If f(x) is a quadratic expression such that `f(x)gt 0 AA x in R`, and if `g(x)=f(x)+f'(x)+f''(x)`, then prove that `g(x)gt 0 AA x in R`.

Q. if f(x) is a quadratic expression such that f(x)>0 AA x epsilon R, and if g(x)=f(x)+f'(x)+f(x), then prove that g(x)>0 x epsilon R. Let f(x)=ax^2+bx+c Given that f(x)>0 so >0, b^2-4ac 0 and discriminant

f(x)=x^(2)-2x,x in R, and g(x)=f(f(x)-1)+f(5-f(x)) Show that g(x)>=0AA x in R

Let f''(x) gt 0 AA x in R and g(x)=f(2-x)+f(4+x). Then g(x) is increasing in

Let f(x)=ax^2+bx+c,a,b,cepsilon R a !=0 such that f(x)gt0AAxepsilon R also let g(x)=f(x)+f\'(x)+f\'\'(x) . Then (A) g(x)lt0AAxepsilon R (B) g(x)gt0AAxepsilon R (C) g(x)=0 has real roots (D) g(x)=0 has non real complex roots

Let f'(sin x) 0AA x in(0,(pi)/(2)) and g(x)=f(sin x)+f(cos x), then

Let f''(x) gt 0 AA x in R and let g(x)=f(x)+f(2-x) then interval of x for which g(x) is increasing is

Given that f(x) gt g(x) for all x in R and f(0) =g(0) then

Let f(x) be polynomial function of defree 2 such that f(x)gt0 for all x in R. If g(x)=f(x)+f'(x)+f''(x) for all x, then

If f(x) is a non-increasing function such that f(x)+f'(x)<=0AA x in R and f(0)=f'(x)=0 then f(x) is not strictly decreasing in