JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 8)
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 :
$$-$$22
5
$$-$$27
0
Explanation
$$f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1)$$ ..... (i)
$$f(x) = 3{x^2} - 6x - {3 \over 2}f''(2)$$ ..... (ii)
$$f''(x) = 6x - 6$$ ..... (iii)
Now, is 3rd 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)
$$f''(x) = 6x - 6$$
$$f''( - 1) = - 12 < 0$$ maxima
$$f''(3) = 12 > 0$$ minima.
Use (i)
$$f(x) = {x^3} - 3{x^2} - {3 \over 2}f''(2)x + f''(1)$$
$$f(x) = {x^3} - 3{x^2} - {3 \over 2} \times 6 \times x + 0$$
$$f(x) = {x^3} - 3{x^2} - 9x$$
$$f(3) = 27 - 27 - 9 \times 3 = - 27$$
$$f(x) = 3{x^2} - 6x - {3 \over 2}f''(2)$$ ..... (ii)
$$f''(x) = 6x - 6$$ ..... (iii)
Now, is 3rd 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)
$$f''(x) = 6x - 6$$
$$f''( - 1) = - 12 < 0$$ maxima
$$f''(3) = 12 > 0$$ 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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