Problem 115 - A Complete Resource Book in Mathematics for JEE Main

Question

The set of points of continuity of the function \(f(x)=\sqrt{\frac{1}{2}-\cos ^{2} x}\) is (A) \(\left\\{x: \frac{\pi}{4}+2 n \pi \leq x \leq \frac{3 \pi}{4}+2 n \pi, n \in I\right\\}\) (B) \(\left\\{x: \frac{5 \pi}{4}+2 n \pi \leq x \leq \frac{7 \pi}{4}+2 n \pi, n \in I\right\\}\) (C) \(\begin{aligned}\left\\{x: \frac{\pi}{4}+2 n \pi \leq\right.& \left.x \leq \frac{3 \pi}{4}+2 n \pi\right\\} \\ & \cup\left\\{x: \frac{5 \pi}{4}+2 n \pi \leq x \leq \frac{7 \pi}{4}+2 n \pi\right\\} \end{aligned}\) (D) None of these

Step-by-Step Solution

Answer

The set of points of continuity is option (C).

Step 1: Identify the Domain

The function is defined as \( f(x) = \sqrt{\frac{1}{2} - \cos^2 x} \). For square roots to be real, the expression inside them must be non-negative. Therefore, we need \( \frac{1}{2} - \cos^2 x \geq 0 \). This simplifies to \( \cos^2 x \leq \frac{1}{2} \).

Step 2: Solve the Inequality

The inequality \( \cos^2 x \leq \frac{1}{2} \) can be rewritten as \( -\frac{1}{\sqrt{2}} \leq \cos x \leq \frac{1}{\sqrt{2}} \). The cosine function achieves these values at specific points. We know that \( \cos x = \frac{1}{\sqrt{2}} \) at \( x = \frac{\pi}{4} \) and \( \cos x = -\frac{1}{\sqrt{2}} \) at \( x = \frac{3\pi}{4} \).

Step 3: Determine the Intervals

Between these points, \( \cos x \leq \frac{1}{\sqrt{2}} \) occurs in intervals. For the interval from \( \frac{\pi}{4} \) to \( \frac{3\pi}{4} \), and similarly in the symmetric intervals about each multiple \( 2n\pi \), the condition holds. Likewise, for \( x = \frac{5\pi}{4} \) to \( x = \frac{7\pi}{4} \), the symmetrical positions will also satisfy the inequality.

Step 4: Consolidate the Valid Intervals

Considering periodicity of \( \cos x \) and the symmetry in trigonometric functions, the function will be continuous where \( -\frac{1}{\sqrt{2}} \leq \cos x \leq \frac{1}{\sqrt{2}} \). This implies two intervals: \( \frac{\pi}{4} \leq x \leq \frac{3\pi}{4} \) and \( \frac{5\pi}{4} \leq x \leq \frac{7\pi}{4} \), which extend with periodicity.

Step 5: Match with Given Options

The valid intervals will be expressed as \( \left\{x: \frac{\pi}{4}+2 n \pi \leq x \leq \frac{3 \pi}{4}+2 n \pi, n \in I\right\} \cup \left\{x: \frac{5 \pi}{4}+2 n \pi \leq x \leq \frac{7 \pi}{4}+2 n \pi\right\} \), which corresponds to option (C).

Key Concepts

continuity of functions

In mathematics, continuity is a fundamental concept that describes a function that does not have any sudden jumps, holes, or interruptions. A function is continuous at a specific point if the following three conditions are satisfied:

• The function is defined at the point.
• The limit of the function as it approaches the point from both sides exists.
• The value of the function at the point is equal to the limit at that point.

This means, intuitively, that you can draw the graph of a continuous function without lifting your pencil off the paper.
The function in the given exercise, represented as \(f(x) = \sqrt{\frac{1}{2} - \cos^2 x}\), is working with trigonometric expressions. Its continuity depends on the domain where the expression inside the square root remains non-negative because square roots of negative numbers are not real. Thus, identifying the domain and ensuring the expression stays greater than or equal to zero is crucial for assessing continuity.
This ensures there are no undefined points or gaps in the graph of the function.

cosine function

The cosine function, denoted \( \cos(x) \), is one of the primary trigonometric functions. It represents the x-coordinate of a point on the unit circle as the angle \(x\) is formed by a line segment from the origin with the positive x-axis.
Cosine is periodic, meaning it repeats values at regular intervals as \(x\) increases or decreases. This periodic nature is defined by its period of \(2\pi\), after which the function values repeat.
When examining inequality conditions involving cosine, such as \(\cos^2 x \leq \frac{1}{2}\), it's crucial to understand where these values occur. In this exercise, \(\cos x\) equals \(\frac{1}{\sqrt{2}}\) at \(x = \frac{\pi}{4}\) and \(-\frac{1}{\sqrt{2}}\) at \(x = \frac{3\pi}{4}\). Understanding these specific points helps in determining where the cosine function lies within the outlined range, both above and below the x-axis, and thus helps find the intervals of continuity for the function \(f(x)\).

mathematical intervals

Mathematical intervals are used to describe a set of numbers between two endpoints. They can be open, closed, or half-open intervals:

• Open intervals do not include endpoint values, denoted \((a, b)\).
• Closed intervals include both endpoints, denoted \([a, b]\).
• Half-open intervals include one endpoint but not the other, denoted \([a, b)\) or \((a, b]\).

When working with trigonometric functions and their continuity, it is often necessary to find specific intervals of the domain that satisfy certain conditions. In this exercise, the solution identified intervals based on where \(-\frac{1}{\sqrt{2}} \leq \cos x \leq \frac{1}{\sqrt{2}}\).
These values define two main intervals for the expression inside the square root to remain non-negative: \(\left[ \frac{\pi}{4}, \frac{3\pi}{4} \right]\) and \(\left[ \frac{5\pi}{4}, \frac{7\pi}{4} \right]\). Each interval takes the periodic nature of cosine into account by extending these values for any integer \(n\) in \(\left[\frac{\pi}{4} + 2n\pi, \frac{3\pi}{4} + 2n\pi\right]\) and \(\left[\frac{5\pi}{4} + 2n\pi, \frac{7\pi}{4} + 2n\pi\right]\). This step is crucial in solving and understanding how the intervals relate to function continuity and the defined domain.