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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

A
β(,13] and γ[127,)
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B
β(,13) and γ[127,)
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C
β(,13] and γ(127,)
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D
None of these
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Solution

The correct option is C β(,13] and γ[127,)
As x1,x2,x3 are in A.P, x1=x2d and x3=x2+d
x1+x2+x3=13x2=1x2=13
Now β=x2(x1+x3)+x1x3=2x22+x1x3
β29=(13d)(13+d)
where d is the common difference of A.P
d2=19+29β
β13 and d20
Next γ=x1x2x3=13(19d2)127γ127
Thus β(,13] and γ[127,)

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