If `xgt0` prove that `x+ 1/x ge2`

Consider the fucntion f(x)=((1)/(x))^(2x^2) , where xgt0 . At what value of x does the function attain maximum value ?

Using the Rolle theorem prove that the derivative f.(x) of the function. {:f(x)={:{(x sinpix,"at",xgt0),(0,"at",x=0):} vanishes on an infinite set of points of the interval (0,1).

(iv)The differentiation of log_e x;xgt0 is 1/x (v) The differentiation of log_a x (agt0;a!=1) with respect to x is 1/(xlog_e a)

If f(x)={{:((sin(a+1)x+2sinx)/x", "xlt0),(2" ,"x=0),((sqrt(1+bx)-1)/x" ,"xgt0):} is continuous at x = 0, then find the values of 'a' and 'b'.

Let f(x)={{:((sin(a+1)x+sinx)/(x)",",xlt0),(c",",x=0),((sqrt(x+bx^(2)))/(bx^((3)/(2)))",",xgt0):} If f(x) is continous at x = 0, then

[ If xgt0 then which of the following is true 1) tan^(-1)xgt(x)/(1+x^(2)), 2) tan^(-1)x=(x)/(1+x^(2)) 3) tan^(-1)xlt(x)/(1+x^(2)), 4) tan^(-1)x!=(x)/(1+x^(2))]