The function


The function

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


  1. increases in $\left[\frac{1}{2}, \infty\right)$

  2. increases in $\left(-\infty, \frac{1}{2}\right]$

  3. decreases in $\left[\frac{1}{2}, \infty\right)$

  4. decreases in $\left(-\infty, \frac{1}{2}\right]$

Correct Option: 1


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

$f^{\prime}(x)=\left(2 x^{2}-x\right)-2 \cos x+2 \cos x-\sin x(2 x-1)$

$=(2 x-1)(x-\sin x)$

for $x>0, x-\sin x>0$

$x<0, x-\sin x<0$

for $x \in(-\infty, 0] \cup\left[\frac{1}{2}, \infty\right), f^{\prime}(x) \geq 0$

for $x \in\left[0, \frac{1}{2}\right], f^{\prime}(x) \leq 0$

$\Rightarrow f(\mathrm{x})$ increases in $\left[\frac{1}{2}, \infty\right) .$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now