61 - Calculus Questions Answers

1+x²+x⁴ = (1+x+x²)(1-x+x²)

now solve

g(x) = 3-5sin(2x-4)

for getting reminder put x³= x so

x¹³⁵ + x¹²⁵ - x¹¹⁵ + x⁵ +1 will give

 x⁴⁵ + x⁴¹*x² - x³⁸*x +x*x² +1

x¹⁵ + x¹³*x²*x² - x¹²*x²*x +x*x² +1

 x¹⁷ + x³ +1

x⁵*x² + x + 1

x⁷ + x + 1

x²*x + x +1 

x³ + x +1 

x+x+1

2x+1

now put x = 10

let lnt = x so dt/t = dx so dt = tdx = e^xdx

now ∫(ln t)dt/(t-1) = ∫xe^xdx/(e^x-1) = ∫x(e^x-1+1)dx/(e^x-1)

now solve 

let  √tanx = t

so sec²xdx/2√tanx = dt

so (1+t⁴)dx/2t = dt

so dx = 2tdt/(1+t⁴)

now solve

2ƒ(x) − 3ƒ(1⁄x) = x²   (1)

so 2ƒ(1/x) − 3ƒ(x) = 1/x²  (2)

multiply (1) to 2 and (2) to 3, then add the two equations, you will get f(x) then calculate f(3)  

it will be 0, assume it by considering that ∞ is a big number 

for ex.   0*(1111111111111111111111111) = 0

f(x)= log₄log₃log₂x 

so 0 < log₃log₂x < ∞

so 1 < log₂x < ∞ 

so 2 < x < ∞

let us consider the case of ellipse with x and y axes as their axes

eq. is x²/a² + y²/b²= 1

on differentiating we get 2x/a² + 2yy^'/b² = 0                                y'is first differential

so yy^'/x = -b²/a²

again differentiate and get answer. 

You should remember that you should differentiate as many times as the number of constants.

for ex. in the case of parabola only first diffrentiation is sufficient.

now complete it for all conics

x=rcosθ and y=rsinθ

so dx=-rsinθdθ  and dy = rcosθdθ

so p = -cotθ

so given eq. will become x-cotθy = (x-y)cosecθ 

on putting values of x and y we get 

0 = r(cosθ-sinθ)/sinθ

so tanθ = 1  so θ = nπ+π/4

so x = rcos(nπ+π/4) and y = rsin(nπ+π/4)