Which of the following is/are correct

Difficult

A

$α, β, γ are roots of 2 x^{3} + x^{2} - 7 = 0 then the value of \sum \frac{α}{β} + \frac{β}{α} is - 3$
B

For all real $x, y$ the minimum value of the expression $x^{2} + 2 x y + 3 y^{2} - 6 x - 2 y cannot be less than - 11$
C

$If ab.c are real and satisfy 2 a + 3 b + 6 c = 0 then the equation a x^{2} + b x + c = 0 h a s r e a l r o o t s$
D

There exists two distinct real values of ‘m’ for which the expression $2 x^{2} + m x y + 3 y^{2} - 5 y - 2 can be expressed as product of linear factors$

Solution

$A) α + β + γ = - \frac{1}{2}$
$α β + β γ + γ α = 0$
$α β γ = \frac{7}{2}$
$(α β γ) (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}) = 0 \Rightarrow \frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = 0$
$Now (α + β + γ) (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}) = 1 + \frac{α}{β} + \frac{α}{γ} + \frac{β}{α} + 1 + \frac{β}{γ} + \frac{γ}{α} + \frac{γ}{β} + 1$
$(\frac{- 1}{2}) (0) = 3 + \frac{α}{β} + \frac{β}{α} + \frac{α}{γ} + \frac{γ}{α} + \frac{γ}{β} + \frac{β}{γ}$
$\sum \frac{α}{β} + \frac{β}{α} = - 3$
$B) x^{2} + 2 x y + 3 y^{2} - 6 x - 2 y \geq k$
$x^{2} + x (2 y - 6) + (3 y^{2} - 2 y - k) \geq 0$
$Δ \leq 0 \Rightarrow (2 y - 6)^{2} - 4 (1) (3 y^{2} - 2 y - k) \leq 0$
$\Rightarrow 4 (y - 3)^{2} - 4 (3 y^{2} - 2 y - k) \leq 0$
$\Rightarrow y^{2} - 6 y + 9 - 3 y^{2} + 2 y + k \leq 0$
$- 2 y^{2} - 4 y + (9 + k) = 0$
$2 y^{2} + 4 y - (9 + k) = 0$
$y \in R \Rightarrow Δ \leq 0$
$\Rightarrow 4^{2} - 4 (2) (- 9 (9 + k)) \leq 0$
$\Rightarrow 8 k + 88 \leq 0$
$\Rightarrow k \leq - 11$
$C) Consider f (x) = 2 a x^{3} + 3 b x^{2} + 6 c x$
$f (0) = 0, f (1) = 0$
$f (x) = 6 a x^{2} + 6 b x + 6 c = 6 (a x^{2} + b x + c)$
$By Rolle's theorem$
$f^{'} (x) has a root between (0, 1)$
$D) 2 x^{2} + m x y + 3 y^{2} - 5 y - 2 = (d x + b y + c) (p x + 2 y + r)$
$\Rightarrow m^{2} = 49$
$\Rightarrow m = 7, - 7$

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