460 reputation
120
bio website alfahub.wordpress.com
location India
age 17
visits member for 1 year, 11 months
seen Aug 17 at 6:51

Hey Stalkers!


I am Saharsh. A high school science student living in Gandhinagar. My basic interest includes Mathematics, Physics, Chemistry and Computer Science.

These are my academic fields and I am here for helping others over their questions and of course also for seeking some help from my side. If you think I can help you on any issue then please feel free to contact me. I am always right here on stackexchange! :)


Jul
3
revised A proof that $1=2$. May I know why it’s false?
mathjax added
Jul
3
comment Why $2x$? Can't it be $x$?
@MJD you're probably pointing to a genuine discussion.
Jul
3
suggested suggested edit on A proof that $1=2$. May I know why it’s false?
Jul
3
asked Why $2x$? Can't it be $x$?
Jul
2
awarded  Curious
Jun
28
awarded  Organizer
Jun
28
revised Evaluating Determinants using elementary operations
tag "determinant" added
Jun
28
suggested suggested edit on Evaluating Determinants using elementary operations
Jun
28
comment Evaluating Determinants using elementary operations
What do you really mean by evaluation? Adding any kind of question like you are asking will help other in explaining you the steps to solve. Also write your approach while you were trying to solve it. It'll not only help you but also others to understand what you really want to know.
Jun
26
revised Values of $a$ for which range of $y=\frac{x+1}{a+x^2}$ contains the interval [0,1]?
mathematical issue reviewed
Jun
26
suggested suggested edit on Values of $a$ for which range of $y=\frac{x+1}{a+x^2}$ contains the interval [0,1]?
Jun
26
comment Values of $a$ for which range of $y=\frac{x+1}{a+x^2}$ contains the interval [0,1]?
I still didn't understood that how you took value of $x$ to be greater than zero. I mean it can belong between $(-1,\infty)$, as if the value of $x$ in negative fraction will also lead to the value of $y$ to be between $[0,1]$. Don't you think?
Jun
26
revised Values of $a$ for which range of $y=\frac{x+1}{a+x^2}$ contains the interval [0,1]?
edited the interval
Jun
26
suggested suggested edit on Values of $a$ for which range of $y=\frac{x+1}{a+x^2}$ contains the interval [0,1]?
Jun
26
comment Values of $a$ for which range of $y=\frac{x+1}{a+x^2}$ contains the interval [0,1]?
I think the interval might be $(-\infty,-1)\cup(1,\frac54]$, check it out.
Jun
26
comment Squeeze theorem
You should edit your question once. Add this into your question. Also specify the limit of $n$, like $\lim_{n\to 0}$
Jun
26
comment Combination Problem on Create Groups !!!
Here you might have to use the application of Permutation and Combination. What you had tried yet?
Jun
26
comment Squeeze theorem
Allsony, Can you give any example question?
Jun
23
awarded  Critic
Jun
16
accepted Determination of $1^\infty$ indeterminate forms