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# If the roots of the equation ${\mathrm{x}}^{3}+{\mathrm{px}}^{2}+\mathrm{qx}-1=0$ form an increasing G.P., where p and q are real, then

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a
p + q = 0
b
p∈(−3, ∞)
c
one of the roots is unity
d
one root is smaller than 1 and one root is greater than 1

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detailed solution

Correct option is A

Let the roots be a/r, a, ar, where a > 0, r > 1. Now,a/r+a+ar=−p                       (1)a(a/r)+a(ar)+(ar)(a/r)=q     (2)(a/r)(a)(ar)=1                           (3)⇒    a3=1⇒    a=1Hence, (3) is correct. From (1), putting a = 1, we get−p−3>0             ∵r+1r>2 ⇒ p<−3Hence, (2) is not correct. Also,1/r+1+r=−p               (4)From (2), putting a = 1, we get1/r+r+1=q                   (5)From (4) and (5), we have−p=q⇒p+q=0Hence, (1) is correct. Now, as r > 1a/r=1/r<1and   ar=r>1Hence, (4) is correct.

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