The graph of a function `y=g(x)` is shown in the following figure. If `f(x)=-3x^2-k x-12 ,k in Ra n df(g(x))>0AAx in R` then least integral value of `k` is equal to a. 13 b.64 c. 15 d.16

If f: RvecR ,g(x)=f(x)+3x-1, then the least value of function y=g(|x|) is a. -9/4 b. -5/4 c. -2 d. -1

Range ofthe function f(x)=cos(Ksin x) is [-1,1] , then the least positive integral value of K will be

If f(x+f(y))=f(x)+yAAx ,y in Ra n df(0)=1, then find the value of f(7)dot

If f(x)=3x-2, g(x)=2x+k and if fog=gof, then find the value of k.

If the range of function f(x) = (x + 1)/(k+x^(2)) contains the interval [-0,1] , then values of k can be equal to

Let g(x)=f(logx)+f(2-logx)a n df^(x)<0AAx in (0,3)dot Then find the interval in which g(x) increases.

If f , R rarr R defined by f (x) = 3x - 6 and g : R rarr R defined by g(x) = 3x + k if fog = gof then the value of k is ........... .

If f(x) is continuous function AAx in R and the range of f(x)i s(2,sqrt(26))a n dg(x)=[(f(x))/c] is continuous AAx in R , then find the least positive integral value of c , where [dot] denotes the greatest integer function.

Let f(x)=x+2|x+1|+2|x-1|dot If f(x)=k has exactly one real solution, then the value of k is (a) 3 (b) 0 (c) 1 (d) 2