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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Mathematics
Chapter
Quadratic Equation

BITSAT 2019 · Mathematics · Quadratic Equation

If x, y, z ∈ ℝ, x + y + z = 5, x^ 2 + y^ 2 + z^ 2 = 9 , then length...

If x,y,z∈R,x+y+z=5,x2+y2+z2=9x, y, z \in \mathbb{R}, x + y + z = 5, x^{2} + y^{2} + z^{2} = 9, then length of interval in which xx lies is

Options

  1. A8/38 / 3
  2. B4/34 / 3
  3. C2/32 / 3
  4. D1/31 / 3

Correct answer

B. 4/34 / 3

Step-by-step solution

We have,x+y+z=5x2+y2+z2=9⇒x2+y2+(5−x−y)2=9⇒y2+x2−5x−5y+xy+8=0⇒y2+y(x−5)+(x2−5x+8)=0⇒x + y + z = 5 x^{2} + y^{2} + z^{2} = 9 ⇒ x^{2} + y^{2} + ( 5 - x - y{)}^{2} = 9 ⇒ y^{2} + x^{2} - 5x - 5y + xy + 8 = 0 ⇒ y^{2} + y ( x - 5 ) + \left(x^{2} - 5x + 8\right) = 0 ⇒ Since y∈Ry \in \mathbb{R}( x - 5{)}^{2} - 4\left(x^{2} - 5x + 8\right)≥0⇒x2−10x+25−4x2+20x−32≥0⇒3x2−10x+7≤0⇒3x2−7x−3x+7≤0⇒(3x−7)(x−1)≤0⇒x∈\ge 0 ⇒ x^{2} - 10x + 25 - 4x^{2} + 20x - 32 \ge 0 ⇒ 3x^{2} - 10x + 7 \le 0 ⇒ 3x^{2} - 7x - 3x + 7 \le 0 ⇒ ( 3x - 7 ) ( x - 1 ) \le 0 ⇒ x \in\left[1, \dfrac{7}{3}\right]LengthofintervalLength of interval= \left(\dfrac{7}{3} -1\right) = \dfrac{4}{3}
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