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The positive values of $a$ for which the equation $\displaystyle \lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right)$ will have no solution is,

where $\lfloor x \rfloor = $ floor function of $x$

Options

$(a)\;\;(0,1)$

$(b)\;\; (1,2)$

$(c)\;\; (0,2)$

$(d)$ None of these

$\bf{My\; Try::}$ Using the formula $x-1<\lfloor x \rfloor \leq x.,$ we get $\displaystyle (x+a)-1< \lfloor x+a \rfloor \leq (x+a)$

Now Given $\displaystyle \lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right).$ So $\displaystyle (x+a) - 1<\sin \left(\frac{\pi x}{2}\right)\leq (x+a)$

Now we will solve $\displaystyle (x+a)-1<\sin \left(\frac{\pi x}{2}\right)$ and $\displaystyle (x+a)\leq \sin \left(\frac{\pi x}{2}\right)$ for Common values of $x$

But I Did Understand How can I solve it

Help me

Thanks

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observe that $\sin(\frac{\pi x}{2})$ only takes integer values for integer $x$'s, hence we can restrict our attention to $x \in \mathbb{Z}$ and so the floor function becomes obsolete. Now, the situation is even better, since there aren't plenty of integer values $\sin$ can take - it's just $-1,0$ and $1$, therefore, since it's a test question the simplest way would be to check all the posibilities. Let me do point a) and I hope you'll finish the rest by yourself following what I do.

first let $a = 0$. we want an integer $x$ such that $x = \sin(\frac{\pi x}{2})$. all values we need to check are $1,0,-1$. $x=1$ works, so we're done, that can't be the answer.

we could now cross answer a) as a wrong one, but nevertheless let me do the case of $a=1$ so that its more instructive

if $a = 1$ then we're looking for $x$ s.t.

$$ x+ 1 = \sin(\frac{\pi x}{2})$$ because of what've said about the range of $\sin$ we only need to check $x = 0$, $x = -1$ and $x = -2$.

$x = 0$ won't work since it leads to $$1 = 0$$

$ x = -1$ doesn't work either cause we get $$ 0 = -1$$

finally $x = -2$ gives $$-1 = 0$$ which isn't true either, so for $a=1$ there are no solutions to your equation.

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  • $\begingroup$ W|A shows two solutions for $a=1$. wolframalpha.com/input/?i=floor%28x%2B1%29%3Dsin%28pi+x%2F2%29 $\endgroup$ – Pranav Arora Jun 7 '14 at 4:15
  • $\begingroup$ it's a numerical computation which in this case yields a wrong result, check it for yourself / try to find a gap in my proof $\endgroup$ – mm-aops Jun 7 '14 at 4:21
  • $\begingroup$ although the answer would change if we defined the floor function in such a way that $[x] < x$ for all $x$, i.e. $[1] = 0$, but that's uncommon and I'm pretty sure it's not the case in wolfram $\endgroup$ – mm-aops Jun 7 '14 at 4:22

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