JEE Main 2014 · 9 Apr · Shift 1 · Mathematics Q64

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MathematicsSingle correct+4 / -1Question 64 of 90

If $\dfrac1{\sqrt\alpha}$ and $\dfrac1{\sqrt\beta}$ are the roots of the equation, $ax^2+bx+1=0$ ( $a\ne0$ , $a,b\in\mathbf R$ ), then the equation, $x\left(x+b^3\right)+\left(a^3-3abx\right)=0$ has roots :

\alpha^{3/2}

and

\beta^{3/2}

\alpha\beta^{1/2}

and

\alpha^{1/2}\beta

\sqrt{\alpha\beta}

and

\alpha\beta

\alpha^{-3/2}

and

\beta^{-3/2}

Answer and worked solution

Official answer: A

Coefficients. From the product of the roots, $\dfrac1{\sqrt{\alpha\beta}}=\dfrac1a$ , so $a=\sqrt{\alpha\beta}$ . From their sum, $\dfrac{\sqrt\alpha+\sqrt\beta}{\sqrt{\alpha\beta}}=-\dfrac ba$ , so $b=-(\sqrt\alpha+\sqrt\beta)$ .

New equation. It is $x^2+(b^3-3ab)x+a^3=0$ , with
$-(b^3-3ab)=(\sqrt\alpha+\sqrt\beta)^3-3\sqrt{\alpha\beta}(\sqrt\alpha+\sqrt\beta)=\alpha^{3/2}+\beta^{3/2},\qquad a^3=(\alpha\beta)^{3/2}.$

The roots are $\alpha^{3/2}$ and $\beta^{3/2}$ .

Topics:Roots and coefficients

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JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 64