Question

Find the intervals in which the function

f

given by

f (x) = \frac{4 s i n x - 2 x - x c o s x}{2 + c o s x}

is decreasing.

Accepted Answer

$f$
$∴ f^{'} (x) = \frac{(2 + c o s x) (4 c o s x - 2 - c o s x + x s i n x) - (4 s i n x - 2 x - x c o s x) (- s i n x)}{{(2 + c o s x)}^{2}}$ [Applying the quotient rule for differentiation]
$= \frac{(2 + c o s x) (3 c o s x - 2 + x s i n x) + s i n x (4 s i n x - 2 x - x c o s x)}{{(2 + c o s x)}^{2}}$
$= \frac{6 c o s x - 4 + 2 x s i n x + 3 c o s^{2} x - 2 c o s x + x s i n x c o s x + 4 s i n^{2} x - 2 x s i n x - x s i n x c o s x}{{(2 + c o s x)}^{2}}$
$= \frac{4 c o s x - 4 + 3 c o s^{2} x + 4 s i n^{2} x}{{(2 + c o s x)}^{2}}$
$= \frac{4 c o s x - 4 + 3 c o s^{2} x + 4 - 4 c o s^{2} x}{{(2 + c o s x)}^{2}}$ $[\sin^{2} x + \cos^{2} x = 1]$
$= \frac{4 c o s x - c o s^{2} x}{{(2 + c o s x)}^{2}}$
$= \frac{c o s x (4 - c o s x)}{{(2 + c o s x)}^{2}}$

It is given that, $f (x)$ is decreasing.
$∴ f^{'} (x) \leq 0$
Now, $(4 - \cos x)$ and ${(2 + \cos x)}^{2}$ is always greater than $0$ (Since $- 1 \leq \cos x \leq 1$ ).
So, for $f^{'} (x) \leq 0 we have to find x for which \cos x \leq 0$ . As, we know $\cos x \leq 0,$ for $x \in . . ., [\frac{- 3 π}{2}, \frac{- π}{2}], [\frac{π}{2}, \frac{3 π}{2}], [\frac{5 π}{2}, \frac{7 π}{2}], . . . . . . .$
This is more clear from the graph of $\cos x$ given below:

Thus in general, $x \in [(2 n + 1) π - \frac{π}{2}, (2 n + 1) π + \frac{π}{2}]$ , where $n \in N$ .
Hence, the intervals in which the function $f (x)$ is decreasing are $[(2 n + 1) π - \frac{π}{2}, (2 n + 1) π + \frac{π}{2}], n \in N .$

Miscellaneous Exercise on Chapter 6

Jharkhand Board Mathematics Solutions for Exercise - Miscellaneous Exercise on Chapter 6

Questions from Miscellaneous Exercise on Chapter 6 with Hints & Solutions

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