The function f(x) = tan⁻¹(sin x + cos x) is an increasing function in (A) (0, π/2) (B) (-π/2, π/2) (

Q: Wait, why do we completely ignore the denominator when checking for the increasing interval?

That is a great observation! In calculus, when we analyze the sign of a derivative $f'(x) = \frac{N(x)}{D(x)}$, if we can prove that $D(x)$ is always positive for all $x$ in the domain, then the sign of $f'(x)$ is entirely determined by the numerator $N(x)$. Here, $D(x) = 1 + (\sin x + \cos x)^2$. Since any real number squared is non-negative, $(\sin x + \cos x)^2 \ge 0$, which means $1 + (\sin x + \cos x)^2 \ge 1$. Because the denominator is strictly positive, it does not affect the sign of the derivative, allowing us to focus solely on the numerator.

Q: Why is the answer a subset of the interval I calculated?

This is a classic JEE Advanced trap! We found that the function is strictly increasing on the interval $(-\frac{3\pi}{4}, \frac{\pi}{4})$. However, the options provided are specific intervals. If a function is increasing on a large interval, it is automatically increasing on any subset of that interval. We simply check which of the given options is contained within our calculated interval $(-\frac{3\pi}{4}, \frac{\pi}{4})$. Since $(-\frac{\pi}{2}, \frac{\pi}{4})$ is a subset of $(-\frac{3\pi}{4}, \frac{\pi}{4})$, it is a valid interval where the function is increasing.

Q: Can I use the identity $\sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$ to solve this?

Absolutely! That is a brilliant way to simplify the problem. By rewriting the function as $f(x) = \tan^{-1}(\sqrt{2} \sin(x + \frac{\pi}{4}))$, you can see that the function is increasing when $\sin(x + \frac{\pi}{4})$ is increasing. This approach is perfectly valid and often leads to the same result faster. It shows you have a deep understanding of trigonometric manipulation!

Question

The function $f(x) = 	an^{-1}(\sin x + \cos x)$ is an increasing function in

Options:

(A) $(0, \pi/2)$

(B) $(-\pi/2, \pi/2)$

(C) $(\pi/4, \pi/2)$

(D) $(-\pi/2, \pi/4)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe. Today, we are going to dissect a function that, at first glance, might seem like a standard trigonometric expression, but beneath the surface lies a beautiful geometric dance.

We are looking at $f(x) = 	an^{-1}(\sin x + \cos x)$. Our mission is to find the interval where this function is strictly increasing. This is not just about crunching numbers; it is about understanding the flow of the function.

### The Bridge of the Chain Rule

To understand if a function is increasing, we must look at its rate of change—its derivative. We want to know where $f'(x) > 0$.

Our function is a composite one: an outer function, the inverse tangent, wrapping around an inner function, the sum of sine and cosine. To differentiate this, we invoke the Chain Rule. Recall that for any differentiable function $u(x)$, the derivative of $	an^{-1}(u)$ is:

$$f'(x) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$$

Imagine the Chain Rule as a bridge connecting the outer layer to the inner core. We differentiate the outer layer first, treating the inner part as a single block, and then multiply by the derivative of that inner core. It is a systematic process, a rhythm we must follow.

### The Anatomy of the Derivative

Let $u = \sin x + \cos x$. Applying our rule, we get:

$$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx}(\sin x + \cos x)$$

Now, let us perform the differentiation of the inner term. The derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$.

Putting it all together, we arrive at our derivative expression:

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

This equation is the heart of our problem. It tells us exactly how the function changes at any point $x$.

### The Denominator Trap

Here is where the intuition of a seasoned mathematician kicks in. Look at the denominator: $1 + (\sin x + \cos x)^2$.

We know that for any real number $y$, $y^2 \ge 0$. Therefore, $(\sin x + \cos x)^2$ is always non-negative. Adding $1$ to this ensures that the denominator is always at least $1$.

It is strictly positive! This is a massive relief. It means the denominator can never be zero, and it can never be negative.

Consequently, the sign of $f'(x)$ is entirely dictated by the numerator: $\cos x - \sin x$. We need $\cos x - \sin x > 0$, or simply $\cos x > \sin x$.

### The Geometric Dance

Now, we shift from algebra to geometry. Where is $\cos x$ greater than $\sin x$?

If you visualize the graphs of $y = \sin x$ and $y = \cos x$, you will see them intersecting where $\sin x = \cos x$, which happens at $	an x = 1$. In the standard interval, this occurs at $x = \frac{\pi}{4}$ and $x = -\frac{3\pi}{4}$.

Between these two points, the cosine curve sits comfortably above the sine curve. Thus, the function $f(x)$ is strictly increasing in the interval:

$$\left(-\frac{3\pi}{4}, \frac{\pi}{4}ight)$$

### Final Calculation

We have our interval, but we must match it with the options provided. The JEE Advanced examiners often test your ability to recognize subsets.

Our calculated interval is $\left(-\frac{3\pi}{4}, \frac{\pi}{4}ight)$. Looking at the options, we see $\left(-\frac{\pi}{2}, \frac{\pi}{4}ight)$.

Since this interval is entirely contained within our solution, the function is indeed increasing here. We have successfully navigated the trap and arrived at the truth. Keep practicing this visualization; it is the key to mastering the beauty of calculus.

Animated Solution for Mathematics - Differentiation: The function $f (x) = tan^{- 1} (sin x + cos x)$ is an increasing function in

Select Answer:

Visualized Solution

Understand the Function $f (x)$

Condition for an Increasing Function

Applying the Chain Rule

Differentiating the Inner Function

Analyzing the Denominator

Solving the Inequality $cos x > sin x$

Finding the General Interval

Matching with the Options

The Sigma Insight: Monotonicity

Analyzing the Setup

The Bridge of the Chain Rule

The Anatomy of the Derivative

The Denominator Trap

The Geometric Dance

Final Calculation

Similar Questions

The function $f (x) = sin^{4} x + cos^{4} x$ increases if

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

For the function $f (x) = sin x + 3 x - \frac{2}{π} (x^{2} + x)$ , where $x \in [0, \frac{π}{2}]$ , consider the following two statements : (I) $f$ is increasing in $(0, \frac{π}{2})$ . (II) $f^{'}$ is decreasing in $(0, \frac{π}{2})$ . Between the above two statements,

Let $g : (0, \infty) \to R$ be a differentiable function such that $\int (\frac{x ( c o s x - s i n x )}{e ^{x} + 1} + \frac{g ( x ) ( e ^{x} + 1 - x e ^{x} )}{( e ^{x} + 1 ) ^{2}}) d x = \frac{xg ( x )}{e ^{x} + 1} + c$ , for all $x > 0$ , where $c$ is an arbitrary constant. Then

The function $f (x) = \frac{l n ( π + x )}{l n ( e + x )}$ is

If $f (x) = \frac{x}{s i n x}$ and $g (x) = \frac{x}{t a n x}$ , where $0 < x \leq 1$ , then in this interval

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

Let $f (x) = x cos^{- 1} (- sin ∣ x ∣)$ , $x \in [- \frac{π}{2}, \frac{π}{2}]$ , then which of the following is true?

Let $f : (- 1, \infty) \to R$ be defined by $f (0) = 1$ and $f (x) = \frac{1}{x} lo g_{e} (1 + x)$ , $x \neq = 0$ . Then the function $f$ :

The function $f (x) = x e^{x (1 - x)}, x \in R$ , is

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=tan−1(sinx+cosx) is an increasing function in