A) No real root is \[\left| a \right|<4\]
B) All roots real if \[\left| a \right|>4\]
C) At most two real roots for all \[a\in R\]
D) None of these
Correct Answer: A
Solution :
| The equation is \[{{({{x}^{3}}+1)}^{2}}+a({{x}^{3}}+1)+4=0\] \[\Rightarrow \]\[{{t}^{2}}+at+4=0\] | ...(i) |
| Where\[t={{x}^{3}}+1.\]If \[D<0\Rightarrow {{a}^{2}}-16<0\] then above equation has no real roots and all six roots are imaginary. If\[D\ge 0\], then equation (i) has two roots. So if \[{{a}^{2}}-16\ge 0,\]then the given equation has two real roots and other imaginary, except when\[t=1\]. That is when\[a=-5\], and equation is\[{{x}^{3}}\left( {{x}^{3\text{ }}}-3 \right)\text{ }=0\], which has three repeated, one real and two imaginary roots. | |
You need to login to perform this action.
You will be redirected in
3 sec
