• edited by
138 views
0 0 votes

The set of all real values of $x$ for which $(x^{2}-\mid x+9 \mid +x)>0$, is

  1. $(-\infty,-3) \cup(3, \infty)$
  2. $(-\infty,-9) \cup(3, \infty)$
  3. $(-9,-3) \cup(3, \infty)$
  4. $(-\infty,-9) \cup(9, \infty)$

Please log in or register to answer this question.

Position:
Show:

Related questions

0 0 votes
0 0 answers
147
147 views
Shubham Sharma 2 asked Jun 2
147 views
If $9^{x^{2}+2 x-3}-4\left(3^{x^{2}+2 x-2}\right)+27=0$, then the product of all possible values of $x$ is$1. 30$$2. 20$$3.5$$4.15$
0 0 votes
0 0 answers
153
153 views
Shubham Sharma 2 asked Jun 2
153 views
The average number of copies of a book sold per day by a shopkeeper is $60$ in the initial seven days and $63$ in the initial eight days, after the book launch. On the ni...
0 0 votes
0 0 answers
153
153 views
Shubham Sharma 2 asked Jun 2
153 views
An item with a cost price of Rs.$1650$ is sold at a certain discount on a fixed marked price to earn a profit of $20 \%$ on the cost price. If the discount was doubled, t...
0 0 votes
0 0 answers
145
145 views
Shubham Sharma 2 asked Jun 2
145 views
If $m$ and $n$ are integers such that $(m+2 n)(2 m+n)=27$, then the maximum possible value of $2 m-3 n$ is
0 0 votes
0 0 answers
135
135 views
Shubham Sharma 2 asked Jun 2
135 views
The sum of digits of the number $(625)^{65} \times(128)^{36}$, is