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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

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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