Q.If $x^3+12x^2+10ax+1999$ definitely has a positive zero, if and only if

Step 1. Examine the sign pattern for $x>0$. For $x>0$: $x^3>0$, $12x^2>0$, $1999>0$ always; the only term whose sign depends on $a$ is $10ax$.

Step 2. Case $a\ge0$. Then $10ax\ge0$ too, so every term is $\ge0$ (with the constant strictly positive) — $P(x)>0$ for all $x>0$. This proves $a\ge0$ makes a positive zero impossible. …

Descartes' Rule of Signs — $a\ge0$ gives zero sign changes (proving no positive zero can exist …

• Assuming $a<0$ alone GUARANTEES a positive zero exists (it only fails to rule one out) — the MCQ tests the necessary-condition direction, matching Descartes' rule's own …