JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q62

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MathematicsSingle correct+4 / -1Question 62 of 90

Let $\alpha$ , $\beta$ be the roots of the equation $x^2-\sqrt2x+\sqrt6=0$ and $\dfrac1{\alpha^2}+1$ , $\dfrac1{\beta^2}+1$ be the roots of the equation $x^2+ax+b=0$ . Then the roots of the equation $x^2-(a+b-2)x+(a+b+2)=0$ are :

non-real complex numbers

real and both negative

real and both positive

real and exactly one of them is positive

Answer and worked solution

Official answer: B

Vieta. $\alpha+\beta=\sqrt2$ , $\alpha\beta=\sqrt6$ . The new roots have sum $2+\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}$ and product $1+\frac{2-2\sqrt6}6+\frac16$ ; together they give $a+b=-\frac56$ .

Equation. $x^2+\frac{17}6x+\frac76=0$ , i.e. $6x^2+17x+7=0$ , with roots $-\frac12$ and $-\frac73$ : real and both negative.

(B)

Topics:Roots and coefficients Nature of roots

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JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 62