Minimum value of x2 - 2xy - 3y2 - 6x - 2y x- y- - is

Let t = x^2 + 2xy + 3y^2 6x 2y, x, y, ϵℝ. ∴ x^2 + 2(y 3) x + 3y^2 2y t = 0 Δ⩾ 0 ⇒ 4(y 3)^2 4(3y^2 2y t) ≥ 0 ⇒ y^2 + 9 6y 3y^2 + 2y + t ≥ ...

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Standard XII

Mathematics

Nature of Roots

Minimum value...

Question

Minimum value of $x^{2} + 2 x y + 3 y^{2} - 6 x - 2 y$ $x, y, ϵ R$ is

- 9

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- 11

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- 12

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Solution

The correct option is B
- 11

$Let t = x^{2} + 2 x y + 3 y^{2} - 6 x - 2 y, x, y, ϵ R . ∴ x^{2} + 2 (y - 3) x + 3 y^{2} - 2 y - t = 0 Δ ⩾ 0 \Rightarrow 4 (y - 3)^{2} - 4 (3 y^{2} - 2 y - t) \geq 0 \Rightarrow y^{2} + 9 - 6 y - 3 y^{2} + 2 y + t \geq 0 \Rightarrow - 2 y^{2} - 4 y + 9 + t \geq 0 i.e., t \geq 2 (y^{2} + 2 y + 1) - 11 t \geq 2 (y + 1)^{2} - 11 ∴ Min is - 11 at y = - 1$

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Minimum value of $x^{2} + 2 x y + 3 y^{2} - 6 x - 2 y$ $x, y, ϵ R$ is

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Minimum value of x2+2xy+3y2−6x−2y x,y,ϵR is

The correct option is B - 11 Let t=x2+2xy+3y2−6x−2y,x,y,ϵR.∴x2+2(y−3)x+3y2−2y−t=0Δ⩾0⇒4(y−3)2−4(3y2−2y−t)≥0⇒y2+9−6y−3y2+2y+t≥0⇒−2y2−4y+9+t≥0i.e.,t≥2(y2+2y+1)−11t≥2(y+1)2−11∴Min is −11 at y=−1

Minimum value of $x^{2} + 2 x y + 3 y^{2} - 6 x - 2 y$ $x, y, ϵ R$ is