Question

If the equation $a{x}^n+3x^5+b{x}^2+c=0,$ $n\in Z^{+}$ has infinite number of real solutions, then $a+b+c+n$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is A $2$

$a{x}^n+3x^5+b{x}^2+c=0,$ $n\in Z^{+}$
When $n\ne 5,$ then for any values of $a,b,c,$ we will not get infinite solutions.
So, $n=5$
Now, $a{x}^n+3x^5+b{x}^2+c=0$
$\Rightarrow (a+3)x^5+b{x}^2+c=0$
This equation has infinite solutions, if it is an identity in $x.$
$b=c=0,a+3=0\Rightarrow a=-3\therefore a+b+c+n=-3+0+5=2$