For x ∈ (0, 3/2), let f(x) = √x, g(x) = tanx and h(x) = (1-x²)/(1+x²). If φ(x) = ((hof)og)(x), then

Q: Wait, why do we simplify the expression first instead of just plugging in $x = \frac{\pi}{3}$ immediately?

Great question! While you *could* plug in the value immediately, it is a trap. Calculating $\tan(\frac{\pi}{3})$ is easy, but then you have to deal with nested square roots and complex fractions. Simplifying algebraically first is the hallmark of a JEE topper—it reduces the chance of arithmetic errors significantly and reveals the underlying structure of the function.

Q: How do I know when to use the identity $\tan(\frac{\pi}{4} - x)$?

Whenever you see the structure $\frac{1-A}{1+A}$ or $\frac{A-B}{1+AB}$, your brain should immediately scream 'Trigonometric Identity!' It is a classic pattern designed to test if you can recognize the expansion of $\tan(A-B)$. If you see a $1$ in the numerator and denominator, it is almost certainly a hint to use $\tan(\frac{\pi}{4}) = 1$.

Q: I got $\tan(-\frac{\pi}{12})$, but the options are all positive. Did I make a mistake?

Absolutely not! You are on the right track. The negative sign just tells you the angle is in the fourth quadrant. Since the options are in terms of positive angles, use the periodicity of the tangent function: $\tan(\pi - \theta) = -\tan \theta$. It is just a different way of expressing the same value, and it is a common way to test your comfort with trigonometric quadrants.

Question

For $x \in (0, 3/2)$, let $f(x) = \sqrt{x}$, $g(x) = 	an x$ and $h(x) = \frac{1-x^2}{1+x^2}$. If $\phi(x) = ((hof)og)(x)$, then $\phi(\pi/3)$ is equal to :

Options:

(A) $	an(\pi/12)$

(B) $	an(7\pi/12)$

(C) $	an(11\pi/12)$

(D) $	an(5\pi/12)$

Sigma Percentile · Accepted Answer

### The Art of the Composite Function: A Journey Through Layers

Welcome, aspiring engineers! Today, we are going to peel back the layers of a problem that, at first glance, might look like a messy algebraic nightmare. In the world of JEE Advanced, complexity is often just a mask for elegance.

When we look at a composite function like $\phi(x) = (h \circ f \circ g)(x)$, we aren't looking at a single, terrifying equation. We are looking at a pipeline—a sequence of transformations. Let us walk through this together.

### Phase 1: The Pipeline Approach

Imagine you are standing at the start of a factory line with an input, $x$. The first machine it encounters is $g(x) = 	an x$. This is our innermost function.

In any composite function, the golden rule is: work from the inside out. We feed $x$ into $g$, and the output is simply $	an x$.

Now, we take that output, $	an x$, and pass it to the next machine, $f(x) = \sqrt{x}$. This machine is a square root operator. Our new value becomes $\sqrt{	an x}$.

Finally, we reach the outermost machine, $h(x) = \frac{1-x^2}{1+x^2}$. We substitute our current value, $\sqrt{	an x}$, into the $x$ of the $h$ function. This gives us the expression:

$$\phi(x) = \frac{1-(\sqrt{	an x})^2}{1+(\sqrt{	an x})^2}$$

### Phase 2: The Algebraic Epiphany

Look closely at the numerator and the denominator. We have the square of a square root, which are inverse operations that cancel each other out with beautiful precision. The $(\sqrt{	an x})^2$ simply becomes $	an x$.

Suddenly, the fog clears. Our complex composite function $\phi(x)$ has collapsed into something much simpler:

$$\phi(x) = \frac{1-	an x}{1+	an x}$$

This is the moment where you should feel a surge of confidence. You have successfully navigated the layers, but we must now see if this expression hides a deeper truth.

### Phase 3: The Trigonometric Identity

Look at $\frac{1-	an x}{1+	an x}$. We know that $	an(\frac{\pi}{4}) = 1$. If we replace the $1$ in the numerator with $	an(\frac{\pi}{4})$, the expression becomes:

$$\frac{	an(\frac{\pi}{4}) - 	an x}{1 + 	an(\frac{\pi}{4})	an x}$$

This is the exact expansion of the tangent subtraction formula: $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$. By setting $A = \frac{\pi}{4}$ and $B = x$, we realize that our entire function $\phi(x)$ is just $	an(\frac{\pi}{4} - x)$.

### Phase 4: The Final Evaluation

Now, we evaluate at $x = \frac{\pi}{3}$. We substitute this into our simplified function:

$$\phi\left(\frac{\pi}{3}ight) = 	an\left(\frac{\pi}{4} - \frac{\pi}{3}ight)$$

Calculating the angle: $\frac{\pi}{4} - \frac{\pi}{3} = \frac{3\pi - 4\pi}{12} = -\frac{\pi}{12}$.

So, we have $	an(-\frac{\pi}{12})$. Since $	an$ is an odd function, this is $-	an(\frac{\pi}{12})$.

### Conclusion

You navigated the composite layers, simplified the algebra, and recognized the trigonometric identity. This is how you conquer JEE Advanced problems—not by brute force, but by understanding the flow of the math. Keep this mindset, and no problem will ever be too complex for you. The final result is **$-	an(\frac{\pi}{12})$**.

Animated Solution for Mathematics - Functions: For $x \in (0, 3/2)$ , let $f (x) = x$ , $g (x) = tan x$ and $h (x) = \frac{1 - x ^{2}}{1 + x ^{2}}$ . If $ϕ (x) = ((h o f) o g) (x)$ , then $ϕ (π /3)$ is equal to :

Select Answer:

Visualized Solution

Introduction to the Functions

The Innermost Function $g (x)$

Applying the Middle Layer $f (x)$

The Outermost Layer $h (x)$

Simplifying the Expression

Applying Trigonometric Identity

Evaluating at $x = \frac{π}{3}$

Adjusting the Angle using Periodicity

The Sigma Insight: Composite Functions

The Art of the Composite Function

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈(0,3/2), let f(x)=x​, g(x)=tanx and h(x)=1+x21−x2​. If ϕ(x)=((hof)og)(x), then ϕ(π/3) is equal to :