Question

Find domain of sin inverse(1-x) it's urgent and pls explain also.

Answer 1

Step-by-step explanation:

comparing both y=1-x

x=y+1

as y€[-1,1 ]

x€[ 0 , 2 ]

Answer 2

Answer:

The domain of the function $sin^{-1}(1-x)$ is $x\epsilon[0,2]$

Step-by-step explanation:

The given inverse trigonometric function is $f(x)=sin^{-1}(1-x)$

We are required to find the domain of this function.For this we have to know between what values the given function exists.

For inverse sine function $sin^{-1} y$ to be defined the values of y are to be lied within the inequality $-1\leq y\leq 1$

Similarly,for our given function the inequality becomes

$-1\leq 1-x\leq 1$

Subtracting 1 from all sides,we get-

$-1-1\leq -x\leq 1-= > -2\leq -x\leq 0$

Multiplying the above inequality with -1,we get

$0\leq x\leq 2$

Therefore,the domain of the function $sin^{-1}(1-x)$ is $x\epsilon[0,2]$

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