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 Rajat Sen's messages in the community 1 2 3 ... 25 26 27 GO Go to Page...
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 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

$xyz= 2^{1}.5^{1}$
Listen if x,y,z are positive :
then let the power of 2 in x,y,z be respectively $a,b,c$
so, by the problem:
a+b+c=1.
Number of non-negative integral solution 0f this = $^{1+3-1}C_{3-1}=3$
similarly, let the powers of 5 in x,y,z be $m,n,o$ respectively.
so, no of solution of this = $^{3}C_{2}$=$3$
so, total number of solutions = $3 \times 3 = 9$
rate if u like it!

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

This is  an INMO problem :

Now ,

$x^{2}+xy+y^{2}=\frac{3}{4}(x+y)^{2} +\frac{1}{4}(x-y)^{2} \geq \frac{3}{4}(x+y)^{2}$
similarly for the other 2 brackets.
so, we get :
$3(x^{2}+xy+y^{2})(y^{2}+yz+z^{2})(x^{2}+xz+z^{2}) \geq \frac{81}{64}(x+y)^{2}(y+z)^{2}(x+z)^{2}$
so, we only have to prove that :
$8(x+y+z)(xy+yz+zx) \leq 9(x+y)(y+z)(z+x)$
after a bit of modification (Try yourself) this yields :
 which is obvious by AM  GM.

 Discussion Forums -> This Post 12 points    (2    in 3 votes )   [?]

let 

let $f(v)=\phi\left(\frac{1}{v}\right)$
Putting back in the equation we get :

so,
$\int \frac{dv}{f(v)}=\int \frac{dx}{x}$
Now if $\int \frac{dv}{f(v)} = \frac{1}{v}$ then we get the required solution!
so,
$\phi\left(\frac{1}{v}\right)=f(v)=-v^{2}$
so,
$\phi\left(\frac{x}{y}\right)=-\frac{y^{2}}{x^{2}}$
!

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

19.
$P(x)=a_{0}x^{n}+a_{1}x^{n-1}+....+a_{n}$
We require  to be prime for any integer.

Thus  is a prime.
But for any integer  :
$P(ka_{n})=a_{0}(ka_{n})^{n}+a_{1}(ka_{n})x^{n-1}+....+a_{n}$
Thus we see that  but this is a contradiction!

 Discussion Forums -> This Post 30 points    (6    in 6 votes )   [?]



so, by ROLLE'S theorem :
 for some  in 
so, by langranje's theorem :
$\frac{f'(x)-f'(c)}{x-c} =f''(a)$ for some  in . () for any  in 
Taking Mod on both sides we get :

so, we proove that  for any x in 
Rate if you like the solution!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

The function is not exactly symmetric with respect to p,q.

So, your assumption is not right, there is l;oss of generality.

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

See , when we generally count 3 digit number we put the restriction that 0 cannot be at the first place, if it is allowed to go to all the places we get the 2 digits and the single digits number aswell.

so, total number of numbers with five at unit place (1digit +2 digit +3digit ) = 10 * 10 = 100

similarly there are 100 each for five at tens place and thousand's place respectively.

So, total number of ways = 100+100+100 = 300

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

I did 2,3 and 6.Actually my solution to the inequality is similar to bhatt sir's solution though there is a better solution as pointed out by anant sir.I thought i had done the 5th one but there is unfortunately a small flow in my solution.I don't think this is enough to get through in class 12.question 3 is quite easy just on quadratic equations.

2 and 6 has already been solved

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

So, that we can start a discussion :

 INMO 2009 Problem 1 Let  be a tringle and let  be an interior point such that $angle BPC = 90 ,angle BAP = angle BCP$.Let  be the mid points of respectively.Suppose .Prove that  are collinear.

 INMO 2009 Problem 2 Define a a sequence  as follows  , if number of positive divisors of  is odd  , if number of positive divisors of  is even  (The positive divisors of  include  as well as .)Let  be the real number whose decimal expansion contains  in the -th place,.Determine,with proof,whether  is rational or irrational.

INMO 2009 Problem 3

Find all real numbers  such that:


INMO 2009 Problem 5

Let  be an acute angled triangle and let  be its ortho centre.Let  denote the largest altitude of the triangle .Prove that:



 INMO 2009 Problem 6 Let  be positive real numbers such that .Prove that:  .

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

$(1+a)(1+b)(1+c)(1+d) = (1+a+b+ab)(1+c+d+cd)$

Now, by A.M>G.M :





Multiply to get :

$(1+a)(1+b)(1+c)(1+d) \geq 16 \sqrt{abcd} = 16$

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

1. Proove $(n-1)^{2}C_{1}+(n-3)^{2}C_{3}+(n-5)^{2}C_{5}+..... = n(n+1)2^{n-3}$

Do not use induction!

