JEE Main 2015MathematicsQuadratic EquationHardMCQ

JEE Main 2015 — Quadratic Equation Question with Solution

JEE Main 2015 (11 Apr Online)

Question

If the two roots of the equation, a-1 x4+x2+1+a+1x2+x+12=0 are real and distinct, then the set of all values of a is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B-12,0 ∪0,12

Step-by-step explanation

a-1x4+x2+1+a+1x2+x+12=0

x2+x+1 a-1x2-x+1+a+1(x2+ x+1)=0

x2+x+12ax2+2x+2a=0

⇒ x2+x+1 ax2+x+a=0

If two Roots are real

Then, roots of ax2+x+a=0 should be real & distinct.

∴1-4a2>0

1-2a1+2a>0

As a≠0 therefore

a∈-12,0 ∪0,12

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About this question

This is a previous-year question from JEE Main 2015, covering the Quadratic Equation chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.