Consider the function `f(x) = In(sqrt(1-x^2)-x)` then which of the following is/are true?

Options

A. f(x) increases in the on `x=(-1,-(1)/(sqrt2))`
B. f has local maximum at `x=-(1)/(sqrt2)`
C. Least value of f does not exist
D. Least value of f exists

Answer

A, B, C

Explanation

`f(x)=log_(e)(sqrt(1-x^(2)-x))`
`sqrt(1-x^(2))-x` is defined for `x in [-1,1]`
If `x in [-1,0], sqrt(1-x^(2))-x gt0`
If `x gt 0`, then `sqrt(1-x^(2))gt x rArr 1 -x^(2)gtx^(2)`
i.e., `x in (0,(1)/(sqrt2))`
Domain of f is `[-1,(1)/(sqrt2)]`
`f'(x)=((-x)/(sqrt(1-x^(2)))-1)/(sqrt(1-x^(2))-x)`
`f'(x)gt0" if "(-x)/(sqrt(1-x^(2))-1) gt 0`
`"or "-x-sqrt(1-x^(2))gt0`
`"or "x in [-1,(-1)/(sqrt2))`
`therefore f(x)` increases in `[-1,-(1)/(sqrt2)]` and decreases in `(-(1)/(sqrt2,(1)/(sqrt2))`
`therefore" f has local max, at x"=-(1)/(sqrt2)`
Also `underset(xrarr(1)/(sqrt2))(lim)log_(e)(sqrt(1-x^(2))-x)=-oo`
`therefore" f has no minima."`