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Question

The real numbers x1,x2,x3 satisfying the equation x3x2+βx+γ=0 are in A.P. Find the intervals in which β and γ lie, respectively:
  1. (12, 13)(127, 127)
  2. (,13][127,)
  3. (,3)
  4. none of these


Solution

The correct option is B (,13][127,)
  x1,x2,x3 are in AP.
   x1=ad, x2=a, x3=a+d
where d is the common difference
Now, since x1,x2,x3 are the roots of the given equation
x3x2+βx+γ=0
So,       α=x1+x2+x3=1
       (a - d) + (a) + (a + d) = 1                  . . . (i)
αβ=x1x2+x2x3+x1x3
     β = (a - d) a + a(a +d) + (a - d) (a + d)      . ..  (ii)
and =αβγ=x1x2x3=γ=(ad)(a)(a+d)                 . . . (iii)
hence from (i) we get a=13
and from eq. (ii) we get
β=3a2d2    β=13d2         (a=13)Thus       β=13d213         [ d20]      β13     βϵ(,13]
Again from eq. (iii)
a(a2d2)=γ     (127)+(d23)=γ      γ=d23127      γ127        ( d20)
Hence option (a) is correct.
 

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