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# Help Needed

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. :)

paper here :)

Note by Brilliant Member
8 months, 2 weeks ago

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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.

Hence the answer is $$(D)$$

- 8 months, 2 weeks ago

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

- 8 months, 2 weeks ago

No problem Bro Just tag me

- 8 months, 2 weeks ago