" 40."x^(2)+(a+(1)/(a))x+1=0

The angles of a triangle are (x-40)^(0),(x-20)^(0) and ((1)/(2)x-10)^(0) .find the value of x

Let x=1+(1)/(1+(1)/(1+(1)/(1+(1)/(1+(1)/(1+alpha))))). Which of the following is correct? x^(2)+x+1=0 (b) x^(2)-x+1=0( c) x^(2)+x-1=0 (d) x^(2)-x-1=0

If (1+x+x^(2))^(20) = a_(0) + a_(1)x^(2) "……" + a_(40)x^(40) , then following questions. The value of a_(0)^(2) - a_(1)^(2) + a_(2)^(2)- "……" - a_(19)^(2) is

Find the coefficient of x^(20) in (x^(2)+2+(1)/(x^(2)))^(-5)(1+x^(2))^(40)

The coefficient of x^(20) in the expansion of (1+x^(2))^(40)*(x^(2)+2+(1)/(x^(2)))^(-5) is:

If f(x)={(ax^2+b ,0lex<1),(4,x=1),(x+3,1ltxge2)) then the value of (a ,b) for which f(x) cannot be continuous at x=1, is (a) (2,2) (b) (3,1) (c) (4,0) (d) (5,2)

Set of all real value of a such that f(x)=((2a-1)+x^(2)+2(a+1)x+(2a-1))/(x^(2)-2x+40) is always negative is -oo,0 b.0,oo c.-oo,1/2 d. none

If f(x) = ""^(40)C_(1).x(1-x)^(39) + 2.""^(40)C_(2).x^(2)(1-x)^(38)+3.""^(40)C_(3).x^(3)(1-x)^(37)+ "….."+40.""^(40)C_(40).x^(40) , then the value of f(3) is