Problem 32 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The function \(f(x)=\left\\{\begin{array}{l}\frac{\pi}{4}+\tan ^{-1} x,|x| \leq 1 \\ \frac{1}{2}(|x|-1),|x|>1\end{array}\right.\) is: (a) continuous on \(\mathbf{R}-\\{1\\}\) and differentiable on \(\mathbf{R}-\\{-1,1\\}\) (b) both continuous and differentiable on \(\mathbf{R}-\\{1\\}\). (c) continuous on \(\mathbf{R}-\\{-1\\}\) and differentiable on \(\mathbf{R}-\\{-1,1\\}\) (d) both continuous and differentiable on \(\mathbf{R}-\\{-1\\}\).

Step-by-Step Solution

Answer

(c) continuous on \( \mathbf{R}-\{-1\} \) and differentiable on \( \mathbf{R}-\{-1, 1\} \).

Step 1: Identify Function Properties

The function \( f(x) \) is defined piecewise. For \( |x| \leq 1 \), the function is \( \frac{\pi}{4} + \tan^{-1}x \). For \(|x| > 1\), the function is \( \frac{1}{2}(|x|-1) \). We need to check continuity and differentiability on different intervals.

Step 2: Continuity Check at \( x = 1 \)

To check if the function is continuous at \( x = 1 \), we calculate \( \lim_{x \to 1^-} f(x) \), \( f(1) \), and \( \lim_{x \to 1^+} f(x) \).\( \lim_{x \to 1^-} f(x) = \frac{\pi}{4} + \tan^{-1}(1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \).\( f(1) = \frac{\pi}{4} + \tan^{-1}(1) = \frac{\pi}{2} \).\( \lim_{x \to 1^+} f(x) = \frac{1}{2}(1 - 1) = 0 \).The limits from the left and right do not match, so the function is discontinuous at \( x = 1 \).

Step 3: Check Differentiability around \( x = 1 \)

Since the function is discontinuous at \( x = 1 \), it cannot be differentiable there. We still check differentiability at points close to but not including \( x = 1 \).

Step 4: Continuity and Differentiability at \( x = -1 \)

At \( x = -1 \), we solve:\( \lim_{x \to -1^-} f(x) = \frac{1}{2}((|-1|-1)) = \frac{1}{2}(1 - 1) = 0 \).\( f(-1) = \frac{\pi}{4} + \tan^{-1}(-1) = \frac{\pi}{4} - \frac{\pi}{4} = 0 \).\( \lim_{x \to -1^+} f(x) = \frac{\pi}{4} + \tan^{-1}(-1) = 0 \).The function is continuous at \( x = -1 \) and differentiable since both segments meet smoothly at \( x = -1 \).

Step 5: Conclusion

Given that the function is discontinuous at \( x = 1 \) and continuous and differentiable at \( x = -1 \), the correct answer is (c): it is continuous on \( \mathbf{R}-\{-1\} \) and differentiable on \( \mathbf{R}-\{-1, 1\} \).

Key Concepts

Continuity

In mathematics, continuity is a crucial property that indicates whether a function behaves predictably without interruptions or jumps at a particular point within its domain.
To check if a function is continuous at a specific point, such as at \(x = 1\), we need to examine the behavior of the function as it approaches that point from both sides.
Here's the basic process to determine continuity:

• Calculate the limit of the function as it approaches the point from the left \(\lim_{x \to 1^-} f(x)\).
• Determine the function's value exactly at the point \(f(1)\).
• Evaluate the limit of the function as it approaches the point from the right \(\lim_{x \to 1^+} f(x)\).

If all three values are identical, the function is continuous at that point.
In our example, while the left-hand limit and the function value at \(x = 1\) both equaled \(\frac{\pi}{2}\), the right-hand limit was \(0\), indicating a discontinuity at \(x = 1\).
This discontinuity suggests that at \(x = 1\), the function experiences a jump, breaking continuity and confirming that it only remains continuous on \(\mathbb{R}-\{-1\}\).

Differentiability

Differentiability is another important property for functions. It describes whether a function has a derivative at a particular point, indicating how smoothly it changes.
For a function to be differentiable at a point, it must first be continuous there, but it also must not have any sharp corners or cusps.
Here's how to think about differentiability:

• First, ensure the function is continuous at the point of interest.
• Next, examine the slope or rate of change as you approach that point from both sides.

If these slopes align and there are no abrupt changes in direction, the function is differentiable.
In our problem, since the function was found to be discontinuous at \(x = 1\), it cannot be differentiable there.
However, at \(x = -1\), the function is continuous and the pieces of the function connect smoothly, confirming differentiability there except for \(x = 1\).
Therefore, the function is differentiable on \(\mathbb{R}-\{-1, 1\}\).

Trigonometric Functions

Trigonometric functions describe relationships in geometry involving angles and lengths. They're ubiquitous in mathematics and appear frequently in calculus problems.
The arctangent function \(\tan^{-1}(x)\), which appears in our piecewise function, is one such trigonometric function.
Here are some notable features of \(\tan^{-1}(x)\):

• It provides the angle whose tangent is \(x\), offering a means to reverse the tangent operation.
• Its range is limited to \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), making it invertible within that interval.

Understanding these properties helps students evaluate limits and find continuity for functions involving trigonometric elements.
In the function given, \(\frac{\pi}{4} + \tan^{-1}(x)\) smoothly defines the function for \(|x| \leq 1\).
This part aligns correctly within its range, aiding our assessment of the continuity and differentiability of the function within the designated intervals.