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 equal to

Correct answer: B. (- (1)/(2), 0) (0, (1)/(2)) (a - 1)( x ^ 4 + x ^ 2 + 1) + (a + 1) ( x ^ 2 + x + 1) ^ 2 = 0 ( x ^ 2 + x + 1)[(a - 1)( x ^ 2 - x + 1) + (a + 1) ( x ^ 2 + x + 1 )] = 0 ( x ^ 2 + x + 1)[2a x ^ 2 + 2x + 2a] = 0 ⇒ ( x ^ 2 + x + 1)[a x ^ 2 + x + a] = 0 If two Roots are real Then, roots of a x ^ 2 + x + a = 0 should be real & distinct. ∴ 1 - 4 a ^ 2 > 0 (1 - 2a)(1 + 2a) > 0 As a ≠ 0 therefore a ∈ (- (1)/(2), 0) (0, (1)/(2))

JEE Main 201511 AprilOnline

Exam: JEE Main
Year: 2015
Date: 11 April
Mode: Online
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2015 · 11 April · Online · Mathematics · Quadratic Equation

If the two roots of the equation, (a - 1)( x ^ 4 + x ^ 2 + 1) + (a...

If the two roots of the equation,

\left(a - 1\right)\left({x}^{4} + {x}^{2} + 1\right) + \left(a + 1\right){\left({x}^{2} + x + 1\right)}^{2} = 0

are real and distinct, then the set of all values of

a

is equal to 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,

Options

A $\left(0, \dfrac{1}{2}\right)$
B $\left(- \dfrac{1}{2}, 0\right) \cup \left(0, \dfrac{1}{2}\right)$
C $\left(- \infty, - 2\right) \cup \left(2, \infty\right)$
D $\left(- \dfrac{1}{2}, 0\right)$

Correct answer

B. $\left(- \dfrac{1}{2}, 0\right) \cup \left(0, \dfrac{1}{2}\right)$

Step-by-step solution

\left(a - 1\right)\left({x}^{4} + {x}^{2} + 1\right) + \left(a + 1\right){\left({x}^{2} + x + 1\right)}^{2} = 0 $ \left({x}^{2} + x + 1\right)\left[\left(a - 1\right)\left({x}^{2} - x + 1\right) + \left(a + 1\right) ( {x}^{2} + x + 1 )\right] = 0$

\left({x}^{2} + x + 1\right)\left[2a{x}^{2} + 2x + 2a\right] = 0 $ $ ⇒ \left({x}^{2} + x + 1\right)\left[a{x}^{2} + x + a\right] = 0 $ If two Roots are real Then, roots of$ a{x}^{2} + x + a = 0 $should be real & distinct. $ ∴ 1 - 4{a}^{2} > 0 $ \left(1 - 2a\right)\left(1 + 2a\right) > 0$
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,

As $a \ne 0$ therefore

$a \in \left(- \dfrac{1}{2}, 0\right) \cup \left(0, \dfrac{1}{2}\right)$

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