Problem 83 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\left|\begin{array}{ccc}x^{n} & \sin x & -\cos x \\ n ! & \sin (n \pi / 2) & \cos (n \pi / 2) \\ a & a^{2} & a^{3}\end{array}\right|\), then \(f^{n}(0)\) for \(n=2 m+1\) is (A) 1 (B) \(-1\) (C) 0 (D) independent of \(a\)

Step-by-Step Solution

Answer

(C) 0

Step 1: Understand the Problem

We are given a function defined by a 3x3 determinant, and we need to evaluate \(f^{n}(0)\) where \(n = 2m + 1\). This means \(n\) is an odd number.

Step 2: Simplify the Determinant

Identify that if \(x = 0\), then the first row of the determinant becomes \([0^{n}, \sin 0, -\cos 0]\) which simplifies to \([0, 0, -1]\). Hence, if there's a row or column of zeros, the determinant evaluates to zero.

Step 3: Evaluate the Determinant at \(x = 0\)

For any \(n \geq 1\), evaluate the determinant using the property that includes a row of zeros: \[\left|\begin{array}{ccc}0 & 0 & -1! & \sin (n\pi / 2) & \cos (n\pi / 2)\a & a^2 & a^3\end{array}\right| = 0\]Since the minor of every element in the first row evaluates to zero.

Step 4: Analyze the Function Behavior

\(f(x)\) evaluated at \(x=0\) simplifies to zero because the first row has all zero values, rendering further evaluation redundant. Thus \(f^n(0) = 0\) regardless of \(m\) because any power of zero remains zero.

Key Concepts

Matrix Algebra

Matrix Algebra is an essential part of understanding determinants, like the one given in the problem. A determinant is a scalar value that can be calculated from the elements of a square matrix. Here, we have a 3x3 matrix:
\[\left|\begin{array}{ccc}x^{n} & \sin x & -\cos x \ n ! & \sin (n\pi / 2) & \cos (n \pi / 2) \ a & a^{2} & a^{3}\end{array}\right|\]
The determinant utilizes mathematical operations like multiplication and addition of the elements and their respective minors and cofactors, helping us solve for unknowns and make various analyses in linear transformations.

• To evaluate the determinant, observe that if a row or column contains all zeros, the determinant is zero.
• In matrix algebra, the order of the matrix (in this case, 3x3) determines the complexity of the determinant calculation.

Understanding these foundational properties supports finding solutions to complex equations, as the zero row in the determinant simplifies the solution here.

Trigonometric Functions

Trigonometric Functions are functions related to angles and must be evaluated carefully in the context of this problem. The terms \( \sin x \) and \( -\cos x \) in the determinant require understanding their specific values at critical points, like when \( x = 0 \).

• The sine function, \( \sin(x) \), at \( x = 0 \) equals zero, and the cosine function, \( \cos(x) \), at \( x = 0 \) equals one.
• The particular elements in the determinant rely on these trigonometric identities, turning complex expressions into simpler ones.

When working with determinants that include trigonometric functions, identifying their values at given points like \( x = 0 \) helps reduce complexity.

Odd and Even Functions

Odd and Even Functions are key concepts needed in understanding the determinants that arise in matrix algebra, especially when terms like \( x^n \) are used. Here's why:

• An **odd function** satisfies \( f(-x) = -f(x) \); for example, \( \sin x \) is an odd function. This property means at symmetrical points opposite to zero, the function behaves similarly but with opposite signs.
• An **even function** satisfies \( f(-x) = f(x) \); examples include \( \cos x \) and \( x^2 \), which look identical on graphs around the y-axis.
• Odd powers resulting from \( n = 2m + 1 \) further ensure that the structure of any resulting expressions maintains symmetry, enhancing simplification when \( n \) is input into the determinant.

Realizing how odd and even properties can simplify evaluations at critical points like zero ensures that the determinant becomes manageable, particularly in contexts where these powers heavily influence outcomes.