106CASE I, x>0
then we have 2*2x ≥ 2√2
=> 2x+1 ≥ 23/2
=> x+1≥1.5
=> x≥ 0.5
CASE II, x<0
then we have
2x + 2-x ≥ 2√2
consider 2x = t
So, t2 - 2√2 t + 1 ≥0
=> (t-(√2-1))(t-(√2+1)) ≥0
=> t ≥ √2+1 OR t≤ √2-1
=> 2x ≥ √2+1 (not possible for x<0) or 2x ≤ √2-1
=> 2x ≤ √2-1
=> xln2 ≤ ln(√2-1)
=> x ≤ ln(√2-1)/ln2
hoping der is no calc error
49x≥0
xx2-x-2<1
or, x^{x^2-x-2}<x^0
or, x^2-x-2<0
or, x^2+x-2x-2=(x+1)(x-2)<0
which gives solution, x ε (-1,2)
x<0
(-x)^{x^2-x-2}<x^0
now, x2-x-2=(x+1)(x-2) which is even...
thus the above expression also boils down to
x^{x^2-x-2}<x^0
thus giving same solution x ε (-1,2)
23Q2 )
|x|^{x^{2}-x-2}< 1
take\; log\;to\;the\;base\;e\;on\;both\;sides
(x^{2}-x-2)(log|x|)< 0
hence 2 cases
Case (1)
(log|x|)< 0 , and\;(x^{2}-x-2) > 0
log|x| < 0
since e > 1 ,
e^{log|x| } < e^{0}
\Rightarrow ,|x|< 1
\Rightarrow x\in (-1,1) \; and\; also, x\in(-∞,-1) U ( 2 ,∞)
wich is true for no x .
CAse (2)
log(|x|) > 0 , (x^{2}-x-2)< 0
hence
\Rightarrow x\in (-infi,-1)\;\bigcup{}\;(1,infi)\;and\;x\in (-1,2)
hence
x\in (1,2) \;is\;thesolution\;set
