166 - Questions Answers

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

two times sign change so 2 positive roots

if ax²+bx+c is positive for real x

then b²< 4ac similarly others

on adding all inequalities we get b² + c² + a²< 4(ac+ab+bc)

now get answer 

general term of the series = r(n+1-r)² = r(n+1)² + r³ - 2(n+1)r²

now put sigma then solve                                       

sovle the two eqs for x and y,

method : 

x^2 + y^2=12 

on putting this in 2nd eq. we get -5x+3y = -10 so x = (3y+10)/5

so [(3y+10)/5]² + y²= 12

now get y then x, then get the tangents in the first circle and then intersection point of the tangents.

  ∑ r ⁿC_rx^ry^n-r

= ∑ r ⁿC_rx^r(1-x)^n-r

= ∑ r n!/r!(n-r)! x^r(1-x)^n-r

= n (n-1)!/(r-1)!(n-r)! x^r(1-x)^n-r 

= n ^(n-1)C_(r-1) x^r(1-x)^n-r

= n ^(n-1)C_r x^r+1(1-x)^n-r-1             let r = r+1

= nx  ^(n-1)C_r x^r(1-x)^n-r-1 

^{now solve} 

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 

for 2<x<4,   f(x) = x-2

so g(x) = f(x-2) = x-2-2 = x-4

so g'(x) = 1 

ƒ(x)=3|2+x|

for x>-2,  f(x) = 3(2+x)  so  f^'(x) = 3 

for x<-2, f(x) = -3(2+x) so f^'(x) = -3