JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q63

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MathematicsSingle correct+4 / -1Question 63 of 90

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 :

\left(-\dfrac12,0\right)

(-\infty,-2)\cup(2,\infty)

\left(-\dfrac12,0\right)\cup\left(0,\dfrac12\right)

\left(0,\dfrac12\right)

Answer and worked solution

Official answer: C

$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ , and $x^2+x+1>0$ . Dividing by it:
$(a-1)(x^2-x+1)+(a+1)(x^2+x+1)=2ax^2+2x+2a=0.$
Real distinct roots need $a\ne0$ and $1-4a^2>0$ : $a\in\left(-\frac12,0\right)\cup\left(0,\frac12\right)$ .

Topics:Nature of roots Equations reducible to quadratic

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JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 63