Correct Answer - (a) `[-1, 0) cup (0,1]`
(b) `[-2,0] cup [2,4]`
(c ) `[-3//2,0]`
(d) `[2,3)`
(e ) `[3,10//3]`
(f) `(-oo,-6] cup [6,oo)`
(a) `f(x)` is defined if `x in [-1,1] " and " x ne 0,`i.e.,
`x in [-1,0)cup (0,1]`
(b) `f(x)=sin^(-1)(|x-1|-2)`
Since the domain of `sin^(-1)x` is `[-1,1],f(x)` is defined if
`-1 le |x-1|-2 le 1`
or `1 le |x-1| le 3`
i.e., `-3 le x-1 le -1 " or " 1 le x -1 le 3`
i.e., `-2 le x le 0 " or " 2 le x le 4`
or domain `=[-2,0]cup [2,4]`
(c ) `-1 le 1+3x+2x^(2) le 1`
or ` 2x^(2)+3x+1 ge -1`
or ` 2x^(2) +3x+2 ge 0 " (1)" `
and `2x^(2) +3x le 0 " (2)" `
From equation (2), `2x^(2) +3x le 0" or " 2x(x+(3)/(2)) le 0`
or `(-3)/(2) le x le 0 " or " x in [-(3)/(2),0]`
In equation (1), we get imaginary root for `2x^(2)+3x+2=0 " and " 2x^(2)+3x+2 ge 0` for all x. Therefore,
domain of function`=[-(3)/(2),0]`
(d) To define `f(x), 9-x^(2) gt 0 " or " -3 lt x lt 3 " (1) " `
`-1 le (x-3) le 1 " or " 2 le x le 4 " (2)" `
From equations (1) and (2), `2 le x lt 3," i.e., " x in [2,3).`
(e ) `f(x)="cos"^(-1)((6-3x)/(4))+"cosec"^(-1)((x-1)/(2))`
For ` "cos"^(-1)((6-3x)/(4)),-1 le (6-3x)/(4) le 1`
or `-4 le 6-3x le 4`
or `-10 le -3x le -2`
or `2//3 le x le 10//3 " (1)" `
For ` "cosec"^(-1)((x-1)/(2)),(x-1)/(2) le -1 " or "(x-1)/(2) ge 1`
i.e., `x le -1 " or " x ge 3 " (2)" `
From equation (1) and (2), ` x in [3,(10)/(3)].`
(f) `f(x)=sqrt("sec"^(-1)((2-|x|)/(4)))`
`sec^(-1)` function always takes positive values which are `[0,pi]-{pi/2}.`
Hence, the given function is defined if
`(2-|x|)/(4) le -1 " or " (2-|x|)/(4) ge 1`
i.e., `|x| ge 6 " or " |x| le -2 i.e., x in (-oo,-6] cup [6,oo)`
