Q. The intervals in which the function $f$ given by $f(x)=\frac{4 \,\sin\, x-2 x-x\, \cos\, x}{2+\cos \,x}$ is increasing in $x \in(0,2 \pi)$ is

Application of Derivatives Report Error

A

$(0, \pi) \cup(2 \pi, 4 \pi)$

B

$\left(0, \frac{\pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right)$

C

$\left(0, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{2}, \pi\right)$

D

None of these

Solution:

$f(x)=\frac{4 \,\sin\, x-2 x-x\, \cos \,x}{2+\cos\, x}$
$=\frac{4\, \sin \,x}{2+\cos\, x}-x$
$\therefore f'(x)=\frac{4 \,\cos \,x(2+\cos\, x)-(-\sin\, x)(4 \\,sin \,x)}{(2+\cos\, x)^{2}}-1$
$=\frac{4+8 \,\cos \,x}{(2+\cos\, x)^{2}-1}$
$=\frac{\cos \,x(4-\cos \,x)}{(2+\cos\, x)^{2}}$
Now, $f'(x)=0 $ or $\cos \,x=0$
$($ As $\cos\, x \neq 4)$
or $x=\frac{\pi}{2}, \frac{3 \pi}{2} ; \cos\, x > 0$
for $x \in(0, \pi / 2) \cup(3 \pi / 2,2 \pi)$
and $\cos \,x < 0$ for $x \in(\pi / 2, \pi) \cup(\pi, 3 \pi / 2)$
Thus, $f ( x )$ is increasing for
$x \in(0, \pi / 2) \cup(3 \pi / 2,2 \pi)$

Questions from Application of Derivatives

1. Let $ 𝑓 \left(x\right) = 𝑥 + log_{e} x − 𝑥 log_{e} 𝑥, 𝑥 ∈ \left(0,∞\right).$
• Column 1 contains information about zeros of $𝑓 (𝑥) , 𝑓'(𝑥)$ and $f ''(𝑥)$.
• Column 2 contains information about the limiting behavior of $𝑓(𝑥) , 𝑓 ' (𝑥)$ and $f ''(𝑥)$ at infinity.
• Column 3 contains information about increasing/decreasing nature of $𝑓(𝑥)$ and $f' (𝑥)$ .

Column 1 Column 2 Column 3

(I) $𝑓 (𝑥) = 0$ for some $x\in\left(1,e^{2}\right)$ (i) $\lim_{x\to\infty}f \left(x\right)=0$ (P) $𝑓$ is increasing in ($0, 1$)

(II) $𝑓(𝑥) = 0$ for some $x\in\left(1, e\right)$ (ii) $\lim_{x\to\infty}f \left(x\right)=-\infty$ (Q) $f$ is decreasing in ($𝑒, 𝑒^2$)

(III) $𝑓(𝑥) = 0$ for some $x\in\left(0, 1\right)$ (ii) $\lim_{x\to\infty}f' \left(x\right)=-\infty$ (R) $f'$ is increasing in ($0, 1$)

(IV) $f ''(𝑥) = 0$ for some $x\in\left(1, e\right)$ (ii) $\lim_{x\to\infty}f'' \left(x\right)=0$ (S) $f'$ is decreasing in ($𝑒, 𝑒^2$)

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