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SuicideFuel Math thread problem (official)

Filthy_Reject

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@trying to ascend what accounts for 80, but only wants 20
 
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Youngcel KHHV
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The sum of the first n terms of an arithmetic progression is equal to 110 and the sum of the first 2n terms is equal to 420. Knowing that the first term is 2, what's the value of the sum of the first 3n terms?
 
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H = (x³ - x² + x) raised to the 9th power

Given that x = 2/(sqrt(5) - 1), find the least positive integer closer to H
 
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Youngcel KHHV
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4x²/(1 - sqrt(1 + 2x))² < 2x + 9.

Find the values of x that satisfy this inequality
 
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Send differentiation questions or limits
Ok:

Given that

Lim f(x)/x = L
X——->0

Find (in terms of L)

Lim f(x^2 - 1)/(x - 1)

X——->0


Are you practicing it?
 
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This is false. Take e.g. f(x) = sin(x). Then sin(x)/x tends to 1 as x tends to 0, yet sin(x^2-1)/(x-1) tends to sin(-1)/(-1) = sin(1) != 2 as x tends to 0.
True, it was actually x ----> 1
 
Fallenleaves

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Yeah, I realized that after I had made the post. In that case 2L is correct of course.

Would you still like to know or can you see it now?
I don't get it
 
Ahnfeltia

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I don't get it
Multiplying both numerator and denominator of f(x^2-1)/(x-1) by (x+1) yields that f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1). Now, as x tends to 1, y = x^2-1 tends to 0, so f(x^2-1)/(x^2-1) = f(y)/y tends to L, whereas x+1 tends to 2. Ergo, f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1) tends to 2*L as x tends to 1.
 
Fallenleaves

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Multiplying both numerator and denominator of f(x^2-1)/(x-1) by (x+1) yields that f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1). Now, as x tends to 1, y = x^2-1 tends to 0, so f(x^2-1)/(x^2-1) = f(y)/y tends to L, whereas x+1 tends to 2. Ergo, f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1) tends to 2*L as x tends to 1.
Thanks
 
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Let ABCDEF be a right triangular prism, with all its edges congruent and its lateral edges AD, BE and CF. Let 0 and 0' be the barycenters of the bases ABC and DEF, respectively, and P be a point belonging to 00' such that P0' = 1/6 00' . Let π be the plane determined by P and the midpoints of AB and DF. The π plane divides the prism into two solids. Determine the ratio of the volume of the smaller solid to the volume of the larger solid, determined by the π plane
 
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Any basic (very) integration or differentiation questions?
 
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Youngcel KHHV
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Any basic (very) integration or differentiation questions?
Find the critical points of (x² - x)/(1 + 3x²).

Simplify the expression: I = (11/8) . e (to the power of 2x) + (3/4) . x² . e (to the power of 2x) - (3/4) . x . e (to the power of 2x) - (3/2) . integral of (x² . e to the power of 2x).

Using first principle, prove that cos'(x) = - sin(x).


Be a = limx→0 of (13x - sin(10x))/(sin(11x) + x).

and b = limx→∞ of x/sqrt(x² + x + 1).

Find a + b
 
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Given that the integral (from 0 to pi) of (f(x) + sin(2x)) = 2.

Find the integral (from 0 to pi) of (f(x) + 9(x/pi)²)
 
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Be f(x) = sqrt(sin(7x + ln(5x))).

Find the first derivative of f(x)
 
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2+2 in spanish when it is tuesday
 
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S = (xy + xz + yz + 2(x + y + z) + 3)/((x + 1)(y + 1)(z + 1)).

Given that x = log ab (in the base c).

y = log ac (in the base b)

z = log bc (in the base a).

Find the value of S
 
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What's the biggest value of n, such that 21 (to the power of n), divides 2020! ?
 
Fallenleaves

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Evaluate the limit

lim [x^2(3+sinx)]/[(x+sinx)^2]

x--->0
 
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3/4?

I just expanded the square in the denominator and then divided everyone by x²
Nice but the question was introduced to me when I was learning how to use sinx/x=1 as x tends to 0 in questions
 
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Find all functions, R ---> R, such that:

f(x)f(y) = f(x + y) + xy.
 
Caesercel

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Using first principle, prove that cos'(x) = - sin(x).

Cos'(x) = lim h ->0 (Cos(x+h) - Cosx) /h

= (CosxCosh - SinxSinh - Cosx) /h

= lim h->0 -SinxSinh/h

= - Sinx
 
Ahnfeltia

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Find all functions, R ---> R, such that:

f(x)f(y) = f(x + y) + xy.
First, plugging in y = 0 yields that f(x)*f(0) = f(x) which means that either f(x) = 0 for all x or f(0) = 1. One can easily check that f = 0 is not a valid solution however.

In case f(0) = 1, plug in x = 1 and y = -1 to find that f(1)*f(-1) = f(0) - 1 = 0. Ergo, either f(1) = 0 or f(-1) = 0. If f(1) = 0, then plugging in y = 1 yields that 0 = f(x+1) + x. Letting z = x+1 and rearranging then yields that f(z) = 1-z. Similarly the case f(-1) = 0 leads to f(z) = 1+z.

All in all, the solutions are f(z) = 1±z.
 
Caesercel

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Simplify the expression: I = (11/8) . e (to the power of 2x) + (3/4) . x² . e (to the power of 2x) - (3/4) . x . e (to the power of 2x) - (3/2) . integral of (x² . e to the power of 2x).
e^2x
 

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