2. If $(1+x+x^{2})^{n} = a_{0}+a_{1}x+.......+a_{2n}x^{2n}$

then proove that :

$^{n}C_{0}a_{r}-^{n}C_{1}a_{r-1}+..........+(-1)^{r}^{n}C_{r}a_{0}$

 when  is not a multiple of 3

 if 

 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]

See as two sides are common so we selet the two consecutive sides which are common .

Number of ways of selecting 2 consecutive sides =  = 

Now, we have to select the fourth point of the quadrilateral.

Out of  are chosen moreover we cannot choose the  points one immediately left and the other right of the chosen points.

so, number of choices for the third point = 

So, number of such quadrilaterals =  = 

Rate if correct!

 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]

Let the circle be : 

Suppose I want to fold the paper in such a way such that a point on the circle

 lie on the point  .

So, the crease will be a line which is the perpendicular bisector af  ,  being the mid - point of  .

So, $X \equiv \left(\frac{x_{0}+Rcos \theta }{2},\frac{y_{0}+Rsin \theta }{2}\right)$

Now it is easy to find the equation of the crease which I am getting as :

$x(Rcos \theta - x_{0}) + y(Rsin \theta - y_{0}) = \frac{R^{2} - a^{2} }{2}$

Moreover we have the condition that :  .

So, by varying  we can obtain the points .

I am trying to obtain the conditions on  explicitly.

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

For this equation to have rational root :

we must have that :

 for some integer p which is odd ( it is easy to see it is odd)

so,  and  for some k and q.

so, we get that 

but this is a contradiction because  is odd and so  does not divide 

so, assertion is the correct explanation!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

Let  be the acceleration of the incline and  be that of the block relative to the wedge.

Suppose the normal reaction = 

so, we ghet the equations :

.........1

$ma_{2} cos \alpha + mg sin \alpha = ma$.....2

$N+ma_{2} sin \alpha = mg cos \alpha$ ....3

solving this we get :

$a_{2} = \frac{mg sin \alpha cos \alpha }{M + msin^{2} \alpha}$

$a = \frac{gsin \alpha (m+M)}{M+msin^{2} \alpha }$

now, by equating the time in both the cases we get :

$v_{wedge} = a_{2}\sqrt{\frac{2h}{asin \alpha}}$

 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]



You may be knowing that :

$1^{3}+2^{3}+.....+2000^{3} = \left( \frac{2000(2000+1)}{2}\right)^{2}$

The R.H.S is a perfect square!

Form the above we get that :

$k^{3} +(2k)^{3} + ....+(2000k)^{3} = k^{3}\left( \frac{2000(2000+1)}{2}\right)^{2}$

now again :

$k^{2}+(2k)^{2}+........+(2000k)^{2} = \left(k^{2}\frac{1}{6}(2000)(2000+1)(4000+1)\right)$

just choose  and we are done !

Rate if correct!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

As you know the G.C.D of  and  is 

so, by Fermat's theorem ,



so, we get :

$2^{1990} \equiv 2^{10}(mod 199)$

so, we see that :



But the last digit of  is 

so, 

so, we get that 

so, the remainder is  .

 Discussion Forums -> This Post 9 points    (1    in 3 votes )   [?]

This is the 'if' part :

suppose 

suppose,  and 

let us assume that 

so,highest power of  in the polynomial  is 

so, degree of  is  itself.

suppose  is a root of 

then we get that $P(x_{o})^{2}=Q(x_{o})^{2}+R(x_{o})^{2}$

From here we arrive that 

but there should be  such real roots of 

But  can have a maximum of  roots and 

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

suppose

case 1 :

 has the prize .

Then the host could have opened any of the doors  and  .

so,

In this case my probability of winning if I change is 

conditional probability of case 1 = $\frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$

case 2 :

Now If  does not have the prize.

Then the host will definitely open door  only if  has the prize.

conditional probability of case 2 = $\frac{\frac{2}{3} * \frac{1}{2}}{\frac{2}{3}}=\frac{1}{2}$

so, probability of winning if I change = $0 \times \frac{1}{2} + 1 \times \frac{1}{2}=\frac{1}{2}$

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]



This implies that : $arg\left(\frac{z+i}{z-i}\right) = \frac{ \pi}{2}$

This implies that  is purely imaginary.

so we can conclude that $\frac{z+i}{z-i}+\frac{\bar{z}-i}{\bar{z}+i}=0$

from here we get that 

This is a circle of radius one having the origin as its center.

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