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1000n/n! = (1000/1).(1000/2).(1000/3)...................(1000/n) For n = 1,2,3...........1000 this value is increasing since every term uptil here is greater than 1. But for n>1000, 1000/n < 1 and the series would start decreasing since the series is multiplied by numbers less than 1. Hence it attains a maximum value at n = 1000. Best Wishes
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Sorry @truly to create confusions.
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konichiwa2x has already writen the expression for the sum of factorials. If you want to gain knowedge on this topic you can refer to the link : @ sachin_batwani, the summation of (r).(r!) gives (n+1)! - 1 and not of r!. Best Wishes
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I think the correct equation is x 2 + (a+b)x + (1-a-b) = 0 where (a,b,x)  R Discriminant >= 0 => (a+b)2 - 4(1-a-b) = 0 => b2 + (2a+4)b + (a2 + 4a - 4) >= 0 Since coefficient of b2 is positive hence discriminant of above equation should be less than zero. => (2a + 4)2 - 4(a2 + 4a - 4) <= 0 => 16 <= 0 which is not possible. Hence no real values of a exist. Best Wishes
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Let 1/x = z Then limit becomes [z] [0] z.sin(1/z) Now sin(1/z) is an oscillating function whose values lies between -1 and 1. Hence as z tends to 0 the limit also becomes zero. Hence the answer is 0. Best Wishes
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If the forward reaction is photochemical then the incident light would favour the reaction towards products side i.e. forward reaction is favoured.
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Friction is necessary for rolling whatever be the surface horizontal or inclined.Friction provides the necessary torque for rolling.
Anubhav, friction does not vanish on pure rolling. Its work done is zero in pure rolling since there is no displacement at the point of friction force acting.
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If the case is observed from the earth then their velocoties are equal.
But from outside the earth one train would be having angular velocity in the direction of earth's angular velocity and the angular velocity for the other would be in opposite direction.
Hence the pressure exerted would be different.
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Its initial horizontal velocity is 20cos600 = 10 m/s
Initailly total speed = 20 m/s
When the projectile becomes horizontal (its vertical velocity becomes zero), the projectile is at maximum height and its speed at this point is half of its initial speed.
Vertical velocity = 20sin600 - gt = 0
t = 20sin600/10 =  3 sec.
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FIrst of all complete H.C.Verma as it is a good concept building book.
After that you can go for Arihant(D.C.Pandey) and Irodov for higher level problems.
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It is not only a function of P but its a function of P/ .Suppose you vary pressure at constant temperature then density, is itself varied in such a way that P/ is constat.
PV = nRT
=> PM = RT
=> P/ = RT/M
Hence it depends on ratio of P/ which in turn is directly proportional to temperature.
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First of all I would like to tell you that clearing JEE in mere one month preparation is difficult.But still if you have beforehand knowledge of some of the topics then what all you have to do is show dedication and make the best possible use of time by sacrificing all your comforts.Make those topics strong.Just see some previous year papers and sample papers and rush on to those topics which you have not covered and have a good weighage in papers.
But simultaneously you should handle your college studies so that at least one way you are going safer.
I must admit that IIT is best even then if you are willing to show dedication and punctuality then you can master in any field even if you are a non-IITian.
Just go ahead
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sin A/2 + sin B/2 + sin C/2 will be minimum.
Applying AM>=GM and it will be minimum when all the entities are equal.
Vish.....when there are 3 negative numbers then its GM will be root of some negative number which is undefined.Hence the inequality is applicable only for positive numbers.
Hope its clear now.
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Due to screening effect the inner electrons shield the outer electrons and the attractive force of the nucleus is somewhat decreased.Thus 5d orbital is filled first instead of 4f orbital thus violating Aufbau Principle.
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Let the leftmost no. be a and its palce value to be a*10b .
After removing the leftmost digit the remaining no. is divisible by 57 hence n must be odd.
As per the question : a*10b + (n/57) = n
=> 57*a*10b = 56n
=> (19).(3a).(5b).(2b-3.23) = (7).(23).n
=> (19).(3a).(5b).(2b-3) = 7n
Since n is odd then RHS will also be odd.
Hence LHS must be odd implies power of 2 should be zero.
Hence b-3 = 0 => b = 3
Putting this above we get
(19).(3a).(125) = 7n
=> 7125a = 7n
Since 7125 is not divisible by 7 and RHS has 7 as a factor then a must be 7.
Hence n comes out to be 7125.
Sum of digits = 15
Hope its clear now.
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Let the leftmost no. be a and its palce value to be a*10b .
After removing the leftmost digit the remaining no. is divisible by 57 hence n must be odd.
As per the question : a*10b + (n/57) = n
=> 57*a*10b = 56n
=> (19).(3a).(5b).(2b-3.23) = (7).(23).n
=> (19).(3a).(5b).(2b-3) = 7n
Since n is odd then RHS will also be odd.
Hence LHS must be odd implies power of 2 should be zero.
Hence b-3 = 0 => b = 3
Putting this above we get
(19).(3a).(125) = 7n
=> 7125a = 7n
Since 7125 is not divisible by 7 and RHS has 7 as a factor then a must be 7.
Hence n comes out to be 7125.
Sum of digits = 15
Hope its clear now.
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You just draw the figure.Since tangent and normal at P are perpendicular to each other,so  TPG = 900.
This is the angle contained in a semicircle and hence TG would be the diameter of the circle passing through P,T and G.
In equation of tangent and normal put y=0 to get T and G.
Then T is (-at2,0) and G is (2a - at2,0)
Hence centre of the circle is (a,0) which is the focus of parabola.
Slope of normal to the circle at P = 2at/(at2-a) = 2t/(t2-1)
Hence slope of tangent to the circle at P = -1/(slope of normal) = (1-t2)/2t
Also slope of tangent at P to parabola = 1/t
Now slopes of two lines are known and you can easily find the angle between them.
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Shakir has done it perfectly.
Amaron's doubt is wat form would |infinity - infinity| would take.
If you take the limiting value of shakir's solution then you will get what shakir has got.
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Substitution : z = (x2+1)
=> x2 = z2 - 1
Hence 2x.dx = 2z.dz => x.dx = z.dz
Now I =  x5/ (x2+1).dx
=> I =  [(x2)2/ (x2+1)].(x.dx)
=> I = [(z2-1)2/z](z.dz)
=> I = [(z2-1)2 dz
=> I = (z4-2z2+1).dz = (z5/5) - (2z3/3) + z + c
Back substitution : z = (x2+1)
Gives I = [(x2+1)5/2/5] - [2(x2+1)3/2/3] + (x2+1)1/2 + c
Please post your other query on a new page as we are instructed to answer one query at a time.Hope you understand.
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If you will draw the graphs of the functions y = cos(x-1) and y=|x-1|/10
then you will find that they intersect at two points,one in the 1st quadrant and the other in the 2nd quadrant.Hence 2 values of x are possible.
Hence answer is (b) 2
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