Let `g(x)=sqrt(x-2k),AA2klt=x<2(k+1),` where `k in ` integer. Check whether `g(x)` is periodic or not.

Let g(x)=sqrt(x-2k), AA 2k le x lt 2(k+1) where, k in l , then

Let f,g : R to R be defined by f(x) = (x-2)|x-2| AA x in R g(x) = sqrt((x-2)^(2)) AA x in R S = {x : x in R " and " f(x) = g(x)} , number of elements in S is………..

Let f (x) =x + sqrt(x^(2) +2x) and g (x) = sqrt(x^(2) +2x)-x, then:

let g(x)=[a sqrt(x+),0

Let f(x)=(sqrt(log(x^(2)+kx+k+1)))/(x^(2)+k) If f(x) is continuous for all x in R then the range of k is

Let f(x)=sqrt(6-x),g(x)=sqrt(x-2) . Find f+g,f-g,f.g and f/g.

Let f(x)="min"{sqrt(4-x^(2)),sqrt(1+x^(2))}AA,x in [-2, 2] then the number of points where f(x) is non - differentiable is

Let g(x)=sqrt(sin^(-1)(cos(tan^(-1)x))+cos^(-1)(sin(cot^(-1)x))) ,then int_(-sqrt((pi)/(2)))^(sqrt((pi)/(2)))g(x)dx equals

Let f (x) = (x+5)/(sqrt(x^(2) +1) ) , then the smallest integral value of k for which f (x) le k AA x in R is

Let f(x)=1+sqrt(x) and g(x)=(2x)/(x^(2)+1)