Let `f^(prime)(x)>0 AA x in R and g(x)=f(2-x)+f(4+x).` Then g(x) is increasing in (i) `(-infty,-1)` (ii) `(-infty,0)` (iii) `(-1,infty)` (iv) none of these

Let f''(x) gt 0 AA x in R and g(x)=f(2-x)+f(4+x). Then g(x) is increasing in

Let f'(x)>0AA x in R and g(x)=f(2-x)+f(4+x) Then g(x) is increasing in (i)(-oo,-1)( ii) (-oo,0)( iii) (-1,oo)( iv ) none of these

The function f(x)=1/(1+x^2) is decreasing in the interval (i) (-infty,-1) (ii) (-infty,0) (iii) (1,infty) (iv) (0,infty)

Let f''(x) gt 0 AA x in R and let g(x)=f(x)+f(2-x) then interval of x for which g(x) is increasing is

Let f''(x) gt 0 AA x in R and let g(x)=f(x)+f(2-x) then interval of x for which g(x) is increasing is

The exhaustive values of x for which f(x) = tan^-1 x - x is decreasing is (i) (1,infty) (ii) (-1,infty) (iii) (-infty,infty) (iv) (0,infty)

Let g(x)=2f(x/2)+f(2-x) and f''(x)<0 , AA x in (0,2) .Then g(x) increasing in

The domain of f(x)=log_x 12 is (i) (0,infty) (ii ) (0,1)cup (1,infty) (iii) (-infty,0) (iv) (2,12)

If the point (1, a) lies in between the lines x + y =1 and 2(x+y) = 3 then a lies in (i) (-infty,0)cup (1,infty) (ii) (0,1/2) (iii) (-infty,0)cup (1/2,infty) (iv) none of these

The solution set of the inequation ||x|-1| < | 1 – x|, x in RR, is (i) (-1,1) (ii) (0,infty) (iii) (-1,infty) (iv) none of these