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Bijective Function

The maximum o...

Question

The maximum of $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 4$ occurs at $x =$

A

1

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B

2

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C

-1

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D

-2

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Solution

The correct option is A 1
$f^{'} (x) = 6 x^{2} - 18 x + 12$
$= 6 (x^{2} - 3 x + 2)$
$= 6 (x - 2) (x - 1)$
$f^{'} (x) > 0$ If $x > 2$ or $x < 1$
$f^{'} (x) < 0$ If $x ϵ (1, 2)$
So $f (x)$ is max at $x = 1$

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2, 4]. The relative maximum occurs at x=

(a)

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(c) 2

(d) 4

Q. The maximum value of

f (x) = 2 x^{3} - 9 x^{2} + 12 x - 3

in the interval

0 \leq x \leq 3

Q. Find the intervals in which the following functions are increasing or decreasing.
(i) f(x) = 10 − 6x − 2x²

(ii) f(x) = x² + 2x − 5

(iii) f(x) = 6 − 9x − x²

(iv) f(x) = 2x³ − 12x² + 18x + 15

(v) f(x) = 5 + 36x + 3x² − 2x³

(vi) f(x) = 8 + 36x + 3x² − 2x³

(vii) f(x) = 5x³ − 15x² − 120x + 3

(viii) f(x) = x³ − 6x² − 36x + 2

(ix) f(x) = 2x³ − 15x² + 36x + 1

(x) f(x) = 2x³ + 9x² + 12x + 20

(xi) f(x) = 2x³ − 9x² + 12x − 5

(xii) f(x) = 6 + 12x + 3x² − 2x³

(xiii) f(x) = 2x³ − 24x + 107

(xiv) f(x) = −2x³ − 9x² − 12x + 1

(xv) f(x) = (x − 1) (x − 2)²

(xvi) f(x) = x³ − 12x² + 36x + 17

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f (x) = \frac{3}{10} x^{4} - \frac{4}{5} x^{3} - 3 x^{2} + \frac{36}{5} x + 11

(xix) f(x) = x⁴ − 4x

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f (x) = \frac{x^{4}}{4} + \frac{2}{3} x^{3} - \frac{5}{2} x^{2} - 6 x + 7

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(xxv)

f (x) = {\{x (x - 2)\}}^{2}

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f (x) = 2 x^{3} - 9 x^{2} + 12 x + 9

, at x = 1 and x = 2 are equal.

Q. The function f(x)=

2 x^{3} - 9 x^{2} + 12 x + 4

is decreasing in the interval(s)

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