If `f(2-x)=f(2+x)` and `f(7-x)=f(7+x)` and `f(0)=0`. If the minimum number of roots of `f(x)=0` where `0lexle100` is `lamda` then `lamda//3` equals

Let f(x) is a continuous function such that f(-1) = -2,f(0) = -1,f(1) = 0,f (2) = 1, f(3) = 0,f (4) = -1 and f(5) = 1 then minimum number of roots of f'(x) + xf " (x) = 0 in the interval (-1,5) are

Let f (x)=(x+1) (x+2) (x+3)…..(x+100) and g (x) =f (x) f''(x) -f'(x) ^(2). Let n be the numbers of real roots of g(x) =0, then:

If f(2(x)/(y))=f(2x)-f(y) for all x,y>0 and f(3)=2, then the value of [f(7)], is

f(x) is a twice differentiable function and g(x) is defined as g(x)=(fprime(x))^2+f prime prime(x)*f(x) on [x_1,x_2]. If x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7,f(x_1)=0,f(x_2)=2,f(x_3)=-3,f(x_4)=4,f(x_5)=-5,f(x_6) and f(x_7)=0, then the minimum number of real roots of g(x)=0 can be

Let f (x)=(x+1) (x+2) (x+3)…..(x+100) and g (x) =f (x) f''(x) -f ('(x)) ^(2). Let n be the numbers of rreal roots of g(x) =0, then:

Let g(x)=2f(x/2)+f(x) and f''(x)lt0 " in "0lexle1 , then g(x)

If f(x+1)=2x+7 then f(0) is equal to

If f(x) is a quadratic expression such that f(1)+f(2)=0, and -1 is a root of f(x)=0, then the other root of f(x)=0 is :

Let f (x) =x ^(2)-bx+c,b is an odd positive integer. Given that f (x)=0 has two prime numbers an roots and b+c =35. If the least value of f (x) AA x in R is lamda, then |(lamda)/(3)| is equal to (where [.] denotes greatest integer functio)