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Question #40639

Difficulty: 8/10 | 🟢 Free

The Question

Let $M(x) = \begin{pmatrix} 0 & 1 & 2x & 0 \\ -1 & 0 & 0 & 2x \\ -2x & 0 & 0 & 1 \\ 0 & -2x & -1 & 0 \end{pmatrix}$ . For $x > \frac{1}{2}$ , a polynomial $P(x)$ is defined as $P(x) = \sqrt{|M(x)|}$ , where $|M(x)|$ denotes the determinant of $M(x)$ . Find the algebraic value of the numerically greatest term in the binomial expansion of $(P(x))^{11}$ evaluated at $x = \frac{3}{4}$ .

Options

A.
$-165 \left(\frac{3}{2}\right)^{16}$
B.
$165 \left(\frac{3}{2}\right)^{16}$
C.
$-55 \left(\frac{3}{2}\right)^{18}$
D.
$55 \left(\frac{3}{2}\right)^{18}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Numerically Greatest Term|Properties of Binomial Coefficients|Skew-Symmetric Determinant|Minors and Cofactors Sign Convention

Why This Question Is Tricky

Common Mistakes

This question tests the properties of skew-symmetric determinants and the concept of the numerically greatest term (NGT) in a binomial expansion.

The primary trap lies in the definition of $P(x) = \sqrt{|M(x)|}$ .

Students often blindly simplify $\sqrt{X^2}$ to $X$ , incorrectly defining $P(x) = 1 - 4x^2$ .

While the magnitude of the NGT remains identical, the alternating signs in the binomial expansion of $(1 - 4x^2)^{11}$ versus the correct $(4x^2 - 1)^{11}$ result in opposite algebraic signs for the final answer.

The domain constraint $x > 1/2$ strictly enforces $P(x) = 4x^2 - 1$ , making the negative sign mathematically mandatory and punishing the missing absolute value.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Numerically Greatest Term
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free