For certain curve `y=f(x)` satisfying `(d^(2)y)/(dx^(2))=6x-4, f(x)` has local minimum value 5 when `x=1`
The number of critical point for `y=f(x)` for `x in [0,2]` is

For certain curves y= f(x) satisfying [d^2y]/[dx^2]= 6x-4 , f(x) has local minimum value 5 when x=1. 9. Number of critical point for y=f(x) for x € [0,2] (a) 0 (b)1. c).2 d) 3 10. Global minimum value y = f(x) for x € [0,2] is (a)5 (b)7 (c)8 d) 9 11 Global maximum value of y = f(x) for x € [0,2] is (a) 5 (b) 7 (c) 8 (d) 9

For certain curve y=f(x) satisfying (d^(2)y)/(dx^(2))=6x-4, f(x) has local minimum value 5 when x=1 Global maximum value of y=f(x) for x in [0,2] is

The curve y =f (x) satisfies (d^(2) y)/(dx ^(2))=6x-4 and f (x) has a local minimum vlaue 5 when x=1. Then f^(prime)(0) is equal to :

A curve y=f(x) satisfies (d^2y)/dx^2=6x-4 and f(x) has local minimum value 5 at x=1 . If a and b be the global maximum and global minimum values of f(x) in interval [0,2] , then ab is equal to…

let f(x)=(x^2-1)^n (x^2+x-1) then f(x) has local minimum at x=1 when

Let y=f(x) satisfies (dy)/(dx)=(x+y)/(x) and f(e)=e then the value of f(1) is

Consider f,g and h be three real valued functions defined on R. Let f(x)={:{(-1", "xlt0),(0", "x=0"),(1", "xgto):} "g(x)(1-x^(2))andh(x) "be such that" h''(x)=6x-4. Also, h(x) has local minimum value 5 at x=1 The area bounded by y=h(x),y=g(f(x))between x=0 and x=2 equals

A curve y=f(x) satisfy the differential equation (1+x^(2))(dy)/(dx)+2yx=4x^(2) and passes through the origin. The function y=f(x)

If y =sin (sin x) and (d^(2)y)/(dx^(2))+(dy)/(dx) tan x + f(x) = 0, then find f(x).

If y sin (sin x) and (d^(2)y)/(dx^(2))+(dy)/(dx) tan x + f(x) = 0, then find f(x).