Mathematics · 2015
JEE Main · 11 April 2015, Shift 1 · Q63
If the two roots of the equation, ( a -1)(x^4+x^2+1)+( a +1)(x^2+x+1)^2=0 are real and distinct, then the set of all values of 'a' is:
If the two roots of the equation, $\displaystyle (\mathrm{a}-1)\left(x^4+x^2+1\right)+(\mathrm{a}+1)\left(x^2+x+1\right)^2=0$ are real and distinct, then the set of all values of 'a' is :
Official answer
(3)
$\displaystyle \left(-\frac{1}{2}, 0\right) \cup\left(0, \frac{1}{2}\right)$
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JEE Main 2015 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.