Question

If \(f(x)=5x^{3}+2x^{2}-3x+1\) has a local maxima and a local minima at \(x=\sin \theta ,\pi \lt \theta \lt \frac{3\pi }{2}\) and \(x=\cos \phi ,\frac{3\pi }{2}\lt \phi \lt 2\pi\) respectively, then \((\tan \theta )(\tan \phi )=\ldots\)

• A. \(\frac{3\sqrt{3}}{2}\)
• B. \(\frac{-3\sqrt{3}}{2}\)
• C. \(\frac{3\sqrt{2}}{2}\)
• D. \(\frac{-3\sqrt{2}}{2}\)

Hint:

Find the stationary points of the cubic, classify them with the second derivative, then locate the angles by quadrant.

Answer 1

The Correct Option is D

Step 1: Identify the Extrema
The maximum is at $x=-\dfrac{3}{5}=\sin\theta$ and the minimum at $x=\dfrac{1}{3}=\cos\phi$.

Step 2: Use Quadrant Signs
In the third quadrant $\tan\theta\gt0$, so $\tan\theta=\dfrac{3/5}{4/5}=\dfrac{3}{4}$.
In the fourth quadrant $\tan\phi\lt0$, so $\tan\phi=-\dfrac{2\sqrt{2}/3}{1/3}=-2\sqrt{2}$.

Step 3: Multiply
$(\tan\theta)(\tan\phi)=-\dfrac{6\sqrt{2}}{4}=-\dfrac{3\sqrt{2}}{2}$, option (D).

Final Answer:
The product equals minus 3 root 2 over 2. \[ \boxed{\text{(D) }\dfrac{-3\sqrt{2}}{2}} \]