The function f(x) $= \frac{4 x^{3} - 3 x^{2}}{6} - 2 s i n x + (2 x - 1) c o s x :$

Option 1 -

Decrease in $(- \infty, \frac{1}{2}]$

Option 2 -

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

Option 3 -

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

Option 4 -

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

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

$f (x) = \frac{4 x^{3} - 3 x^{2}}{6} - 2 s i n x + (2 x - 1) c o s x$

$f' (x) = 2 x^{2} - x - 2 c o s x + 2 c o s x - (2 x - 1) s i n x$

= $(2 x - 1) (x - s i n x) \geq 0 f o r x \geq 0$

$f' (x) \geq 0 \forall x \geq 1 / 2$

$∴ f (x)$ is increasing in $[\frac{1}{2}, \infty)$

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The function f(x) $= \frac{4 x^{3} - 3 x^{2}}{6} - 2 s i n x + (2 x - 1) c o s x :$