1) The value of 50C6 - 5C1.40C6 + 5C2.30C6 - 5C3.20C6 + 5C4.10C6 is
2) The minimum value of (8x^{2}+y^{2}+z^{2})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2} is
3) In a ΔABC, tanB/tanC = 2/3 and tanA.tanB.tanC=6 then the value(s) of tanA will satisfy the equation
A) 2x3-13x2+36x-25 = 0 B) x3 -12x2+36x-25 = 0
C) x3-12x2+35x-24 = 0 D) 2x3-13x2+35x-24 = 0
[I've doubt with the answers, so i'm not posting the given answers as ur solution may tend to achieve the final answer by some way (:D). please have patience to post the complete solution (minimizing the no. of steps). If ur solution is correct, then obviously ur answer will also be correct :) ]
please help.!
Thanq :)
31st que :
given expression is same as coefficient of x^6 in the expansion of :
((1+x)10 - 1)5
Expand and get the answer...
341yeah. the equality condition in AM-GM requires the summands to be equal.
Here's a method that uses only Cauchy Schwarz only
(1^2+1^2+1^2+1^2)(4x^2+4x^2+y^2+z^2) \ge (2x+2x+y+z)^2 by CS
Further (2x+2x+y+z) \left(\frac{1}{2x} + \frac{1}{2x} + \frac{1}{y} + \frac{1}{z} \right)\ge (1+1+1+1)^2 = 4^2 again by CS [u can also use AM-HM or AM-GM)
Therefore we have
4(4x^2+4x^2+y^2+z^2) \left(\frac{1}{2x} + \frac{1}{2x} + \frac{1}{y} + \frac{1}{z} \right)^2 \ge \left[ \left(2x+2x+y+z) \left(\frac{1}{2x} + \frac{1}{2x} + \frac{1}{y} + \frac{1}{z} \right) \right]^2 \ge 4^4
and hence the inequaliyty holds
3othrwise i had thought abt sumthing else...
if u c carefully, in the first expression, equality holds whn 8x2 = y2 = z2 = some k
in the second one, equality holds whn x-1 = y-1 = z-1 = some constant
now 54 is the min value only if both are satisfied simultaneously, ie for the same values of x, y and z...which is clearly not possible...
so i think 64 is 100 % correct ans..
11I think i found it
in nihit's solution
Second part
squaring the inequality can reverse the inequality because its value can be <1 also
so that solution will not be valid
:D
1hm ! even the explanation given in solution booklet matches with the one given by prophet sir. :)
But still, why is AM-GM not working here !?
11I think he meant at what values do we get the value 54?
the values of x,y,z
1well, i'm sorry. Even the condition x,y,z>0 is given in the question. But still, answer is given as 64, not 54 ! dont know why?
prophet sir, please explain ur statement : "Also the equality condition is not satisfied, so that 54 is definitely a lower bound , but not the greatest lower bound for this expression."
i think this statement has everything that declares 64 as correct answer. !
but i couldn't get it clearly. please explain sir !
3surely the inequality u've used is far more sound mathematically...but whr is the concept going wrng with AM - GM thing ??
3see...
(8x2 + y2 + z2)/3 ≥ (8x2 * y2 * z2 )1/3
now calculating..v get :
(8x2 + y2 + z2) ≥ 6 (xyz)2/3 ... (1)
Second part :
(x-1 + y-1 + z-1 )/3 ≥ (1/xyz)1/3
again shfting 3 to rite and squaring...v get
(x-1 + y-1 + z-1 )2 ≥ 9 (1/xyz)2/3 ... (2)
multiplying (1) and (2) we get the result naa...strangely it comes out to be independent of x, y and z...
plzz sum1 temme wats wrong...
341Well, I dont know if there is any other method. the answer 54 is got by applying AM-GM on the two factors. But that needs x,y and z to be positive. Also the equality condition is not satisfied, so that 54 is definitely a lower bound , but not the greatest lower bound for this expression.
3hey...u welcum..but wait !...
i m still not convinced y that ans cannot b 54...
i mean, even i got that whn i workd with AM GM inequality...i cant find my mistake...
1thanx nihit :)
well, it's done then.
Thanq everyone who involved in this forum and also the one's who tried these problems. Thanx once again :)
11Q3
my answer B(:D)
tanAtanBtanC=6
tanA+tanB+tanC=6
tanB=2/3 tanC
Substitute tanB in the first two equations
and solve for tanA
1thanx a lot sir. :)
but if that's not applicable for jee syllabus, how to solve such problems within jee level.?
