Messages posted by jishnudas1991

Yes Rajat is right that the solution is wrong.

But now it only remains to prove the inequality using the assumption that p<q.......

Re:Solving an equation

Re:A question on limits:

Re:A nice result

Compute this limit:

$\mathop{\lim}\limits_{n \to \infty}\sqrt{n}\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right)$

Re:quadratic.........again a gud one

Re:the hardest sum

This is without calculus:

Re:A primitive electromagnetic gun.

I hope there is no conceptual mistake:

Edit. Sorry My solution was not O.K. as bhatt sir ha pointed out. But are you sure that you have not misused the term 'rational' in place of 'linear'. And ,is nothing specified about whether a,b,c are rational or not??

Re:How do u prove this -->

Re:BAloon

$b-1=k(a-1)$

$b^2-1=k^2(a-1)^2+2k(a-1)=k^2a^2-2a(k^2-k)+(k^2-2k)$

$a^2-1|b^2-1$

or, $a^2-1|k^2a^2-2a(k^2-k)+(k^2-2k)$

or, $a^2-1|-2a(k^2-k)+(2k^2-2k)$

or, $a^2-1|2(a-1)k(k-1)$

or, $2(a-1)k(k-1)=t(a^2-1)$

or, $a+1=\frac{2k(k-1)}{t}$

Let us take $t=k-1$

$a=2k-1$

or, $b=2k^2-2k+1$

see that a and b satisfy the second property i.e. $a<b<a^2$ for k>1:

$(2k^2-2k+1)-(2k-1)=2(k^2-2k+1)>0$

or, $b>a$

again $(4k^2-4k_1)-(2k^2-2k+1)=2k(k-1)>0$ for $k>1$

or, $b<a^2$

so,we can generate the sequence $a_n$ as $a_n=2(n+1)-1=2n+1$

and $b_n$ as $b_n=2n^2+2n+1$

(This is not the only way we can generate the sequence $a_n$ we can use any function of n $f(n)$ in place of $n+1$ taking care that the function is increasing for all $x>1$ and $f(1)>1$ . Also the function $f$ must attain integral values for all +ve integral argument. Here since we only have to show that such a sequence exists.So we are done by finding such a sequence). 10/10

differentiation of sec2x=2sec2x.tan2x..Not 2tan²2x.

Please wait

$xy\left(\left(\frac{dy}{dx}\right)^2-1\right)=(x^2-y^2)\frac{dy}{dx}$

$xy\left(\frac{dy}{dx}\right)^2-(x^2-y^2)\frac{dy}{dx}-xy=0$

$=>\left(y\left(\frac{dy}{dx}\right)-x\right)\left(x\left(\frac{dy}{dx}\right)+y\right)=0$

now, $\frac{dy}{dx}=\frac{x}{y}$ or $\frac{dy}{dx}=-\frac{y}{x}$

i.e.

solutions are: $x^2-y^2=C_1$ and $xy=C_2$

The first part I have done exactly as rajat has done. My solution to the second part is as follows.

see that $b=\frac{pq+qr+rp}{p+q+r}=\frac{pq+qr+rp}{pqr}$

so, $b^2=\frac{(pq+qr+rp)^2}{pqr(p+q+r)}$

or, $b^2=\frac{(pq+qr+rp)\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)}{(p+q+r)}$

$=\frac{2(p+q+r)+\frac{pq}{r}+\frac{qr}{p}+\frac{rp}{q}}{p+q+r}$

$=2+\frac{\frac{pq}{r}+\frac{qr}{p}+\frac{rp}{q}}{p+q+r}$

if we assume $p\ge q\ge r$

then $pq \ge rp \ge qr$

and $\frac{1}{r}\ge \frac{1}{q}\ge \frac{1}{p}$

By Chebychev's ineq.:

$\frac{\frac{pq}{r}+\frac{qr}{p}+\frac{rp}{q}}{3}\ge\frac{(pq+qr+rp)}{3}.\frac{\left(\frac{1}{r}+ \frac{1}{q}+ \frac{1}{p}\right)}{3}$

or, $\frac{\frac{pq}{r}+\frac{qr}{p}+\frac{rp}{q}}{p+q+r}\ge\frac{(pq+qr+rp)}{p+q+r}.\frac{\frac{1}{r}+ \frac{1}{q}+ \frac{1}{p}}{3}=\frac{1}{3}b^2$

or, $b^2\ge2+\frac{1}{3}b^2$

or, $b\ge\sqrt{3}$

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