If `f(x)=x^3+4x^2+lambdax+1` is a monotonically decreasing function of `x` in the largest possible interval `(-2,-2/3)dot` Then `lambda=4` (b) `lambda=2` `lambda=-1` (d) `lambda` has no real value

Explanation

Here `f'(x) le 0`
`rArr" "3x^(2)+ 8x + lamda0 AA x in (-2, -(2)/(3))`
Then the situations for `f'(x)` are as follows :

Given that `f(x)` decreases in the largest possible interval `(-2, -(2)/(3))`, then `f'(x)=0` must have roots -2 and `-2//3`.
So the product of the roots is `(-2)(-(2)/(3))= (lamda)/(3) rArr lamda=4`.