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

Increases in $(-\infty, 1 / 2)$

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

JEE MATHEMATICS MCQ: INCREASING & DECREASING

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Analyzing +4 | -1

The function

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

View answer and explanation

Correct option: A

Increases in

(1 / 2, \infty)

Option A is correct. The function increases in

(1 / 2, \infty)

because its derivative is positive in that interval.

Practice All Chapter MCQs →

f\left( x \right) = \frac{{K\sin x + 2\cos x}}{{\sin x + \cos x}}

x

f\left( x \right) = {x^2} + ax + 1

a

f(x)=

\sin x-\cos x-a x+b

f(x)=x e^{x(1-x)}, x \in R

f

g

fog

f\left( x \right) = {x^3} + b{x^2} + cx + d

0 < {b^2} < c

\left( { - \infty ,\;\infty } \right)