This Questions seems daunting but let’s put our thinking in order.
Let’s suppose that for some (n) there exist the desired numbers. we also assume that S1 <S2 <S3<…<Sn
surely S1 > 1 else (1- 1/S1) = 0
so we have 2 ≤ S1 ≤ S2–1 ≤ S3–2≤…Sn -(n-1) , hence Si ≥i+1 for each i = 1,2…n. Therefore
51/2010 = 17/670 =
So how do we approach this Q? one way is to have it as a function of (x) only and express it as
g(x) = x^x + (1-x)^(1-x) and then try to find the f’(x) =0 and solve the usual way… and you can see you’re headed to trouble. So let’s try another way!
Using the AM-GM inequality, we can write
x^x + y^y ≥ 2 sqrt( x^x * y^y) = 2* e ^f(x)
now let’s e^f(x) = sqrt( x^x * y^y) so by taking log both sides we get
f(x) = x log x + (1-x) log(1-x) and
Q: Lets a and b be real numbers for which the equation x⁴ + ax³ + bx² +ax +1 = 0 has at least one real solution. For all such pairs (a, b), find the minimum value of a²+b².
A: As there is symmetry is the coefficients, first consider the eq. y = x+ 1/x and this equation is equivalient to x² -yx +1 =0, a quadratic in x which has real roots iff its discriminant y² - 4 ≥0 i.e iff |y|≥2
Now to solve x⁴ + ax³ + bx² +ax +1 = 0 we divide by x² and…
Shifu: Hello Po, it seems you’ve got a good challenge to solve… how do you plan to go about it?
Po: Master shifu, thanks but I guess there must be some tricks to solve it! I guess I should first get the expression in simplest form and then apply the famous AM/GM. I guess this will work?
Shifu: Why to guess, what you can KNOW! Go ahead do it.
Po: so let 1/a = x then the expression be comes,
1/(1+x) + 1/(1+y) + 1/(1+z) = 1 then prove that xyz ≥8
This is much simple equation to solve, Why…
Hey guys, it’s the time of your lives when everyone around you turns into an IIT-JEE-maverick, offering you advice about “How to crack it”.
The Internet gets flooded with people giving out a bulky list of dos and dont during the exam.
We, at SolveitNow, believe that all this complexity further adds to your skyrocketing anxiety and so we bring to you a few simple words of wisdom, powered by Brain Science principles to instill growth mindset confidence so as to bring out the best in you!
The exam is just a process:
This exam is nothing more than a…
Shifu: How do we find the electric field?
Po: if it’s a point charge we use the formula KQ/r² otherwise we divide the charge into little pieces, find the electric field from each (as point charge), and add the electric fields (as vectors).
Shifu: So what’s the case here? is it a point charge or what?
Po: if the spehere would be non-conducting then the charge placed on surface will remain as point charge but here it’s a conducting sphere, so charge will spread. One more thing will it make any difference if the conducting spehere is hollow or solid?
Think of it as a WhatsApp chat with a Physics ninja friend of yours or like a LinkedIn connect with your senior who was Maths champ and cracked IIT last year!
You also want to go to IIT and later to Stanford. Right? Yeah
We’ll show you how you can use SolveitNOW for learning to solve problems elegantly and how mentorship from Grandmasters will give you confidence to keep sailing smoothly towards your IIT/Stanford goals.
After the basic setup…Just click or upload the picture of your problem and send it as challenge to your friends who are also preparing for…
Solve ⊕ Challenge-your-friend with IIT/Olympiad level problems
Once upon a time, Ryan, a 12 grade student who aspires to study at IIT/Stanford got vexingly STUCK on a Physics problem. Ryan is NOT alone. Some 6.2 mn students in India, 8.1 mn students in US and many more in other countries feel the same way.
Everyday Ryan would study hard-&-slow and had many questions to ask to clear his doubts. Simple doubts were resolved by further reading and discussing with friends. It’s the complex doubts that bothered him a lot. His school teachers said his concepts were not clear and his…