If x is a non−zero number, is x

Question

If x is a non−zero number, is  x < |x|^x ?

(1) x < 1
(2) x > 0

nick1816 · Answer

We can break the curve in 4 parts. 1. x<0 |x|^x is always positive Hence, x < |x|^x 2. 01 The above inequality is true as 01 Hence, x^x>x 3. x>1 (x>1)^{+ve.... number} is always greater than 1 hence x^{x-1}>1 \frac{x^x}{x}>1 |x|^x=x^x>x 4. At x=1 |x|^x=x^x=x=1 We can clearly observe that except x=1, |x|^x is always greater than x. A Or we can do it graphically.

chetan2u · Answer

x < |x|^x will always be true except at x=0, when 0^0 cannot be defined, and at x=1, when x=1... |x|^x=1^1=1 , so x = |x|^x ... (1) x <0< |x|^x for all x<0 (2) x < |x|^x when x>0 except at 1.. for example \frac{1}{2}<\sqrt{\frac{1}{2}} or 2<2^2 So, the question is basically 'Is x eq{1} ?'.. (1) x < 1 Yes.. x eq{1} (2) x > 0 x can be 2, and x can be 1.. A

Oppenheimer1945 · Answer

only point of touch is x=1.

x>1 or x<1 would be suff to tell

Hence A)

If x is a non−zero number, is x <|x|^x?

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