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Solution for The Function F(X) = 2 X 3 − 15 X 2 + 36 X + 4 is Maximum at X = (A) 3 (B) 0 (C) 4 (D) 2 - CBSE (Science) Class 12 - Mathematics

Question

The function f(x) = $2 x^3 - 15 x^2 + 36x + 4$ is maximum at x =
(a) 3
(b) 0
(c) 4
(d) 2

Solution

(d) 2

$\text { Given }: f\left( x \right) = 2 x^3 - 15 x^2 + 36x + 4$

$\Rightarrow f'\left( x \right) = 6 x^2 - 30x + 36$

$\text { For a local maxima or a local minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow 6 x^2 - 30x + 36 = 0$

$\Rightarrow x^2 - 5x + 6 = 0$

$\Rightarrow \left( x - 2 \right)\left( x - 3 \right) = 0$

$\Rightarrow x = 2, 3$

$\text { Now },$

$f''\left( x \right) = 12x - 30$

$\Rightarrow f''\left( 2 \right) = 24 - 30 = - 6 < 0$

$\text { So, x = 1 is a local maxima } .$

$\text { Also },$

$f''\left( 3 \right) = 36 - 30 = 6 > 0$

$\text { So,x=2 is a local maxima }.$

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Solution for question: The Function F(X) = 2 X 3 − 15 X 2 + 36 X + 4 is Maximum at X = (A) 3 (B) 0 (C) 4 (D) 2 concept: Graph of Maxima and Minima. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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