Q.
If the roots of the equation, x3+px2+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 root is unity
d
one root is smaller than 1 and one root is greater than 1
answer is A.
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Detailed Solution
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=1 Hence, (c) is correct. From (1), putting a=1, we get−p−3>0 (∵ r+1r>2) ⇒−p>3⇒ p<−3 Hence, (b) is not correct. Also,(1/r)+1+r=−p ------------- (4) From (2), putting a=1, we get(1/r)+r+1=q ----------------(5) From (4) and (5), we have −p=q ⇒ p+q=0 Hence, (a) is correct. Now, asr>1 a/r=1/r<1 and ar=r>1
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