1x= \frac{3}{\sqrt{2}}
$Find value of $x$ that satisfy the equation $x^{\lfloor \ x\rfloor}=\frac{9}{2}$\\\\ Where $\lfloor x \rfloor $= G.I.F=greatest integer function.
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5 Answers
1firstly i just did it by hit and trial and x = 3√2(which is almost 2.13..) satisfied .Now for any value of x>3 you'll get a value of 'x' which will be lesser than 3√2 and thus the solution for 'x' that you obtain will be lesser that 3 ,which is not possible because x is greater than 3 already;so no solution for any x>3
Similarly u can prove that there's no solution for any x<2 also
1pritishmast answer seems to be correct.
see
f(x) = 1 0<x<1
x 1<x<2
x2 2<x<3
thus the inequlity has a solution in 2<x<3 x = 3/√2