Multi correct answer ,
previous year jee advance question.
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If x + y + z = 5 and xy + yz + zx = 3 (x,y,z€R) then
(a) maximum value of x,y , and z are same
(b) minimum value of x, y, and z are same .
(c) maximum value of x = 13/3 and minimum value of x = -1
(d) Probability for x is positive , is 13/16 .
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Answer :-(a,b,c,d)
Step-by-step explanation:
We have , x+ y + z = 5
z = 5-y-x ____(1)
And xy+ yz + zx = 3 (given )
From )1) , xy + y (5-y-x) + (5-y-x) x = 3
=> xy +5y -y² - xy +5x -yx -x² = 3
=> y²+y(x-5) +x²-5x+3 =0
we know that ' D = b²-4ac ≥0 for real roots
=> (x-5)²-4(x²-5x+3)≥ 0
=> x²-10x +25-4x²+20 -12 ≥0
=> 3x² -10x -13 ≤0
=>3x²-13x+3x-13≤0
=> (3x-13) (x+1) ≤0
=> -1≤ x 1≤3/3
similarly , -1≤ y≤ 13/3
and- 1 ≤z ≤13/3
Thus option (a) ,(b) and (c) are correct .
now required Probability
= ₀∫¹³ˡ³dx/_₁∫¹³ˡ³dx =13/3/13/3+1=>13/16 Answer
Hence option (D) is also correct
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