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

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=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