If x, y ∈ R satisfy the equation 3 x2+2 y2-6 y-3 x+(21/4)=0, then the value of displaystyle∑r=010(2 y+r x)= k then find the value of [(k/11)]. [Note: [x] denotes greatest integer less than or equal to x.]

Question

If x, y ∈ R satisfy the equation 3 x2+2 y2-6 y-3 x+(21/4)=0, then the value of displaystyle∑r=010(2 y+r x)= k then find the value of [(k/11)]. [Note: [x] denotes greatest integer less than or equal to x.] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

3(x-(1/2))2+2(y-(3/2))2=0 ∴ x =(1/2) and y =(3/2) Hence, displaystyle∑r=010(2 y+r x)=k=(121/2)

Q. If $x, y \in R$ satisfy the equation $3 x^{2} + 2 y^{2} - 6 y - 3 x + \frac{21}{4} = 0$ , then the value of $r = 0 \sum 10 (2 y + r x) =$ $k$ then find the value of $[\frac{k}{11}]$ .

[Note: $[x]$ denotes greatest integer less than or equal to $x$ .]

$3 (x - \frac{1}{2})^{2} + 2 (y - \frac{3}{2})^{2} = 0$
$∴ x = \frac{1}{2}$ and $y = \frac{3}{2}$
Hence, $r = 0 \sum 10 (2 y + r x) = k = \frac{121}{2}$

Q. If x,y∈R satisfy the equation 3x2+2y2−6y−3x+421​=0, then the value of r=0∑10​(2y+rx)= k then find the value of [11k​]. [Note: [x] denotes greatest integer less than or equal to x.]

3(x−21​)2+2(y−23​)2=0 ∴x=21​ and y=23​ Hence, r=0∑10​(2y+rx)=k=2121​

Q. If $x, y \in R$ satisfy the equation $3 x^{2} + 2 y^{2} - 6 y - 3 x + \frac{21}{4} = 0$ , then the value of $r = 0 \sum 10 (2 y + r x) =$ $k$ then find the value of $[\frac{k}{11}]$ .

[Note: $[x]$ denotes greatest integer less than or equal to $x$ .]

$3 (x - \frac{1}{2})^{2} + 2 (y - \frac{3}{2})^{2} = 0$
$∴ x = \frac{1}{2}$ and $y = \frac{3}{2}$
Hence, $r = 0 \sum 10 (2 y + r x) = k = \frac{121}{2}$