Given: \[ x(x + 3) < 0 \] Solve the inequality:
\[ -3 < x < 0 \] Now, can we conclude \( |x| < 3 \)? Yes, because:
BUT: We are told to assess if this statement **alone** is enough to **guarantee** \( |x| < 3 \) in general. Yes — in fact, from this alone we can conclude \( |x| < 3 \), since \( x \) is between \(-3\) and \(0\). (So this contradicts the original claim in your comment — see clarification below.)
Given: \[ x(x - 3) > 0 \] Solve the inequality:
This is a union of two disjoint intervals. On its own, this does **not** help us determine whether \( |x| < 3 \), because:
So Statement B **alone** is **not sufficient**.
Hence, both statements together are sufficient to conclude \( |x| < 3 \).
\[ \boxed{\text{Both statements are needed to conclude that } |x| < 3} \]

In \(\triangle ABC\), \(DE \parallel BC\). If \(AE = (2x+1)\) cm, \(EC = 4\) cm, \(AD = (x+1)\) cm and \(DB = 3\) cm, then the value of \(x\) is

In the adjoining figure, PA and PB are tangents to a circle with centre O such that $\angle P = 90^\circ$. If $AB = 3\sqrt{2}$ cm, then the diameter of the circle is
In the adjoining figure, TS is a tangent to a circle with centre O. The value of $2x^\circ$ is
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: