Draw the graph of `f(x)=log_(e)(sqrt(1-x^(2)))`.Find the range of the function .Also find the values of k if (x) has two distinct real roots.

Explanation

we have y=`f(x)=log_(e)sqrt(1-x^(2)-x)`
f(x) is defined if `sqrt(1-x^(2))-xgt0 and 1 -x^(23)ge0`
For `1-x^(2)ge0,-1lexle1`
For `[-1,0]`
`sqrt(1-x^(2))-xgt0`
For (0,1)we have `1-x^(2)gtx^(2)`
or `d^(2)lt(1)/(2)`
or `0ltxlt(1)/sqrt(2)`
Thus domina of the function is `-1,(1)/sqrt(2)`
f(x)=0
So, curve passes through origin
`-(x)/sqrt(1-x^(2))-1`
f(x) =`sqrt(1-x^(2)-x)`
`=-x-sqrt(1-x^(2))/sqrt(1-x^(2))`
`=-x-sqrt(1=-x^(2))/sqrt(1-x^(2)/sqrt(1-x^(2)))sqrt(1-x^(2)-x)`
f(x)=0
`-x=sqrt(1-x^(2))`
`x=-(1)/(2)`
`f(x)gt0` for x in `(-1)/sqrt(2)`
`f(x)lt0` for x in `-(1)/(2),(1)/sqrt(2)`
`x=-(1)/sqrt(2)` is point of maxima
f(-1)=0
and `underset(xrarr)(1)/sqrt(2)limlog_(e)sqrt(1-x^(2))=-00`
Thus `x=(1)/sqrt(2)` is an asymptote.
From the above discussion the graph of the function is as shown in the following figure

Clearly range of f(x) is `(-00,f(-1))/sqrt(2) or -00, log_(e)sqrt(2)`
For two distict real roots of `f(x)=k, k in 0,log_(e)sqrt(2)`