The following functions are continuous on (0, π). (a) tan x (b) ∫₀ˣ t sin(1/t) dt (c) {1 if 0

Q: Wait, why do we reject $\tan x$? It's continuous everywhere else!

In JEE Advanced, 'continuous on an interval' means *every* point in that interval. Since $\pi/2$ is inside $(0, \pi)$ and $\tan x$ explodes there, the whole function fails the test.

Q: How can $\int_0^x t \sin(1/t) dt$ be continuous at $x=0$ if $1/t$ is undefined?

Great observation! While $1/t$ is undefined at $0$, the *product* $t \sin(1/t)$ approaches $0$ because $\sin(1/t)$ is bounded. It's a removable singularity, making the integrand well-behaved.

Q: Do I always have to check the junction point for piecewise functions?

Yes! The pieces themselves are usually continuous polynomials or trig functions. The only place they can 'break' is where they meet. That junction is the most critical point in the entire problem.

Question

The following functions are continuous on $(0, \pi)$.

Options:

(A) $	an x$

(B) $\int_0^x t \sin \frac{1}{t} dt$

(C) $\begin{cases} 1, & 0 < x \leq \frac{3\pi}{4} \ 2 \sin \frac{2x}{9}, & \frac{3\pi}{4} < x < \pi \end{cases}$

(D) $\begin{cases} x \sin x, & 0 < x \leq \frac{\pi}{2} \ \frac{\pi}{2} \sin(\pi + x), & \frac{\pi}{2} < x < \pi \end{cases}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a path defined by the interval $(0, \pi)$. In the world of calculus, continuity is the ultimate test of a function's integrity. It means you can walk from one end of the interval to the other without ever having to jump, teleport, or fall off a cliff.

Today, we are going to investigate four different functions to see which ones allow for this smooth, uninterrupted journey.

### The Trap of Trigonometry: Analyzing $f(x) = 	an x$

Our first candidate is the tangent function, $f(x) = 	an x$. At first glance, it seems innocent enough. But remember, $	an x$ is defined as:

$$f(x) = \frac{\sin x}{\cos x}$$

The moment the denominator, $\cos x$, hits zero, our path is destroyed. In our interval $(0, \pi)$, this happens exactly at $x = \frac{\pi}{2}$.

As we approach $\frac{\pi}{2}$ from the left, the function shoots up to positive infinity. From the right, it plunges to negative infinity. This is a classic vertical asymptote—a cliff that makes the function discontinuous. Thus, $	an x$ fails the test.

### The Beauty of the Integral Function: $f(x) = \int_0^x t \sin \frac{1}{t} dt$

Next, we encounter an integral function. This looks intimidating, but let's break it down. The Fundamental Theorem of Calculus tells us that if the integrand, $g(t) = t \sin \frac{1}{t}$, is continuous, then its integral function is smooth.

But what about $t=0$? The term $\sin \frac{1}{t}$ oscillates wildly as $t$ approaches zero. However, it is always bounded between $-1$ and $1$.

When we multiply this by $t$, the $t$ term acts like a squeeze, forcing the entire expression to zero. Because the limit exists and equals the function value, we have a removable singularity. The integrand is effectively bounded and continuous, meaning our integral function remains **perfectly continuous** on $(0, \pi)$.

### The Piecewise Puzzle: Building Bridges

Finally, we look at our piecewise functions. These are like bridges built in sections. The only place they can fail is at the junction point.

For our third option, the junction is at $x = \frac{3\pi}{4}$. We calculate the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL). The LHL is simply the constant $1$.

For the RHL, we plug $\frac{3\pi}{4}$ into $2 \sin \frac{2x}{9}$, which gives us:

$$2 \sin\left(\frac{3\pi}{4} \cdot \frac{2}{9}ight) = 2 \sin\left(\frac{\pi}{6}ight) = 2 \cdot \frac{1}{2} = 1$$

Since $1 = 1$, the bridge is perfectly aligned! This function is **continuous**.

Conversely, for our fourth option, the junction is at $x = \frac{\pi}{2}$. The LHL gives us $\frac{\pi}{2}$, but the RHL gives us $-\frac{\pi}{2}$. They don't meet! There is a massive jump discontinuity of size $\pi$.

### Conclusion

By carefully checking the asymptotes, the behavior of integrals, and the alignment of piecewise junctions, we have successfully identified that the integral function and the first piecewise function are our winners.

Keep this methodical approach in your toolkit, and no continuity problem will ever stand in your way!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The following functions are continuous on $(0, π)$ .

Select Answer:

Visualized Solution

Understanding the Interval $(0, π)$

Analyzing Option A: $f (x) = tan x$

The Asymptote at $x = \frac{π}{2}$

Analyzing Option B: The Integral Function

Behavior of $g (t) = t sin \frac{1}{t}$

Analyzing the Piecewise Function at $x = \frac{3 π}{4}$

Limits at $x = \frac{3 π}{4}$

Option C is Continuous

Analyzing the Piecewise Function at $x = \frac{π}{2}$

Limits at $x = \frac{π}{2}$

Jump Discontinuity at $x = \frac{π}{2}$

Summary of Continuous Functions

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Trap of Trigonometry

The Beauty of the Integral Function: $f (x) = \int_{0}^{x} t sin \frac{1}{t} d t$

The Piecewise Puzzle

Conclusion

Similar Questions

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The following functions are continuous on (0,π).

Select Answer:

Visualized Solution

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

Analyzing Option A: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=tanx

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

Analyzing Option B: The Integral Function

Behavior of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(t)=tsint1​

Analyzing the Piecewise Function at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=43π​

Limits at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=43π​

Option C is Continuous

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

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

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