Let f: (0, π) → R be a twice differentiable function such that lim (t→x) (f(x)sin t - f(t)sin x)/(t-

Q: Why can't I differentiate with respect to $x$ in the limit?

Great question! In the limit $\lim_{t\to x}$, $t$ is the variable approaching $x$. $x$ is acting as a fixed parameter. If you differentiate with respect to $x$, you are treating the limit as a function of $x$ rather than evaluating the limit itself. Always differentiate with respect to the variable that is 'moving'—in this case, $t$.

Q: How did you know to divide by $\sin^2 x$?

It's all about pattern recognition. When you see $f'(x)\sin x - f(x)\cos x$, it screams 'Quotient Rule'. The standard form of the Quotient Rule has a $v^2$ in the denominator. Since $v = \sin x$, $v^2 = \sin^2 x$. Dividing by $\sin^2 x$ is the natural step to reveal the derivative of $\frac{f(x)}{\sin x}$.

Q: Why does the inequality sign flip in Option B?

This is a classic trap! We know $\sin x > x - \frac{x^3}{6}$ for $x > 0$. When we multiply both sides by $-x$, we are multiplying by a negative number. In algebra, multiplying an inequality by a negative number reverses the direction of the inequality. Never forget this simple rule—it's where many students lose marks!

Question

Let $f: (0, \pi) 	o R$ be a twice differentiable function such that $\lim_{t	o x} \frac{f(x)\sin t - f(t)\sin x}{t-x} = \sin^2x$ for all $x \in (0, \pi)$. If $f\left(\frac{\pi}{6}ight) = -\frac{\pi}{12}$, then which of the following statement(s) is (are) TRUE ?

Options:

(A) $\int_{0}^{\pi/2} f(x) dx = \frac{\pi^2}{4\sqrt{2}}$

(B) $f(x) < -\frac{x^2}{6}$ for all $x \in (0, \pi)$

(C) There exists $\alpha \in (0, \pi)$ such that $f'(\alpha) = 0$

(D) $f''\left(\frac{\pi}{2}ight) + f\left(\frac{\pi}{2}ight) = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with the expression:
$$\lim_{t	o x} \frac{f(x)\sin t - f(t)\sin x}{t-x} = \sin^2 x$$

At first glance, this looks like a nightmare. But pause and take a deep breath. If we plug $t = x$ into this expression, the numerator becomes $f(x)\sin x - f(x)\sin x = 0$, and the denominator becomes $x - x = 0$. We have a $\frac{0}{0}$ indeterminate form.

Whenever you see $\frac{0}{0}$ in a limit, your mind should immediately jump to **L'Hopital's Rule**. Since we are taking the limit as $t 	o x$, $t$ is our variable and $x$ is merely a constant parameter. We must differentiate with respect to $t$.

Applying the derivative to the numerator:
$$\frac{d}{dt} [f(x)\sin t - f(t)\sin x] = f(x)\cos t - f'(t)\sin x$$

The derivative of the denominator, $\frac{d}{dt} [t - x]$, is simply $1$. Evaluating the limit by substituting $t = x$, the expression simplifies beautifully to:
$$f(x)\cos x - f'(x)\sin x = \sin^2 x$$

We have successfully tamed the beast and transformed a limit into a differential equation.

### The Hidden Quotient Rule

Now, we have $f(x)\cos x - f'(x)\sin x = \sin^2 x$. To isolate the terms, let us divide both sides by $\sin^2 x$:
$$\frac{f(x)\cos x - f'(x)\sin x}{\sin^2 x} = 1$$

Look closely at the left-hand side. Recall the **Quotient Rule** for differentiation:
$$\frac{d}{dx}\left(\frac{u}{v}ight) = \frac{v u' - u v'}{v^2}$$

If we set $u = f(x)$ and $v = \sin x$, then the derivative is $\frac{f'(x)\sin x - f(x)\cos x}{\sin^2 x}$. Our expression is exactly the negative of this. Therefore, we can rewrite our equation as:
$$-\frac{d}{dx}\left(\frac{f(x)}{\sin x}ight) = 1$$

This is the "Aha!" moment. We have reduced a complex differential equation to a simple derivative equal to a constant.

### The Function Unveiled

Integrating both sides with respect to $x$, we get:
$$\frac{f(x)}{\sin x} = -x + C$$

To find the constant $C$, we use the initial condition provided: $f(\frac{\pi}{6}) = -\frac{\pi}{12}$. Substituting $x = \frac{\pi}{6}$ into our equation:
$$\frac{-\pi/12}{\sin(\pi/6)} = -\frac{\pi}{6} + C$$

Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, the left side becomes $\frac{-\pi/12}{1/2} = -\frac{\pi}{6}$. Thus, $-\frac{\pi}{6} = -\frac{\pi}{6} + C$, which implies $C = 0$. Our mystery function is revealed:
$$f(x) = -x \sin x$$

### Final Analysis

Now that we have $f(x) = -x \sin x$, we can evaluate the properties of the function.

For the integral $\int_{0}^{\pi/2} -x \sin x \, dx$, using integration by parts, we find the result is $-1$. This confirms that specific integral values do not match arbitrary constants.

For the inequality analysis, we use the Taylor series expansion $\sin x = x - \frac{x^3}{6} + \dots$. For $x > 0$, we know $\sin x > x - \frac{x^3}{6}$. Multiplying by $-x$ (and flipping the inequality sign because $-x$ is negative), we get:
$$-x \sin x < -x^2 + \frac{x^4}{6}$$

Regarding **Rolle's Theorem**, since $f(0) = 0$ and $f(\pi) = -\pi \sin(\pi) = 0$, and the function is continuous and differentiable, there must exist some $\alpha \in (0, \pi)$ where $f'(\alpha) = 0$. This is mathematically guaranteed.

Finally, calculating the derivatives: $f'(x) = -\sin x - x \cos x$ and $f''(x) = -2 \cos x + x \sin x$. Evaluating at $x = \frac{\pi}{2}$, we find $f''(\frac{\pi}{2}) = \frac{\pi}{2}$ and $f(\frac{\pi}{2}) = -\frac{\pi}{2}$. Their sum is zero.

Calculus is not about memorizing formulas; it is about recognizing patterns and following the logic wherever it leads.

Select Answer:

Visualized Solution

The Limit Equation for $f (x)$

Applying L'Hopital's Rule

Evaluating the Limit at $t = x$

Rearranging the Equation

Recognizing the Quotient Rule

Integrating the Equation

Finding the Constant $C$

The Function $f (x)$

Checking Option A

Evaluating Option A

Checking Option B

Proving Option B

Checking Option C

Checking Option D

Final Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Hidden Quotient Rule

The Function Unveiled

Final Analysis

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