The function f(x)=4x3−3x26−2sin⁡x+(2x−1)cos⁡x

f(x)=4x3−3x26−2sin⁡x+(2x−1)cos⁡x f1(x)=2x2−x−2cos⁡x−(2x−1)sin⁡x=(2x−1)(x−sin⁡x) for x>0,x−sin⁡x>0 for x<0,x−sin⁡x<0 for x∈[−∞,0]∪12,∞⇒f1(x)≥0 for x∈0,12⇒f1(x)≤0 Thus, f(x) increases in 12,∞

The function f(x)=4x3−3x26−2sin⁡x+(2x−1)cos⁡x

f(x)=4x3−3x26−2sin⁡x+(2x−1)cos⁡x f1(x)=2x2−x−2cos⁡x−(2x−1)sin⁡x=(2x−1)(x−sin⁡x) for x>0,x−sin⁡x>0 for x<0,x−sin⁡x<0 for x∈[−∞,0]∪12,∞⇒f1(x)≥0 for x∈0,12⇒f1(x)≤0 Thus, f(x) increases in 12,∞

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The function $f (x) = \frac{4 x^{3} - 3 x^{2}}{6} - 2 \sin x + (2 x - 1) \cos x$

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Decreases in $(- \infty, 1 / 2]$

Increases in $[\frac{1}{2}, \infty)$

Increases in $(- \infty, \frac{1}{2})$

Decreases in $[\frac{1}{2}, \infty)$

answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{4 x^{3} - 3 x^{2}}{6} - 2 \sin x + (2 x - 1) \cos x f^{1} (x) = (2 x^{2} - x) - 2 \cos x - (2 x - 1) \sin x \\ = (2 x - 1) (x - \sin x) (for x > 0, x - \sin x > 0 for x < 0, x - \sin x < 0) \end{array}$
$for x \in [- \infty, 0] \cup [\frac{1}{2}, \infty] \Rightarrow f^{1} (x) \geq 0 for x \in [0, \frac{1}{2}] \Rightarrow f^{1} (x) \leq 0 Thus, f (x) increases in [\frac{1}{2}, \infty)$