If `f(x)=x^2+2b x+2c^2 and g(x)= -x^2-2c x+b^2` are such that min `f(x)> m a x g(x)` , then the relation between `b and c` is

If f(x)=A x^2+B x+C then find f '(x)

f(x)=min({x,x^(2)} and g(x)=max{x,x^(2)}

If f(x)=a x^2+b x+c ,g(x)=-a x^2+b x+c ,where ac !=0, then prove that f(x)g(x)=0 has at least two real roots.

Let f(x)=x^2+xg^2(1)+g^''(2) and g(x)=f(1).x^2+xf'(x)+f''(x), then find f(x) and g(x).

If f(x)=x^2+x+3/4 and g(x)=x^2+a x+1 be two real functions, then the range of a for which g(f(x))=0 has no real solution is (A) (-oo,-2) (B) (-2,2) (C) (-2,oo) (D) (2,oo)

Let f(x) = max{|x^2 - 2 |x||,|x|} and g(x) = min{|x^2 - 2|x||, |x|} then

If f"(x) > 0 and f(1) = 0 such that g(x) = f(cot^2x + 2cotx + 2) where 0 < x < pi , then g'(x) decreasing in (a, b). where a + b + pi/4 …

If f(x) = x^(2) and g(x) = (1)/(x^(3)) . Then the value of (f(x)+g(x))/(f(-x)-g(-x)) at x = 2 is

If f(x)=(a x^2+b)^3, then find the function g such that f(g(x))=g(f(x))dot

Statement-1: If a, b, c, A, B, C are real numbers such that a lt b lt c , then f(x) = (x-a)(x-b)(x-c) -A^(2)(x-a)-B^(2)(x-b)-C^(2)(x-c) has exactly one real root. Statement-2: If f(x) is a real polynomical and x_(1), x_(2) in R such that f(x_(1)) f(x_(2)) lt 0 , then f(x) has at least one real root between x_(1) and x_(2)