1. Tardigrade
  2. Question
  3. Mathematics
  4. 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.]

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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 $\displaystyle\sum_{r=0}^{10}(2 y+r x)=$ $k$ then find the value of $\left[\frac{k}{11}\right]$.

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

Complex Numbers and Quadratic Equations

Solution:

$ 3\left(x-\frac{1}{2}\right)^2+2\left(y-\frac{3}{2}\right)^2=0$
$\therefore x =\frac{1}{2}$ and $y =\frac{3}{2}$
Hence, $\displaystyle\sum_{r=0}^{10}(2 y+r x)=k=\frac{121}{2}$