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Practicing Success
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If $(x)=x^3+a x^2+b x$ has a maximum at x = -2 and minimum at x = 1, then (a, b) is: |
$\left(\frac{2}{3}, 6\right)$ $\left(4, \frac{1}{5}\right)$ $\left(6, \frac{3}{2}\right)$ $\left(\frac{3}{2},-6\right)$ |
$\left(\frac{3}{2},-6\right)$ |
The correct answer is Option (4) → $\left(\frac{3}{2},-6\right)$ |
