let `f(x)=(lncosx)/{(1+x^2)^(1/4)-1}` if `x > 0` and `f(x)=(e^(sin4x)-1)/(ln(1+tan2x))` if `x < 0` Is it possible to difine `f(0)` to make the function continuous at `x=0`. If yes what is.the value of `f(0)`, if not then indicate the nature of discontinuity.

Find k, if possible, so that f (x)= [{:((ln (2- cos 2x))/(ln ^(2) (1+ sin 3x)),,, x lt 0),(k,,, x=0),((e ^(sin 2x )-1)/(ln (1+ tan 9x)),,, x gt 0):} is continious at x=0.

If f(x)=(log)_e((x^2+e)/(x^2+1)) , then the range of f(x)

Let f(x+(1)/(x))=x^(2)+(1)/(x^(2)),(x ne 0) then f(x) equals

If 3f(x)-f((1)/(x))= log_(e) x^(4) for x gt 0 ,then f(e^(x))=

Let f(x)=x-1/2 log (x^(2)+1) . Then f' (x) is :

Let f(x)=int x^2/((1+x^2)(1+sqrt(1+x^2)))dx and f(0)=0 then f(1) is

Let f(x)={(|x+1|)/(tan^(- 1)(x+1)), x!=-1 ,1, x!=-1 Then f(x) is

Column I: Function, Column II: Value of x for which both the functions in any option of column I are identical f(x)=tan^(-1)((2x)/(1-x^2)),g(x)=2tan^(-1)x , p. x in {-1,1} f(x)=sin^(-1)(sinx)a n dg(x)="sin"(sin^(-1)x) , q. x in [-1,1] f(x)=(log)_(x2)25a n dg(x)=(log)_x5 , r. x in (-1,1) f(x)=sec^(-1)x+cos e c^(-1)x ,g(x)=sin^(-1)x+cos^(-1)x , s. x in (0,1)

The primitive of the function f(x)=(1-1/(x^2))a^(x+1/x)\ ,\ a >0 is (a) (a^(x+1/x))/((log)_e a) (b) (log)_e adota^(x+1/x) (c) (a^(x+1/x))/x(log)_e a (d) x(a^(x+1/x))/((log)_e a)

Let f (x) be a continous function in [-1,1] such that f (x)= [{:((ln (ax ^(2)+ bx +c))/(x ^(2)),,, -1 le x lt 0),(1 ,,, x =0),((sin (e ^(x ^(2))-1))/(x ^(2)),,, 0 lt x le 1):} Then which of the following is/are corrent