I was doing problems when I encountered this: IIT-JEE adv. 201 paper-II problem no. 50 (maths), I think the answer should be a) but answer given is d) , share your thoughts and help me please. :)

Q: The The quadratic equation \(f\left( x \right) =0\) with real coefficients has purely imaginary roots. Then the equation \(f\left( f\left( x \right) \right) =0\) has

(A) only purely imaginary roots

(B) all real roots

(C) two real and two purely imaginary roots

(D) neither real nor purely imaginary roots

Answer
Let \(f\left( x \right) =a{ x }^{ 2 }+c\) Where \(a\) and \(c\) are Real and \(\alpha i\) and \(-\alpha i\) be the roots of the given equation.

we can write from above equation that \(f\left( \alpha i \right) =f\left( -\alpha i \right) =0\)

We need to find the roots of \(f\left( f\left( x \right) \right) =0\)

For \(f\left( f\left( x \right) \right) =0\), \(f\left( x \right) \) Should be Purely Imaginary which gives us

\(f\left( x \right) =\alpha i=a{ x }^{ 2 }+c\) \(--------(1)\)

Case 1 : x is Purely Real

Now since a and c are Real then x cannot be real to satisfy the Equation \((1)\) => No real Roots.

Case 2 : x is Purely Imaginary

since x is purely imaginary \({ x }^{ 2 }\) is Purely Real which gives us No Imaginary Roots.

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## Comments

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TopNewestQ: The The quadratic equation \(f\left( x \right) =0\) with real coefficients has purely imaginary roots. Then the equation \(f\left( f\left( x \right) \right) =0\) has(A) only purely imaginary roots

(B) all real roots

(C) two real and two purely imaginary roots

(D) neither real nor purely imaginary roots

AnswerLet \(f\left( x \right) =a{ x }^{ 2 }+c\) Where \(a\) and \(c\) are Real and \(\alpha i\) and \(-\alpha i\) be the roots of the given equation.we can write from above equation that \(f\left( \alpha i \right) =f\left( -\alpha i \right) =0\)

We need to find the roots of \(f\left( f\left( x \right) \right) =0\)

For \(f\left( f\left( x \right) \right) =0\), \(f\left( x \right) \) Should be Purely Imaginary which gives us

\(f\left( x \right) =\alpha i=a{ x }^{ 2 }+c\) \(--------(1)\)

Case 1:x is Purely RealNow since a and c are Real then x cannot be real to satisfy the Equation \((1)\) =>

No real Roots.Case 2:x is Purely Imaginarysince x is purely imaginary \({ x }^{ 2 }\) is Purely Real which gives us

No Imaginary Roots.Hence the answer is \((D)\)

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thank you bro very much , now i think i get it :) , you are awesome :) thnx again .

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No problem Bro Just tag me

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@Prakhar Bindal @Aniket Sanghi @Rishabh Deep Singh

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