JEE Main 2016 (13)

Algebra Level 5

If $a,b$ are the roots of $x^{2}+\omega x-\frac{1}{2\omega ^{2}}=0$ , where $\omega$ is a positive constant, find the minimum value of $a^{4}+b^{4}$ .

Try my set JEE Main 2016 .

2-\sqrt{2}

2

2+\sqrt{2}

3+\sqrt{2}

2+2\sqrt{2}

2 solutions

Rishabh Jain
Mar 21, 2016

$a+b=-ω, ab=\dfrac{-1}{2ω^2}$ $\begin{aligned}a^4+b^4&=((a+b)^2-2ab)^2-2a^2b^2\\&=\underbrace{ω^4+\dfrac{1}{2ω^4}}_{\color{#D61F06}{\text{Applying AM-GM}}}+2\\&\geq \sqrt 2+2\end{aligned}$ Equality occurs when $ω=\dfrac{1}{\sqrt[8]{2}}$ .

Happy Holi Bro

Kushagra Sahni - 5 years, 2 months ago

Yeah.. Same to you... Thank god for the finish of tomorrow's game otherwise India would have been out of the race for world cup... :-) Nothing better can happen on Holi than this...!!

Rishabh Jain - 5 years, 2 months ago

Ah yes! Thrilling match. But India didn't win people's hearts because we were expecting them to beat Bangladesh by a big margin.Today is Australia vs Pakistan.

Kushagra Sahni - 5 years, 2 months ago

@Kushagra Sahni – Oh exactly..... Now last one Ind VS Aus will be thriller.... The team which wins goes forward.. As simple as that... Will be a nice match..

Rishabh Jain - 5 years, 2 months ago

Hi i reached till 3rd step but can u elaborate Am>=Gm step plz

Nimit Dave - 5 years, 2 months ago

Yep. Sure... Applying $AM\geq GM$ . $\dfrac{ω^4+\dfrac{1}{2ω^4}}{2}\geq \require {cancel}\sqrt{\cancel{ω^4}\times\dfrac{1}{2\cancel{ω^4}}}$ $\implies ω^4+\dfrac{1}{2ω^4}\geq \sqrt 2$ Now add $2$ on both sides to get: $ω^4+\dfrac{1}{2ω^4}+2\geq \sqrt 2+2$

Rishabh Jain - 5 years, 2 months ago

@Shivam Jadhav Why must $\omega$ be a positive value that we can apply AM-GM to?

It is only stated that it's a constant, which possibly could be complex valued.

Calvin Lin Staff - 5 years, 2 months ago

Yeah It must be mentioned that $ω$ is a real number

Shivam Jadhav - 5 years, 2 months ago

Chew-Seong Cheong
Mar 21, 2016

Since $a$ and $b$ are roots of $x^{2}+ωx-\dfrac{1}{2ω^{2}}=0$ , then $a+b = -ω$ and $ab = -\dfrac{1}{2ω^2}$ . And we know that:

$\begin{aligned} a^2 + b^2 & = (a+b)^2 - 2ab \\ & = ω^2 + \frac{1}{ω^2} \\ a^4 + b^4 & = (a^2+b^2)^2 - 2(ab)^2 \\ & = \left(ω^2 + \frac{1}{ω^2}\right)^2 - \frac{2}{4ω^4} \\ & = ω^4 + \frac{1}{2ω^4} + 2 \\ & = \left(ω^2 - \frac{1}{\sqrt{2}ω^2}\right)^2 + 2 + \sqrt{2} \end{aligned}$

Therefore, minimum $a^4+b^4 = \boxed{2 + \sqrt{2}}$ , when $ω^2 - \dfrac{1}{\sqrt{2}ω^2}= 0$ .

Can we do it this way:

$a^4 - b^4 = (a^2 - b^2)^2 + 2a^2b^2$

So, minimum value occurs when $a =b$ , on further solving we get $w^4 = -2$ . On putting values we get value of expression $\dfrac{-1}{4}$

By this method roots and $w$ turns out to be imaginary but there is no constraint of them to be real.

Please tell where i am going wrong.

neelesh vij - 5 years, 2 months ago

$(a^2-b^2)^2 + 2a^2b^2 = a^4+b^4$

Chew-Seong Cheong - 5 years, 2 months ago

oohhh seems my brain is getting fused off !!

neelesh vij - 5 years, 2 months ago

JEE Main 2016 (13)

Try my set JEE Main 2016 .

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