What is the number of integral values of $x$ satisfying the inequality:

$$\frac{(e^x-1)(\sin(x)-2)(x^2-5x+4)}{x^2(-x^2+x-2)(2x+3)}\le 0$$

I was able to find three solutions: $0$, $1$ and $4$. Is there any other solution?

  • $\begingroup$ Is it correct now? $\endgroup$ – Dr. Sonnhard Graubner May 20 at 16:58
  • $\begingroup$ Yes , thanks sir $\endgroup$ – Abhishek Kumar May 20 at 16:58
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    $\begingroup$ Function is not defined at $0$ $\endgroup$ – Andrei May 20 at 17:09
  • $\begingroup$ But 0/0 is zero I think it is defined $\endgroup$ – Abhishek Kumar May 20 at 17:13
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    $\begingroup$ Why $0$? Try plotting just $(e^x-1)/x^2$ The function has a discontinuity at $0$. On one side it goes to $+\infty$, on the other side it goes to $-\infty$ $\endgroup$ – Andrei May 20 at 17:21

Note the $x^2$ in the denominator ensures the function is not defined at $x=0$, otherwise has no effect on the inequality, so we may ignore it. Similarly, $\sin x -2$ is always negative, and so is $-x^2+x-2$, so both may be together ignored. As $e^x-1$ has the same sign as $x$, essentially we can substitute that, and equivalently solve for $$\frac{x(x-1)(x-4)}{2x+3} \leqslant 0$$

The intervals to check are $x< -\frac32,x \in (-\frac32, 0), x \in (0, 1), x \in [1, 4]$ and $x> 4$, which is easily done to get $x \in (-\frac32,0) \cup [1, 4]$, so integral solutions are $x\in \{-1, 1, 2, 3, 4\}$.


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