Question

Solving quation?

Answer 1

`sgn(1-x) < [2-x]` where `x in (-2,-1)`

Explanation:

`sgn(1-x)` where `x in (-2,-1) = +1`
Explain: [According to Wikipedia] "sgn is an odd mathematical function that extracts the sign of a real number".
if `x in (-2,-1)` it means `x` can get any real number between -2 and -1, and obviously it will be a negative number.
Because sgn is an ... that extracts the sign of a real number, in our case `sgn(1-x)` where `x in (-2,-1) = sgn(1-(-))= +1`

`f_(x)=[2-x]` where `x in (-2,-1) iff f in (3,4) iff min_{x=-1}=3`

`3>+1 => sgn(1-x) < [2-x]` where `x in (-2,-1)`

Answer 2

`sgn(1-x) color(red)lt [3-x]`.

Explanation:

Recall that, the Signum Function `sgn : RR-{0} to RR^+` is defied by,

`sgn(x)=x/|x|, x in RR, x ne 0.`

Let us first modify the defn. of `sgn`.

Now, `x in RR, x ne 0 rArr x gt 0, or x lt 0.`

If `x gt 0, |x|=x," so that, "sgnx=x/|x|=x/x=1, x gt 0......<<1>>`.

On the similar lines, `sgnx=-1, if x lt 0......<<2>>`.

`<<1 & 2>> rArr sgn(x)=1, if x gt 0; sgn(x)=-1, x lt 0...(star)`.

For `x in (-2,-1), -2 lt x lt -1`.

Multiplying this inequality by `-1 lt 0,` we have to reverse it, & get,

`2 gt -x gt 1...................(star^0)`.

Now adding `1, 1+2 gt 1-x gt 1+1, i.e., 2 lt 1-x lt 3`.

Thus, since

`AA x in (-2,-1), (1-x) gt o, :. sgn(1-x)=1........(star^1)`.

Further, `(star^0)rArr 2+2 gt 2-x gt 2+1rArr3 lt 2-xlt4`.

Clearly, `[2-x]=3..........................................................(star^2)`.

We compare `(star^1) and (star^2),` and find that,

`sgn(1-x) color(red)lt [3-x]`.

Enjoy Maths.!

Answer 3

`abs(2-x) > "sign"(1-x)`

Explanation:

In blue the `"sign"(1-x)` function and in red the `abs(2-x)` function.

As can be depicted, `abs(2-x) > "sign"(1-x)` because at `x = 1` the function `"sign"(1-x)` is not defined.