Messages posted by Rajat Sen

$xyz= 2^{1}.5^{1}$
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!
! Idea

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 :
$(x+y)(y+z)(z+x) \geq 8xyz$ which is obvious by AM $\geq$ GM.

let $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
let $f(v)=\phi\left(\frac{1}{v}\right)$
Putting back in the equation we get :
$x\frac{dv}{dx}=f(v)$
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,
$\phi\left(\frac{x}{y}\right)=-\frac{y^{2}}{x^{2}}$
Idea !

$f(0)=f(1)$

so, by ROLLE'S theorem :
$f'(c)=0$ for some $c$ in $(0,1)$
so, by langranje's theorem :
$\frac{f'(x)-f'(c)}{x-c} =f''(a)$ for some $a$ in $(x,c)$ . ( $f'(c)=0$ ) for any $x$ in $(0,1)$
Taking Mod on both sides we get :
$|f'(x)-0|=|x-c||f''(a)|<1$
so, we proove that $|f'(x)|<1$ for any x in $(0,1)$
Rate if you like the solution!

Idea

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

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

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

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

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

Now, by A.M>G.M :

$(1+a+b+ab)\geq 4\sqrt{ab}$

$(1+c+d+cd)\geq 4\sqrt{cd}$

Multiply to get :

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

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

Number of ways of selecting 2 consecutive sides = $10-1$ = $9$

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

Out of $10$ , $3$ are chosen moreover we cannot choose the $2$ points one immediately left and the other right of the chosen points.

so, number of choices for the third point = $5$

So, number of such quadrilaterals = $9 \times 5$ = $45$

Rate if correct! Idea

For this equation to have rational root :

we must have that :

$b^{2}-4ac = p^{2}$ for some integer p which is odd ( it is easy to see it is odd)

so, $b^{2} = 8k+1$ and $p^{2} = 8q+1$ for some k and q.

so, we get that $4ac = 8(q-k)$

but this is a contradiction because $ac$ is odd and so $8$ does not divide $4ac$ .

so, assertion is the correct explanation!

please rate!

Let $a_{2}$ be the acceleration of the incline and $a$ be that of the block relative to the wedge.

Suppose the normal reaction = $N$

so, we ghet the equations :

$Nsin \alpha = Ma_{2}$ .........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 :

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

So, we get the answer

$2 :$

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 $k = (667)^{4}.(4001)^{4}$ and we are done !

Rate if correct! Idea

As you know the G.C.D of $2$ and $199$ is $1$ .

so, by Fermat's theorem ,

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

so, we get :

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

so, we see that :

$2^{1990} = 199k + 1024$

But the last digit of $2^{1990}$ is $4$ .

so, $10|2^{1990}-1024$

so, we get that $1990|2^{1990}-1024$

so, the remainder is $1024$ . Idea

suppose

case 1 :

$A$ has the prize .

Then the host could have opened any of the doors $B$ and $C$ .

so,

In this case my probability of winning if I change is $0$ .

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

case 2 :

Now If $A$ does not have the prize.

Then the host will definitely open door $C$ only if $B$ 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 = $0 \times \frac{1}{2} + 1 \times \frac{1}{2}=\frac{1}{2}$

please correct if wrong ! Pilot

$3.$

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

This implies that $\frac{z+i}{z-i}$ 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 $|z|^{2} = 1$

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

please rate ! Pilot

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