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

Concept: Monotonic Function Analysis — we examine the sign of $f'(x)$.

Step 1 – Simplify $f(x)$

Write $f(x) = \frac{4\sin x - 2x - x\cos x}{2 + \cos x}$.

Notice that $4\sin x - x(2 + \cos x) = 4\sin x - x(2 + \cos x)$. So

Step 2 – Differentiate

The numerator simplifies: $4\cos x(2 + \cos x) + 4\sin^2 x = 8\cos x + 4(\cos^2 x + \sin^2 x) = 8\cos x + 4$.

Thus

Step 3 – Combine into a single fraction

Expand $(2 + \cos x)^2 = 4 + 4\cos x + \cos^2 x$. The numerator becomes

Since $2 + \cos x > 0$ for all $x$, the sign of $f'(x)$ is the sign of $\cos x(4 - \cos x)$.

But $4 - \cos x > 0$ always (because $\cos x \le 1$). So $f'(x) \ge 0$ exactly when $\cos x \ge 0$.

Step 4 – Determine intervals

$\cos x \ge 0$ on $[-\frac{\pi}{2} + 2n\pi,\; \frac{\pi}{2} + 2n\pi]$, $n \in \mathbb{Z}$.

$\cos x < 0$ on $(\frac{\pi}{2} + 2n\pi,\; \frac{3\pi}{2} + 2n\pi)$.

To decide where a function increases or decreases, we look at its derivative. If $f'(x) > 0$, the function is rising; if $f'(x) < 0$, it is falling. The trick here is that $f(x)$ looks messy, but its derivative simplifies beautifully — a classic sign of a well-designed exam problem.

The denominator $2 + \cos x$ is always positive (since $\cos x \geq -1$, so $2 + \cos x \geq 1 > 0$). That means the sign of $f'(x)$ depends only on the numerator after differentiation. Let’s work through it.

1. Differentiate $f(x)$ using the quotient rule.

   Let $u = 4\sin x - 2x - x\cos x$ and $v = 2 + \cos x$. Then $f'(x) = \frac{u'v - uv'}{v^2}$.

   First, $u'$:

   • Derivative of $4\sin x$ is $4\cos x$.
   • Derivative of $-2x$ is $-2$.
   • Derivative of $-x\cos x$: use product rule — $(-1)(\cos x) + (-x)(-\sin x) = -\cos x + x\sin x$. So $u' = 4\cos x - 2 - \cos x + x\sin x = 3\cos x - 2 + x\sin x$.

   Next, $v' = -\sin x$.

   Now compute $u'v - uv'$:

So the numerator $N$ of $f'(x)$ is:

2. Expand and simplify $N$. Expand the first product:

Now add the second part: $(4\sin x - 2x - x\cos x)\sin x = 4\sin^2 x - 2x\sin x - x\sin x \cos x$.

Add them:

Notice $2x\sin x$ cancels with $-2x\sin x$, and $x\sin x \cos x$ cancels with $-x\sin x \cos x$. So all $x$-terms vanish!

Use $\sin^2 x = 1 - \cos^2 x$:

So $f'(x) = \frac{\cos x (4 - \cos x)}{(2 + \cos x)^2}$.

3. Analyze the sign of $f'(x)$.

   Denominator: $(2 + \cos x)^2 > 0$ always (since $2 + \cos x \geq 1$).

   Factor $(4 - \cos x)$: since $\cos x \leq 1$, we have $4 - \cos x \geq 3 > 0$. So this factor is always positive.

   Therefore, the sign of $f'(x)$ is exactly the sign of $\cos x$.

   • $f'(x) > 0$ when $\cos x > 0$ → function is increasing.
   • $f'(x) < 0$ when $\cos x < 0$ → function is decreasing.
   • $f'(x) = 0$ when $\cos x = 0$ → at $x = \frac{\pi}{2} + n\pi$, these are critical points (where monotonicity may change).

4. Write the intervals.

   $\cos x > 0$ on intervals $\left( -\frac{\pi}{2} + 2n\pi,\ \frac{\pi}{2} + 2n\pi \right)$ for $n \in \mathbb{Z}$.

   $\cos x < 0$ on intervals $\left( \frac{\pi}{2} + 2n\pi,\ \frac{3\pi}{2} + 2n\pi \right)$ for $n \in \mathbb{Z}$.

   The function is defined for all real $x$ (denominator never zero), so these intervals cover the entire domain.

Method: Testing Monotonicity of a Function via the Sign of $f'(x)$

This method finds the intervals on which a function increases or decreases, and is the standard approach whenever a question asks "find the intervals in which $f$ is increasing/decreasing."

Steps

Step 1: Differentiate $f(x)$

Use the appropriate rule (quotient rule, product rule, chain rule) to find $f'(x)$. If $f$ is a quotient, $f'(x)=\dfrac{u'v-uv'}{v^2}$.

Step 2: Simplify the derivative as far as possible, aiming to factor it

Expand the numerator and look for trigonometric identities (like $\sin^2x+\cos^2x=1$) that cancel terms — a "messy" derivative from a board-exam question almost always collapses into a clean product of simple factors after using such an identity.

Step 3: Determine the sign of each factor separately

Check whether the denominator (or any factor) is always positive — e.g. $2+\cos x \ge 1 > 0$ for all $x$ — so it can be ignored when deciding the overall sign. This narrows the sign analysis down to only the factors that can actually change sign.

Step 4: Solve $f'(x)=0$ and locate where $f'(x)$ is undefined

These points are where monotonicity can switch.

Step 5: State the increasing/decreasing intervals

Match the sign of the surviving factor(s) to standard trig-sign intervals (e.g. $\cos x \ge 0$ on $[-\pi/2+2n\pi,\ \pi/2+2n\pi]$) and write the final answer, including the general $n\in\mathbb{Z}$ if the domain is all real numbers.

Common Mistakes

Mistake 1: Giving up on simplifying a messy derivative and guessing the sign numerically at a few points

Why it's wrong: without simplifying, it's easy to miss cancellations (like the $x$-dependent terms cancelling here) and to test too few points to correctly capture a periodic sign pattern. Correct approach: always fully expand and simplify $f'(x)$ symbolically first — a clean factorised form usually falls out, and the sign can then be read off algebraically rather than guessed.

Mistake 2: Restricting the answer to $[0,2\pi]$ when the domain is all real numbers

Why it's wrong: trigonometric functions are periodic, so if no domain restriction is given in the question, the increasing/decreasing behaviour repeats every $2\pi$ and must be expressed with a general integer $n$. Correct approach: check the question's stated domain; if it says "for all $x$" or gives no restriction, express intervals using $2n\pi$ shifts, $n\in\mathbb{Z}$.

Mistake 3: Assuming a factor that can be negative (like $\cos x$) is always positive

Why it's wrong: only genuinely sign-definite pieces (such as $2+\cos x$ or $4-\cos x$, both bounded away from zero) can be dropped from the sign analysis — dropping a factor that actually changes sign gives a completely wrong set of intervals. Correct approach: check the range of each factor explicitly (e.g. $\cos x\in[-1,1]$) before deciding it never changes sign.