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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Basics of Mathematics

BITSAT 2020 · Mathematics · Basics of Mathematics

If x, y and z are real numbers, the x^ 2 +4 y^ 2 +9 z^ 2 -6 y z-3 z...

If x,yx, y and zz are real numbers, the x2+4y2+9z2−6yz−3zx−2xyx^{2}+4 y^{2}+9 z^{2}-6 y z-3 z x-2 x y is always

Options

  1. Apositive
  2. Bnon-positive
  3. Czero
  4. Dnon-negative

Correct answer

D. non-negative

Step-by-step solution

x2+4y2+9z2−6yz−3zx−2xy=x2+(2y)2+(3z)2−(2y)(3z)−(3z)(x)−x(2y)≥0∀x,y,zx^{2}+4 y^{2}+9 z^{2}-6 y z-3 z x-2 x y=x^{2}+(2 y)^{2}+(3 z)^{2}-(2 y)(3 z)-(3 z)(x)-x(2 y) \geq 0 \forall x, y, z\left[\because \mathrma^{2}+\mathrmb^{2}+\mathrmc^{2}-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq0\right]
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