Question

If $f(x)=x^3+4x^2+\lambda x+1$ is a monotonically decreasing function of $x$ in the largest possible interval $(-2,-\frac{2}{3})$ , then

• A. $\lambda$ = 4
• B. $\lambda$ = 2
• C. $\lambda$ = -1
• D. $\lambda$ has no real value

Answer 1

Answer: (A)

Here, $f^{\prime }(x)\le 0$
$\Rightarrow 3(x{)}^2+8x+\lambda \le 0,\forall x\in \left(-2,-\frac{2}{3}\right)$
Then, situations for $f^{\prime }(x)$ is as follow:

Given that $f(x)$ decreases in the largest possible interval $\left(-2,-\frac{2}{3}\right)$ then $f^{\prime }(x)=0$ must have roots $-2$ and $-\frac{2}{3}$
$\Rightarrow$ Product of roots is $(-2)\left(-\frac{2}{3}\right)=\frac{\lambda }{3}$
$\Rightarrow \lambda =4$