This question was asked in our "iit-jee grand test".!
Did they expect us to solve it by this method or is there any other simple method too (which is within jee level) ?
341For the inequality what is needed is an inequality known as Hölder Inequality. It is a generalisation of Cauchy Schwarz inequality
I will give it in a form that is useful in this problem:
\left(x_{11}^3 + x_{12}^3 + x_{13}^3 \right)^{\frac{1}{3}} \left(x_{21}^3 + x_{22}^3 + x_{23}^3 \right)^{\frac{1}{3}} \left(x_{31}^3 + x_{32}^3 + x_{33}^3 \right)^{\frac{1}{3}} \ge \left(x_{11}x_{21}x_{31} + x_{12}x_{22} x_{32} + x_{13}x_{23}x_{33} \right)
So when you apply it, we get
\left(8x^2+y^2+z^2 \right) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \ge (2+1+1)^3 = 64
Note: In JEE Cauchy Schwarz itself may not find any application. So Hölder may be overkill as far as JEE preparation is concerned.
1The answer given is 64 itself, but most of my friends argued it to be 54.!
i couldn't have a detailed discussion with them, so posted it here if anyone could solve it in an easier/compatible way.!
brother, u post ur generalization thought, it may be right as the answer is correct :)
but please, post it in a detailed way, as i'm newly exposed to this cauchy schwarz !
62The Race...
I seem to have lost my own arguement..
I think what i have done is wrong..
what is the right answer??
For a moment i thought that the generalization of cauchcy schwarz wud work.. but now i am not feeling the same..
1@Nishant bro. Thanx for the links. I got the inequality now. :)
but, i didn't get ur soluton yet ! please post 2-3 intermediate steps at the point where u wrote "now apply cauchy schwarz". that will be helpful for everyone, especially me ! please
11=50!/6!44! - 5!/4!.40!/6!34! + 5!/2!3!.30!/6!24! - 5!/3!2!.20!/6!14! + 5!/4!.10!/6!4!
=1/6!(50!/44!- 5!/4!.40!/34! +5!/2!3! . 30!/24! - 5!/3!2!.20!/14! + 5!/4!.10!/4!)
=1/6!([50*49*48*47*49*45]-[40*39*38*37*36*34] + [2*5*30*29*28*27*26*25] -[5*2*20*19*18*17*16*15] + [10*9*8*7*6*5])
10/6!([5*49*48*47*49*45]-[4*39*38*37*36*34] +[*30*29*28*27*26*25] -[*20*19*18*17*16*15] + [*9*8*7*6*5]
50/6!([49*48*47*46*45]-[4*39*38*37*7.2*34] +[6*29*28*27*26*25] -[4*19*18*17*16*15] + [*9*8*7*6])
200/6!([49*12*47*46*45]-[39*38*37*7.2*34] +[6*29*7*27*26*25] -[19*18*17*16*15] + [*9*2*7*6])
1800/6!([[49*12*47*46*5]-[13*38*37*2.4*34] +[6*29*7*3*26*25] -[19*6*17*16*5] + [**2*7*6])
=14400/6!([49*3*47*23*5]-[13*38*37*0.3*34] +[3*29*7*3*13*12.5] -[19*6*17*2*5] + [3.5*3])
i am getting mad
11bhaiyya
how do you take it from there in post two
im not that good at this inequality
but isnt that a whole square in the second term
so explain a bit more :)
62http://targetiit.com/iit_jee_forum/posts/22nd_december_2008_1146.html
http://targetiit.com/iit_jee_forum/posts/inequalities_3234.html
read these two..
62sorry i explained the above very clumsily.. not much was clear..
but let me try to find some good explanation..
1@Nishant bro, can u briefly explain the basic definition of cauchy schwarz and explain it's applications in jee level.
I've done a lot of google search but couldn't get any conclusion. so please explain me it in ur words.!
62(8x^{2}+y^{2}+z^{2})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2} \\=(4x^{2}+4x^2+y^{2}+z^{2})(\frac{1}{2x}+\frac{1}{2x}+\frac{1}{y}+\frac{1}{z})^{2}
>=
now apply cauchy schwarz..
8x^3=y^3=z^3
2x=y=z=t
4t^2 x (4/t)^2 = 64