If f(x) = x / sin x and g(x) = x / tan x, where 0

Q: Wait, why do we define a new function $h(x)$? Can't we just look at the derivative?

That is a fantastic question! Often in calculus, the derivative itself is a complex expression whose sign isn't immediately obvious. By defining $h(x)$ as the numerator, we isolate the part that determines the sign. Differentiating $h(x)$ allows us to see its 'trend' or 'rate of change', which helps us determine its sign relative to its starting value. It is a powerful technique for proving inequalities!

Q: Why does $h(0) = 0$ matter so much?

Think of $h(x)$ as a car starting its journey at $x=0$. We know $h(0) = 0$. If we prove that the derivative $h'(x)$ is always positive for $x > 0$, it means the car is always moving forward (increasing). If it starts at zero and is always increasing, it must be positive for all $x > 0$. This is the fundamental logic behind using derivatives to prove that a function is positive or negative.

Q: Is there a shortcut using Taylor series?

Yes! For advanced students, Taylor series expansions are a brilliant shortcut. For small $x$, $\sin x \approx x - \frac{x^3}{6}$ and $\tan x \approx x + \frac{x^3}{3}$. Substituting these into $f(x)$ and $g(x)$ gives you approximations that clearly show the direction of change. However, for a rigorous JEE proof, the derivative method is the standard and most robust approach.

Question

If $f(x) = \frac{x}{\sin x}$ and $g(x) = \frac{x}{	an x}$, where $0 < x \le 1$, then in this interval

Options:

(A) both $f(x)$ and $g(x)$ are increasing functions

(B) both $f(x)$ and $g(x)$ are decreasing functions

(C) $f(x)$ is an increasing function

(D) $g(x)$ is an increasing function

Sigma Percentile · Accepted Answer

### The Dance of Functions: A Calculus Odyssey Welcome, my dear students! Today, we are embarking on a journey into the heart of calculus. We are not just solving a problem; we are learning to read the language of change. We have two functions, $f(x) = \frac{x}{\sin x}$ and $g(x) = \frac{x}{ an x}$, and we want to know how they behave in the interval $(0, 1]$. Do they climb like a mountain or descend like a valley? Let us find out. ### Phase 1: The Power of the Quotient Rule To understand if a function is increasing or decreasing, we must look at its rate of change—its first derivative. If $f'(x) > 0$, the function is climbing; if $f'(x) < 0$, it is falling. Since both our functions are ratios, we summon the mighty Quotient Rule: $$\frac{d}{dx}\left(\frac{u}{v} ight) = \frac{v u' - u v'}{v^2}$$ Let us start with $f(x) = \frac{x}{\sin x}$. Applying the rule, we get: $$f'(x) = \frac{(\sin x)(1) - (x)(\cos x)}{\sin^2 x}$$ Look at the denominator, $\sin^2 x$. It is a square, so it is always positive. The fate of our function rests entirely on the numerator: $h(x) = \sin x - x \cos x$. Is this numerator positive or negative? It is not immediately obvious, is it? This is where the magic happens. ### Phase 2: The Nested Function Strategy When we face a mystery like the sign of $h(x)$, we do not guess; we investigate. Let us define $h(x) = \sin x - x \cos x$ and find its derivative to see how it evolves. Using the product rule on the second term, we get: $$h'(x) = \cos x - (1 \cdot \cos x + x \cdot (-\sin x))$$ Watch closely as the $\cos x$ terms cancel out, leaving us with $h'(x) = x \sin x$. In our interval $(0, 1]$, both $x$ and $\sin x$ are positive. Therefore, $h'(x) > 0$. This means $h(x)$ is strictly increasing! Since $h(0) = \sin(0) - 0 \cdot \cos(0) = 0$, and the function is always increasing, $h(x)$ must be positive for all $x > 0$. Consequently, $f'(x) > 0$, and **$f(x)$ is strictly increasing**. The blue curve rises! ### Phase 3: The Descent of $g(x)$ Now, let us turn our attention to $g(x) = \frac{x}{ an x}$. We apply the same logic. The derivative is: $$g'(x) = \frac{( an x)(1) - (x)(\sec^2 x)}{ an^2 x}$$ Again, the denominator is positive. We focus on the numerator, $p(x) = an x - x \sec^2 x$. Let us differentiate $p(x)$ to see its behavior: $$p'(x) = \sec^2 x - [1 \cdot \sec^2 x + x \cdot 2 \sec x (\sec x an x)]$$ $$p'(x) = -2x \sec^2 x an x$$ In the interval $(0, 1]$, $x$, $\sec^2 x$, and $ an x$ are all positive. But notice that negative sign! It makes $p'(x) < 0$. This means $p(x)$ is strictly decreasing. Since $p(0) = 0$, and the function is decreasing, $p(x)$ must be negative for all $x > 0$. Thus, $g'(x) < 0$, and **$g(x)$ is strictly decreasing**. The green curve dips! ### Conclusion We have peeled back the layers of these functions. We found that $f(x)$ is increasing and $g(x)$ is decreasing. This is the beauty of calculus—it allows us to see the hidden geometry of functions. Keep this analytical mindset, and no problem will ever be too difficult for you. Keep practicing, and I will see you in the next challenge!

Animated Solution for Mathematics - Differentiation: 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

Select Answer:

Visualized Solution

Defining the Functions

Differentiating $f (x)$

Derivative of $f (x)$

Analyzing the Numerator $h (x)$

Conclusion for $f (x)$

Differentiating $g (x)$

Analyzing the Numerator $p (x)$

Conclusion for $g (x)$

Final Conclusion

The Sigma Insight: Monotonicity

The Dance of Functions

Phase 1

Phase 2

Phase 3

Conclusion

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=sinxx​ and g(x)=tanxx​, where 0<x≤1, then in this interval