The sum of all the local minimum values of the twice differentiable function $f: R \rightarrow R$ defined by $f(x)=x^{3}-3 x^{2}-\frac{3 f^{\prime \prime}(2)}{2} x+f \prime \prime(1)$ is :

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Application of Derivatives

The sum of all the local minimum values of the twice differentiable function

f : R \to R

defined by

f (x) = x^{3} - 3 x^{2} - \frac{3 f ^{''} ( 2 )}{2} x + f'' (1)

is :

Answer: (D)

Differentiating twice removes every unknown constant, since

f^{'} (x) = 3 x^{2} - 6 x - \frac{3 f ^{''} ( 2 )}{2}, f^{''} (x) = 6 x - 6

f^{''} (2) = 6

and

f^{''} (1) = 0

. Substituting these values back,

f (x) = x^{3} - 3 x^{2} - 9 x

Then

f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1)

, which changes from negative to positive only at

x = 3

, so that is the one point of local minimum. Its value is

f (3) = 27 - 27 - 27 = - 27

, so the sum of all local minimum values is

- 27

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option