Question

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

• A. $2$
• B. $-1$
• C. $4$
• D. Has no real value.

Answer 1

Answer: (C)

$4$

The given equation is:

$f\left(x\right)=x^3+4x^2+\lambda x+1$

Differentiating once we get,

$\Rightarrow f^{\prime }\left(x\right)=3x^2+8x+\lambda$

Now the interval where the function is monotonically decreasing is $(-2,-\frac{2}{3})$

$\Rightarrow f^{\prime }\left(x\right)=(x+2)(x+\frac{2}{3})$

$\Rightarrow f^{\prime }\left(x\right)=\frac{3x^2+8x+4}{3}$

Now $f^{\prime }\left(x\right)<0$ for monotonically decreasing function so

$\Rightarrow 3x^2+8x+4<0$

On comparison we get the value of $\lambda =4$ ......Answer