Let f : [0, ∞) → [0, 3] be a function defined by f(x) = { max{sin t : 0 ≤ t ≤ x}, 0 ≤ x ≤ π; 2 + cos

Q: Wait, why does the function become constant after $x = \frac{\pi}{2}$? Shouldn't it just be $\sin x$ everywhere?

That is a fantastic question! Remember, $f(x)$ is defined as the *maximum* value of $\sin t$ in the interval $[0, x]$. Once $x$ passes $\frac{\pi}{2}$, the sine function starts to decrease, but the *maximum* value we have encountered so far is still the peak value of $1$ that occurred at $t = \frac{\pi}{2}$. Therefore, the function 'remembers' that peak and stays constant at $1$ until $x$ reaches $\pi$.

Q: Do I really need to check differentiability at $x = \frac{\pi}{2}$ and $x = \pi$? Can't I just assume it's differentiable?

In JEE Advanced, never assume! Piecewise functions are notorious for having 'kinks' or sharp corners at the points where the definition changes. Even if the function is continuous, the slope might jump abruptly. We must explicitly verify that the derivative from the left matches the derivative from the right at these transition points.

Q: Is the derivative of a constant always zero? That seems too simple.

It is simple, but it is also powerful! Yes, the derivative of any constant function is zero because a constant function has a slope of zero—it never changes. When we check the differentiability at $x = \frac{\pi}{2}$ and $x = \pi$, we find that the derivatives of the adjacent functions also approach zero, which is exactly why the function remains smooth.

Question

Let $f : [0, \infty) 	o [0, 3]$ be a function defined by $f(x) = \begin{cases} \max\{\sin t : 0 \le t \le x\}, & 0 \le x \le \pi \ 2 + \cos x, & x > \pi \end{cases}$ Then which of the following is true ?

Options:

(A) $f$ is continuous everywhere but not differentiable exactly at one point in $(0, \infty)$

(B) $f$ is differentiable everywhere in $(0, \infty)$

(C) $f$ is not continuous exactly at two points in $(0, \infty)$

(D) $f$ is continuous everywhere but not differentiable exactly at two points in $(0, \infty)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The hiker's altitude is defined by the function $f(x) = \max\{\sin t : 0 \le t \le x\}$. This function acts as a "memory" of the maximum value attained by the sine function over the interval $[0, x]$.

To understand the behavior of $f(x)$, we must partition the domain into distinct phases based on the properties of the sine function.

### Phase 1: The Ascent

For the interval $0 \le x \le \frac{\pi}{2}$, the sine function is strictly increasing. Because the hiker is constantly reaching new heights, the running maximum is always equal to the current altitude.

Thus, for $0 \le x \le \frac{\pi}{2}$, we have:
$$f(x) = \sin x$$

### Phase 2: The Plateau

As $x$ moves from $\frac{\pi}{2}$ to $\pi$, the sine function begins to descend. However, the device only updates when a new record is set.

Since the hiker reached a maximum altitude of $1$ at $x = \frac{\pi}{2}$, and the altitude decreases thereafter, the record remains fixed. Therefore, for $\frac{\pi}{2} < x \le \pi$:
$$f(x) = 1$$

### Phase 3: The New Path

For $x > \pi$, the function definition transitions to $f(x) = 2 + \cos x$. We must verify the continuity and differentiability at the transition points to ensure the path is seamless.

At $x = \frac{\pi}{2}$, the left-hand limit is $\sin(\frac{\pi}{2}) = 1$, and the right-hand limit is $1$. Since these match, the function is continuous.

For differentiability at $x = \frac{\pi}{2}$, the left-hand derivative is $\cos(\frac{\pi}{2}) = 0$, and the right-hand derivative of the constant $1$ is $0$. The function is perfectly smooth at this point.

### Phase 4: The Final Transition

At $x = \pi$, we check the continuity again. The left-hand limit is $1$, and the right-hand limit is:
$$2 + \cos(\pi) = 2 - 1 = 1$$

Because the limits match, the function is continuous at $x = \pi$.

To check differentiability, we compare the derivatives. The left-hand derivative is $0$, and the right-hand derivative is the derivative of $2 + \cos x$, which is $-\sin x$. At $x = \pi$:
$$-\sin(\pi) = 0$$

Since the derivatives match, the transition is seamless.

### Conclusion

We have analyzed the entire journey of the hiker. From the initial climb to the plateau and the final path, the function $f(x)$ remains continuous and differentiable.

**The function $f(x)$ is differentiable everywhere in $(0, \infty)$.** This problem demonstrates that breaking a complex function into its geometric components allows the underlying calculus to reveal its inherent elegance.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : [0, \infty) \to [0, 3]$ be a function defined by $f (x) = {max {sin t : 0 \leq t \leq x}, 2 + cos x, 0 \leq x \leq π x > π$ Then which of the following is true ?

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Analyzing the Running Maximum for $0 \leq x \leq \frac{π}{2}$

Analyzing the Running Maximum for $\frac{π}{2} < x \leq π$

Defining the Full Piecewise Function

Continuity at $x = \frac{π}{2}$

Differentiability at $x = \frac{π}{2}$

Continuity at $x = π$

Differentiability at $x = π$

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Analyzing the Running Maximum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0≤x≤2π​

Analyzing the Running Maximum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2π​<x≤π

Defining the Full Piecewise Function

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​

Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=π

Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=π