QJEEM21Ma1aa9: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1aa9 Solution: The sum of all the local minimum values of the twice…

Question

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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} - {{3f''(2)} \over 2}x + f''(1)

is :

Accepted Answer

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

..... (i)

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

..... (ii)

f''(x) = 6x - 6

..... (iii)

Now, is 3^rd equation

f''(2) = 12 - 6 = 6

f''(11 = 0)

Use (ii)

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

f'(x) = 3{x^2} - 6x - {3 \over 2} \times 6

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

f'(x) = 0

3{x^2} - 6x - 9 = 0

\Rightarrow

x =

-

1 & 3

Use (iii)

f''(x) = 6x - 6

f''( - 1) = - 12 &lt; 0

maxima

f''(3) = 12 &gt; 0

minima.

Use (i)

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

f(x) = {x^3} - 3{x^2} - {3 \over 2} \times 6 \times x + 0

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

f(3) = 27 - 27 - 9 \times 3 = - 27

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