recategorized by
540 views
0 votes
0 votes

The set $\{x \: : \begin{vmatrix} x+\frac{1}{x} \end{vmatrix} \gt6 \}$ equals the set

  1. $(0,3-2\sqrt{2}) \cup (3+2\sqrt{2}, \infty)$
  2. $(- \infty, -3-2\sqrt{2}) \cup (-3+2 \sqrt{2}, \infty)$
  3. $(- \infty, 3-2\sqrt{2}) \cup (3+2\sqrt{2}, \infty)$
  4. $(- \infty, -3-2\sqrt{2}) \cup (-3+2 \sqrt{2},3-2\sqrt{2}) \cup (3+2 \sqrt{2}, \infty )$
recategorized by

1 Answer

0 votes
0 votes
We know,IF $|f(x)| \geqslant a, \quad ;a>0$

then,

$$f(x) \geqslant a \quad \cup \quad f(x) \leqslant-a$$

$\frac{x^{2}+1}{x}> 6$  &  $\frac{x^{2}+1}{x}<-6$

$$\frac{x^{2}-6 x+1}{x}>0..........(1)$$

$$\frac{x^{2}+6 x+1}{x}<0............(2)$$

 

From (1) ,I can find,
$$x \in(0,3-2 \sqrt{2}) \cup(3+2 \sqrt{2}, \alpha)$$

From (2) ,I can find,
$$x \in(-\alpha,-3-2 \sqrt{2}) \cup(-3+2 \sqrt{2},0)$$

 

So the domain is :

$$(-\infty,-3-2\sqrt2)\cup(-3+2\sqrt2,0)\cup(0,3-2\sqrt2)\cup(3+2\sqrt2,+\infty)$$

Related questions

3 votes
3 votes
2 answers
1
Arjun asked Sep 23, 2019
1,641 views
If $a,b$ are positive real variables whose sum is a constant $\lambda$, then the minimum value of $\sqrt{(1+1/a)(1+1/b)}$ is$\lambda \: – 1/\lambda$$\lambda + 2/\lambda...
3 votes
3 votes
3 answers
2
Arjun asked Sep 23, 2019
1,369 views
Let $x$ be a positive real number. Then$x^2+\pi ^2 + x^{2 \pi} x \pi+ (\pi + x) x^{\pi}$$x^{\pi}+\pi^x x^{2 \pi} + \pi ^{2x}$$\pi x +(\pi+x)x^{\pi} x^2+\pi ^2 + x^{2 \...
1 votes
1 votes
1 answer
3
Arjun asked Sep 23, 2019
733 views
Consider the polynomial $x^5+ax^4+bx^3+cx^2+dx+4$ where $a,b,c,d$ are real numbers. If $(1+2i)$ and $(3-2i)$ are two two roots of this polynomial then the value of $a$ i...
1 votes
1 votes
1 answer
4
Arjun asked Sep 23, 2019
1,186 views
Consider the following system of equivalences of integers,$$x \equiv 2 \text{ mod } 15$$$$x \equiv 4 \text{ mod } 21$$The number of solutions in $x$, where $1 \leq x \leq...