Letfx=x3+ax2+bx+c be a cubic polynomial with real coefficients and all real roots. Also fi=1 where : .i=−1Statement-I: All 3 roots of fx=0 are zero.Statement-II: a+b+c=0.

Let x1,x2,x3∈R be the roots fx=0 of ∴fx=x−x1x−x2x−x3 fi=i−x1i−x2i−x3 fi=x1−ix2−ix3−i=1 ∴x12+1x22+1x32+1=1This is possible only if x1=x2=x3=0 ⇒fx=x3⇒a=0=b=c⇒a+b+c=0 (Sum of coefficients zero can not imply that all zero roots).

Letfx=x3+ax2+bx+c be a cubic polynomial with real coefficients and all real roots. Also fi=1 where : .i=−1Statement-I: All 3 roots of fx=0 are zero.Statement-II: a+b+c=0.

Let x1,x2,x3∈R be the roots fx=0 of ∴fx=x−x1x−x2x−x3 fi=i−x1i−x2i−x3 fi=x1−ix2−ix3−i=1 ∴x12+1x22+1x32+1=1This is possible only if x1=x2=x3=0 ⇒fx=x3⇒a=0=b=c⇒a+b+c=0 (Sum of coefficients zero can not imply that all zero roots).

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Let $f (x) = x^{3} + a x^{2} + b x + c$ be a cubic polynomial with real coefficients and all real roots. Also $|f (i)| = 1$ where : . $i = \sqrt{- 1}$
Statement-I: All 3 roots of $f (x) = 0$ are zero.
Statement-II: $a + b + c = 0.$

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Statement I is True, Statement II is True and Statement II is correct explanation of Statement I

Statement I is True, Statement II is True but Statement II is not correct explanation of Statement I

Statement I is True, Statement II is False

Statement I is False, Statement II is True

answer is C.

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Detailed Solution

Let $x_{1}, x_{2}, x_{3} \in R$ be the roots $f (x) = 0$ of
$∴ f (x) = (x - x_{1}) (x - x_{2}) (x - x_{3})$
$f (i) = (i - x_{1}) (i - x_{2}) (i - x_{3})$
$|f (i)| = |x_{1} - i| |x_{2} - i| |x_{3} - i| = 1$
$∴ \sqrt{x_{1}^{2} + 1} \sqrt{x_{2}^{2} + 1} \sqrt{x_{3}^{2} + 1} = 1$
This is possible only if $x_{1} = x_{2} = x_{3} = 0$
$\Rightarrow f (x) = x^{3}$ $\Rightarrow a = 0 = b = c$ $\Rightarrow a + b + c = 0$
(Sum of coefficients zero can not imply that all zero roots).