If \(xy ≠ 0\), is \(x^y \leq y^x\)?
(1) \(\sqrt{x} = y\)
\(x^y \leq y^x\)
Substitute the value of y...
\(x^{\sqrt{x}} \leq \sqrt{x}^x.............\sqrt{x}^x-x^{\sqrt{x}
]\geq{0}............\sqrt{x}^x-(\sqrt{x^}^2)^{\sqrt{x}}\geq{0}............\sqrt{x}^{\sqrt{x}}-(\sqrt{x^})^{2\sqrt{x}}\geq{0}.\)..
\(\sqrt{x}^{\sqrt{x}}(1-(\sqrt{x^})^{\sqrt{x}}\geq{0}\)
Now \(x^{\sqrt{x}}\)>0, so we have to check whether \(1-x^{\sqrt{x}}\geq{0}\) or \(1\geq{x^{\sqrt{x}}}\)..
If x <1, say \(x=\frac{1}{4}.....\sqrt{x}=\frac{1}{2} ...1>(\frac{1}{2})^{\frac{1}{2}}\)...Yes
If \(x=4....\sqrt{x}=2.....1>2^2\)..No
Insuff
(2) \(x = 2y\)
Substitute the value of x...
\((2y)^y \leq y^{2y}............. y^{2y}-(2y)^y\geq{0}............(y^y)^{2}-2^yy^y\geq{0}............(y^y)(y^y-2^y\geq{0}\)..
\(\sqrt{x}^{\sqrt{x}}(1-(\sqrt{x^})^{\sqrt{x}}\geq{0}\)
If y =1, say \(1(1-2)\geq{0}\)...No
If \(y=2....2^2(2^2-2^2)\geq{0}\)..yes
Insuff
Combined
\(\sqrt{x} = y.......x=y^2=2y....y(y-2)=0\)
So y=0 or y=2, but since \(xy ≠ 0\), y=2..
when y=2, x=4...\(x^y \leq y^x.......4^2\leq 2^4...16\leq 16\)..Yes
suff
C
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