Messages posted by Soumik

try out jenson's inequality

Post the entire qsn.

No matter in whichever pattern u choose, the nos. shall always add upto 260.

There's a wonderful proof for this. Try it else I shall post it.

Reflect the cosine graph about the origin to get that of -cosx....shift it downwards bmg times, and shift the sinx graph upwards mg times, now superimpose them strecthing it k times.....Hopefully got it now...

This is the graph of the above function, so even though becomes integer for 2 values of x other than 0, it posses only 2 roots...

If [x] represents my frac part of x, then the above exp takes an integral value at [x]=0.8 as well......What about that....

1)

that gives xyz can have a max of 5.....thus we need to check for (2,1,1), (3,1,1), (4,1,1), (5,1,1) & (2,2,1)....

Thus no solutions....

But how does Pole-strength become halved in 2nd case?

$(am_i,\frac{a}{m_i}})$

Do there exists a point equidistant from the set of points represented by the above exp ?

$1\le i\le 4$

Consider the situation shown...

The eqn is $mgsin\theta-bmgcos\theta=kx$

b is the coefficient of friction,

When $\theta\le 4$

$sin\theta \approx \theta\\ \ cos\theta \approx 1$

So eqn becomes that of a sraight line...

Later on,

$\boxed{x=\frac{mgsin\theta-bmgcos\theta}{k}}$

Hpefully u can do this graph, if u still face prob, nudge me, I'll be there for help again!

$int_0^{pi} rac{sinnx}{sinx(2^x+1)}dx$

Edit :- Limits are from -pi to pi

Is it not a std fact that extremes of the trig values occur at x=y=z=60 deg....

The qsn asks us to determine the no of reals, not necessarily the reals satisfying the above eqn....

S the graph turns out as a compressed parabola , and the line y=16, as the min value of f(x)=(6-x)⁴+(8-x)⁴turns out to be less than that....

Just an intuition....

Check for the derivative of f(x)=x²+7x-13....

f'(x)=2x+7, g'(x)=2x-3....

Clearly from the derivatives, f(x) becomes steeper, meaning after both the parabolas cross the X-axis, f(x)>g(x), while the opposite happens when they are below the X-axis.....

Will this do?

Well, I played up my part as an idiot.....I proved no soln exists for xy+yz+zx=5xyz.....and not 5(xy+yz+zx)=xyz......

In fact there are finitely many solutions, (more than 8).....

@ Abhinav, according to the qsn, x,y,z are naturals, since when did 1/15 become a natural no.....?

Rudra, go ahead with ur solution.....

For the first 1, D>0 meaning a<-1 or a>4.....Prod of roots >9 meaning a>-4, sum of roots >6 meaning a<-2, thus range = (-4,-2)

$x_1^2+y_1^2+z_1^2<x^2+y^2+z^2\ (x_i , y_i, z_i)<(x_{i+1} , y_{i+1}, z_{i+1})$ ....................(1) x>x₁ y>y₁ z>z₁

Now observe

3xyzle x^2+y^2+z^2le xy+yz+zx

For proving this use the fact

x^2+y^2+z^2 >(x-y)^2+(y-z)^2+(z-x)^2 together with (x,y,z) ge 2 ...(Try it, I shall give it if u can't).

This proved suffices the case for x=y=z=1, by virtue of (1) above...

Thus no solutions exist!

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