161 - Mathematics Questions Answers

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

let  √tanx = t

so sec²xdx/2√tanx = dt

so (1+t⁴)dx/2t = dt

so dx = 2tdt/(1+t⁴)

now solve

exact 0 and lim tends to 0 are different

180 degree

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6)

= (π-2) + (π-4) + (6-2π) 

= 0

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)