediate stations. Find the no. of ways in which this can happen provided that no two of the haulting stations are consecutive?? ( haulting station means the intermediate station where the train has stopped)
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Hiiii, A function, say f(x), is continuous at x = a, if Lt x  a+ f(x) = Ltx a- f(x) = f(a) Here, Ltx 0+ sinx/x = Ltx 0- sinx/x = Ltx 0 sinx/x = 1 But the function sinx / x is not defined at x = 0. Hence the limitting value of the function at x=0 is not equal to the functional value at x=0. So, the function sinx/x is not continuous at x=0. When a function is not continuous at any pt, its derivative also does not exist at that pt. Hence sinx/x is not differentiable at x = 0. I hope that satisfies you !! 
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Hiiii, Let 1/(1+x^3) = A/(1+x) + (Bx+C)/(x^2 - x + 1) i.e 1 = (A+B)x^2 + x(B+C-A) + (A+C) Now, comparing the coefficients of x^2, x & x^0, we have, A+B=0 ...........eqn (1) A+C=1 ...........eqn (2) B+C-A = 0 .........eqn.(3) Solving the above three eqns, A = 1/3, B = - 1/3 & C = 2/3 Therefore, [ ] [ ] 1/(1+x^3) dx = [ [ ][ ] 1/(1+x)dx + [ ][ ] (-x+2)/(1+x^2-x)dx ]/3 The first integral is in the standard form. Break up the 2nd integral into two parts as follows: [ ][ ] (-x+2)/(1+x^2-x)dx = - [ [ ][ ] (2x-1)/(1+x^2-x)dx]/2 + 3[ [ ][ ]1/(1+x^2-x)dx]/2 For the first part, write 1+x^2-x = t, then (2x-1)dx = dt For the 2nd part, 1 - x + x^2 = (x-1/2)^2 + 3/4 Take x-1/2 = z and reduce the 2nd integral part to standard form. If you still face any problem, plz let me know. 
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Hiii dfgdf, Take some examples when some of the quantities are positive or negative, all of them are positive or negative. I will just try to give you an outline, a/c - b/d > 0 i.e (ad - bc )/cd > 0 Take the following cases and the criteria that a>b , c>d 1. ad - bc > 0 with c,d of the same sign ( i.e c,d both positive or negative) 2. ad - bc < 0 with c,d of the opposite sign (i.e c >0 and d < 0, the converse is not true according to our criteria) I'm sure you will be able to proceed further on yourself. If you face any difficulties plz let me know !!!1  |
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Hiiiiii, Although this problem lies in the Algebra section, I would like to solve this problem using calculus. Let us define a function, f(x) = [(x-c)/(x+c)]^x where x>c and x,c>0 It is obvious f(x)>0 for all x>c>0 If now, I can show f ' (x) >0 i.e f(x) is an increasing function, then it is obvious that f(a) > f(b) where a>b>c>0 i.e [(a-c)/(a+c)]^a > [(b-c)/(b+c)]^b where a>b>c>0 ln f(x) = x[ln(x-c)-ln(x+c)] Differentiating both sides w.r.t x, we have, f ' (x) = [(x-c)/(x+c)]^x [ln{(x-c)/(x+c)} + 2cx/(x^2 - c^2)] [(x-c)/(x+c)]^x >0 ; so our proof reduces to proving only that [ln{(x-c)/(x+c)} + 2cx/(x^2 - c^2)] > 0 ln{(x-c)/(x+c)} = ln{(1-y)/(1+y)} where y=c/x, so 0<y<1 Now, ln(1+y) = y - y^2 / 2 + y^3 / 3 ..... (using logarithmic series formula) Ln(1-y) = -y - y^2 / 2 - y^3 / 3 ...... Therefore, ln(1-y)-ln(1+y) = ln{(1-y)/(1+y)} = -2[y+y^3 / 3 + y^5 / 5 + ..... ] i.e. ln{(1-y)/(1+y)} > - 2 [ y + y^3 + y^5 + ...... ] i.e. ln{(1-y)/(1+y)} + 2y/(1-y^2) >0 Putting y = c/x, we have, [ln{(x-c)/(x+c)} + 2cx/(x^2 - c^2)] > 0 Hence, f ' (x) > 0 i.e f(x) is an increasing function. So, f(a) > f(b) for a>b>c>0 i.e. [(a-c)/(a+c)]^a > [(b-c)/(b+c)]^b I hope I have explained all the steps properly. Also plz tell me if I went wrong anywhere.
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Hiiii, I don't think KAB'S method is corrrect. This is because componendo-dividendo method in inequalities can give misleading results. Take the following e.g. : 4/5>1/2 Applying componendo-dividendo,(5-4)/(5+4)>(2-1)/(2+1) i.e. 1/9>1/3 is not true. Plz tell me if I am wrong any where!! 
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Sir/Madam, I have not given the pre-jee test that has been conducted in the early December of 2006 as I didn't know about the site then. I would like to know whether any such more tests will be conducted soon? How can I give that test?
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