Solving quation?

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3 Answers
Dec 26, 2017

#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)#

Dec 26, 2017

#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.!

Dec 26, 2017

#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.

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