" (f) "x^(2)-(sqrt(3)+1)x+sqrt(3)=0

Solve: x^(2)-(sqrt(3)+1)x+sqrt(3)=0

Solve x^(2)-(sqrt3+1)x+sqrt3=0 .

Solve x^(2)-(sqrt3+1)x+sqrt3=0 .

x^(2)-(sqrt(3)+1)x+sqrt(3)=0 2x^(2)+x-4=0

x^2+(f(x)-2)x-sqrt(3).f(x)+2sqrt(3)-3=0 , then the value of f(sqrt(3))

The tangent to the graph of the function y=f(x) at the point with abscissae x=1, x=2, x=3 make angles pi/6,pi/3 and pi/4 respectively. The value of int_1^3f\'(x)f\'\'(x)dx+int_2^3f\'\'(x)dx is (A) (4-3sqrt(3))/3 (B) (4sqrt(3)-1)/(3sqrt(3)) (C) (4-3sqrt(3))/2 (D) (3sqrt(3)-1)/2

x^(2)+(f(x)-2)x-sqrt(3).f(x)+2sqrt(3)-3=0, then the value of f(sqrt(3))

intf(x)dx=2(f(x))^3+C ,and f(0)=0 then f(x) is (A) x/2 (B) x^2/2 (C) sqrt(x/3) (D) 2 sqrt(x/3)

If the function f(x) = x^(3) - 6x^(2) + ax + b satisfies Rolle's theorem in the interval [1,3] and f'[(2sqrt(3)+1)/sqrt(3)]=0 , then a=

If the function f(x)= x^(3)-6x^(2)+ax+b satisfies Rolle's theorem in the interval [1,3] and f'((2sqrt(3)+1)/(sqrt(3)))=0 , then