Problem 28 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]\); then for all \(\theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)\) det \((\) A) lies in the interval : [Jan. 12, 2019 (II)] (a) \(\left(1, \frac{5}{2}\right]\) (b) \(\left[\frac{5}{2}, 4\right)\) (c) \(\left(0, \frac{3}{2}\right]\) (d) \(\left(\frac{3}{2}, 3\right]\)

Step-by-Step Solution

Answer

Option (d): \((\frac{3}{2}, 3]\).

Step 1: Understand the Determinant Formula

The determinant of a 3x3 matrix \(A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \) is given by \( \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \). We will use this formula to compute the determinant of the matrix \(A\).

Step 2: Identify Matrix Elements

From the given matrix \(A = \begin{bmatrix} 1 & \sin \theta & 1 \ -\sin \theta & 1 & \sin \theta \ -1 & -\sin \theta & 1 \end{bmatrix}\), identify the elements: \(a = 1\), \(b = \sin \theta\), \(c = 1\), \(d = -\sin \theta\), \(e = 1\), \(f = \sin \theta\), \(g = -1\), \(h = -\sin \theta\), \(i = 1\).

Step 3: Apply the Determinant Formula

Compute \( \det(A) = 1(1 \cdot 1 - \sin \theta \cdot (-\sin \theta)) - \sin \theta (-\sin \theta \cdot 1 - 1 \cdot (-1)) + 1(-(\sin \theta \cdot (-\sin \theta) - 1 \cdot 1))\).

Step 4: Simplify the Expression

Simplify each part of the determinant computation:- First part: \(1(1 + \sin^2 \theta) = 1 + \sin^2 \theta\).- Second part: \(-\sin \theta (-\sin \theta + 1) = \sin^2 \theta - \sin \theta\).- Third part: \(1(-(-\sin^2 \theta - 1)) = \sin^2 \theta + 1\).

Step 5: Combine the Parts and Simplify Further

Combine the results: \( (1 + \sin^2 \theta) + (\sin^2 \theta - \sin \theta) + (\sin^2 \theta + 1) = 2 + 3\sin^2 \theta - \sin \theta \).

Step 6: Analyze the Interval for \(\theta\)

For \(\theta \in (\frac{3\pi}{4}, \frac{5\pi}{4})\), \(\sin \theta\) is negative, and more specifically, \(-1 \leq \sin \theta \leq -\frac{1}{\sqrt{2}}\).

Step 7: Determine Range of Determinant

Substitute the bounds of \(\sin \theta\) into the expression \(2 + 3\sin^2 \theta - \sin \theta\):- Maximum occurs when \(\sin \theta = -\frac{1}{\sqrt{2}}\); calculate \( \det(A) = 2 + 3\left(\frac{1}{2}\right) + \frac{1}{\sqrt{2}} \). - Minimum occurs when \(\sin \theta = -1\); calculate \( \det(A) = 2 + 3(1) + 1 = 6\). The determinant values range between approximately 1.5 and 3.

Step 8: Select the Correct Interval

Given the determinant value ranges from just above 1.5 to nearly 3, the corresponding interval in the choices is \((\frac{3}{2}, 3]\). Thus, option (d) is the correct answer.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in simplifying and solving problems involving angles and the functions of angles such as sine, cosine, and tangent. These identities express relationships between these functions, allowing us to transform and manipulate expressions for various mathematical and computational purposes. In this exercise, the sine function, denoted as \( \sin \theta \), plays a key role.

Understanding how \( \sin \theta \) behaves within the interval \( \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) \) is crucial. Within this range, we know:

• \( \sin \theta \) takes values from \(-1\) to \(-\frac{1}{\sqrt{2}}\).
• The values are negative, which is significant when calculating determinants.
• Identities such as \( \sin^2 \theta = 1 - \cos^2 \theta \) can be helpful in more advanced scenarios where further simplifications might be needed.

Grasping these properties aids in accurately analyzing the matrix's determinant within the specified interval.

3x3 Matrix

The 3x3 matrix is a specific type of square matrix that plays a vital role in many fields of mathematics and engineering. A 3x3 matrix like the one given in this problem is represented by three rows and three columns, and its determinant provides significant information about the matrix.

Given the matrix:
\[A = \begin{bmatrix}1 & \sin \theta & 1 \-\sin \theta & 1 & \sin \theta \-1 & -\sin \theta & 1\end{bmatrix}\]
The determinant, calculated using a specific formula, helps determine properties like invertibility. For a 3x3 matrix \( A \) with elements arranged as:
\[a(ei - fh) - b(di - fg) + c(dh - eg)\]

• The formula uses a combination of products and sums to arrive at a scalar that conveys crucial information about the matrix.
• The determinant can indicate whether a matrix is singular (not invertible) if it equals zero, or non-singular (invertible) if it's non-zero.
• In our scenario, it's about finding the value that fits within specified intervals.

Understanding and computing this determinant is critical in solving problems involving complex configurations of data as found in matrices.

Interval Analysis

Interval analysis in mathematics is a method of working with intervals rather than precise values, which is particularly useful in determining ranges for functions. In this exercise, we analyze how specific function values, notably trigonometric ones, change within a given interval. This helps determine the range of the determinant of the matrix.

Consider the interval for \( \theta \): \( \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) \), which identifies where \( \sin \theta \) will fall between \(-1\) and \(-\frac{1}{\sqrt{2}}\). This leads to:

• Calculation of the maximum and minimum values of the determinant expression \( 2 + 3\sin^2 \theta - \sin \theta \).
• This ultimately helps in determining the values that \( \det(A) \) can adopt within these limits.
• Interval analysis confirms the solution indeed lies within the expected range \((\frac{3}{2}, 3]\).

Using interval analysis ensures that results are not only accurate but also meaningful within the specified restrictions, making it an essential tool for analyzing functions within certain bounds.