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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

f’(x)…


How to use Quadratic here?

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…


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Q: A charge of Q of + 10 micro Columb is placed on a hollow conducting sphere of radius R 5 cm. Find the electric field magnitude everywhere. What-if the spehere is SOLID? What if Non-conducting?

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?


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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.

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