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siddharthkapoor
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11 Oct 2019, 01:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
38% (01:49) correct
62%
(02:01)
wrong
based on 196
sessions
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Not Attempted Yet
11 Oct 2019, 06:02
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We can break the curve in 4 parts.
1. x<0
\(|x|^x\) is always positive
Hence, \(x < |x|^x\)
2. 0<x<1
for positive values of x, \(|x|^x= x^x\)
\(x^{x-1}>1\)
The above inequality is true as 0<x<1, x-1<0 and 1/x>1
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.
1. x<0
\(|x|^x\) is always positive
Hence, \(x < |x|^x\)
2. 0<x<1
for positive values of x, \(|x|^x= x^x\)
\(x^{x-1}>1\)
The above inequality is true as 0<x<1, x-1<0 and 1/x>1
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.
iamsiddharthkapoor
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11 Oct 2019, 22:40
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iamsiddharthkapoor
\(x < |x|^x\) 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 \(x\neq{1}\)?'..
(1) x < 1
Yes..\(x\neq{1}\)
(2) x > 0
x can be 2, and x can be 1..
A
