Question

Let \(f(x)=(a+2)x^{3}-3ax^{2}+9ax-5\). If \(f(x)\) is strictly decreasing for all \(x\in R\), then the interval in which \(a\) lies is

• A. \((-\infty ,-3)\)
• B. \((-\infty ,-2)\)
• C. \((0,\infty )\)
• D. \((-\infty ,-5)\)

Hint:

Make f'(x) negative for all x: leading coefficient negative and discriminant negative.

Answer 1

The Correct Option is A

Step 1: Leading Coefficient
Option (C) is impossible because $a\gt0$ makes $a+2\gt0$, so $f'$ opens upward and cannot stay negative.

Step 2: Test a Value
For $a=-4$: $f'(x)=-6(x^{2}-4x+6)$ with discriminant $16-24\lt0$, so $f'\lt0$ for all $x$. For $a=-2.5$: $f'(x)=-1.5(x^{2}-10x+15)$ has discriminant $100-60\gt0$, so it fails.

Step 3: Conclusion
The valid set must include $-4$ but not $-2.5$, and the exact condition $a(a+3)\gt0$ with $a\lt-2$ gives $a\lt-3$, option (A).

Final Answer:
The valid range is a less than -3. \[ \boxed{\text{(A) }(-\infty,-3)} \]