If P is a constant, fx=x2x3x4236PP2P3 andf"x=0 has roots α,β ,then

f"x=26x12x2236pp2p3=0Expand the determinant⇒ 122p2−3px2−6x2p3−6p+6p2p−2=0⇒ 12p2p−3x2−12pp2−3+6p2p−2=0At P=0, its an identityAt P=3,ba=0,so roots are opposite in sign and equal in magnitudeAt P=2,ca=0, so product of the roots is zero. At P=−3,ca=6p2p−212p2p−3 negativeSo product of the roots is negative

If P is a constant, fx=x2x3x4236PP2P3 andf"x=0 has roots α,β ,then

f"x=26x12x2236pp2p3=0Expand the determinant⇒ 122p2−3px2−6x2p3−6p+6p2p−2=0⇒ 12p2p−3x2−12pp2−3+6p2p−2=0At P=0, its an identityAt P=3,ba=0,so roots are opposite in sign and equal in magnitudeAt P=2,ca=0, so product of the roots is zero. At P=−3,ca=6p2p−212p2p−3 negativeSo product of the roots is negative

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$I f P i s a c o n s \tan t$ , $f (x) = |\begin{matrix} x^{2} & x^{3} & x^{4} \\ 2 & 3 & 6 \\ P & P^{2} & P^{3} \end{matrix}|$ $a n d$ $f " (x) = 0$ $h a s r o o t s$ $α, β$ , $t h e n$

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$α$ and $β$ have opposite sign and equal magnitude at $p = \sqrt{3}$

at p = 0, $f " (x) = 0$ represents an identity

$A t P = 2, p r o d u c t o f t h e r o o t s i s u n i t y$

at $p = - \sqrt{3}$ product of roots of $f " (x) = 0$ is negative

answer is A, B, D.

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Detailed Solution

$\begin{array}{l} f " (x) = |\begin{matrix} 2 & 6 x & 12 x^{2} \\ 2 & 3 & 6 \\ p & p^{2} & p^{3} \end{matrix}| = 0 \\ E x p a n d t h e d e t e r m i n a n t \\ \Rightarrow 12 (2 p^{2} - 3 p) x^{2} - 6 x (2 p^{3} - 6 p) + 6 p^{2} (p - 2) = 0 \\ \Rightarrow 12 p (2 p - 3) x^{2} - 12 p (p^{2} - 3) + 6 p^{2} (p - 2) = 0 \\ At P = 0, its an identity \\ At P = \sqrt{3}, \frac{b}{a} = 0, so roots are opposite in sign and equal in magnitude \\ At P = 2, \frac{c}{a} = 0, so product of the roots is zero. \\ At P = - \sqrt{3}, \frac{c}{a} = \frac{6 p^{2} (p - 2)}{12 p (2 p - 3)} negative \\ So product of the roots is negative \end{array}$