f(x)=log x" is on "(0,oo)

Show that f(x) = log_a x "on" (0,oo ),(a gt 0 "and" a != 1) functions are injective.

The domain of f(x)=log|log x|is(0,oo)(b)(1,oo)(c)(0,1)uu(1,oo)(d)(-oo,1)

int sqrtx log x dx on (0,oo).

Let f(x) lt 0 AA x in (-oo, 0) and f (x) gt 0 ,AA x in (0,oo) also f (0)=0, Again f'(x) lt 0 ,AA x in (-oo, -1) and f '(x) gt 0, AA x in (-1,oo) also f '(-1)=0 given lim _(x to -oo) f (x)=0 and lim _(x to oo) f (x)=oo and function is twice differentiable. If f'(x) lt 0 AA x in (0,oo)and f'(0)=1 then number of solutions of equation f (x)=x ^(2) is : (a) 1 (b) 2 (c) 3 (d) 4

Let f(x) lt 0 AA x in (-=oo, 0) and f (x) gt 0 AA x in (0,oo) also f (0)=o, Again f'(x) lt 0 AA x in (-oo, -1) and f '(x) gt AA x in (-1,oo) also f '(-1)=0 given lim _(x to oo) f (x)=0 and lim _(x to oo) f (x)=oo and function is twice differentiable. If f'(x) lt 0 AA x in (0,oo)and f'(0)=1 then number of solutions of equatin f (x)=x ^(2) is :

The domain of f(x)="log"|logx|i s (0,oo) (b) (1,oo) (c) (0,1)uu(1,oo) (d) (-oo,1)

The domain of f(x)="log"|logx|i s (0,oo) (b) (1,oo) (c) (0,1)uu(1,oo) (d) (-oo,1)

The domain of definition of the function f(x)="log"|x| is R b. (-oo,0) c. (0,oo) d. R-{0}

Let f(x) lt 0 AA x in (-oo, 0) and f (x) gt 0 AA x in (0,oo) also f (0)=0, Again f'(x) lt 0 AA x in (-oo, -1) and f '(x) gt AA x in (-1,oo) also f '(-1)=0 given lim _(x to oo) f (x)=0 and lim _(x to oo) f (x)=oo and function is twice differentiable. The minimum number of points where f'(x) is zero is:

Evaluate the integerals. int (e ^(log x))/(x ) dx on (0,oo).