1
$\begingroup$

We are required to find the no. of ordered pairs $(x,y)$ satisfying the equation

$13+12[tan^{-1}x]=24[ln x]+8[e^x]+6[cos^{-1}y]$. ($[.]$ is the Greatest integer function, e.g.$[2.3]=2, [5.6]=5, [-1.6]=[-2]$ etc)

The answer happens to be zero. I tried to arrange the terms so that I can show that the ranges on either side of the equation don't overlap, but the logarithmic and exponential terms always make the range the set of real numbers, so that doesn't work. Also, the constraint on the domain is that $x$ must be positive because of the logarithm term. I then tried to study two cases $x>1$ and then $x$ between zero and one. But I haven't made any progress. Any help would be appreciated; thanks in advance!

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Note that the L.H.S is always odd and the R.H.S is always even, so there is no solution.

$\endgroup$
5
  • $\begingroup$ is that a sufficient answer? I mean can we not rigorously 'show' that there are no solutions? $\endgroup$
    – GRrocks
    Jun 3, 2016 at 6:06
  • $\begingroup$ What's the problem in this answer? $\endgroup$
    – Nikunj
    Jun 3, 2016 at 6:07
  • $\begingroup$ No I don't say that there is a problem; I haven't heard about no solutions if one side is even and other is odd; so I thought there was maybe a more ranges-based answer. Thanks anyways. $\endgroup$
    – GRrocks
    Jun 3, 2016 at 6:08
  • $\begingroup$ One side is always even and other is always odd, the word 'always' is what matters here. An even number can never be equal to an odd one. $\endgroup$
    – Nikunj
    Jun 3, 2016 at 6:11
  • $\begingroup$ Thanks a lot @Nikunj ...but I'd still like to look at a rigorous solution... $\endgroup$
    – GRrocks
    Jun 3, 2016 at 7:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .