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Question #70522

Difficulty: 8/10 | 🟢 Free

The Question

Let M(x,t)=(txxxtxxxt)M(x, t) = \begin{pmatrix} t & x & x \\ x & t & x \\ x & x & t \end{pmatrix} and N(x,t)=xM(x,t)N(x, t) = x M(x, t). Let f(x)=∫0x∣N(x,t)∣(x−t)2dtf(x) = \int_{0}^{x} \frac{|N(x, t)|}{(x-t)^2} dt for x>0x > 0. Let k=225f′(2)k = \frac{2}{25} f'(2). If PP is a 3×33 \times 3 matrix such that ∣P∣=k|P| = k, find the value of log⁡2(∣adj(adj(2P))∣)\log_2 \left( |\text{adj}(\text{adj}(2P))| \right).

Options

  • A.

    28

  • B.

    24

  • C.

    32

  • D.

    20

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Leibniz Rule for Differentiation under Integral|Determinant Scaling Property|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

The question tests the 'Zero-Factor Cancellation Trap' in the context of the Leibniz rule and matrix singularities.

If a student attempts to differentiate the determinant ∣N(x,t)∣|N(x,t)| using Jacobi's formula or applies the Leibniz rule without first cancelling the (x−t)2(x-t)^2 factor, they encounter a singularity at the upper limit t=xt=x where the matrix becomes non-invertible (yielding an indeterminate 0/00/0 boundary term).

By simplifying the integrand algebraically before applying calculus operations, the removable singularity is resolved.

The problem further layers determinant scaling properties (∣cA∣=cn∣A∣|cA| = c^n|A|) and nested adjoint determinants to ensure comprehensive mastery.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptLeibniz Rule for Differentiation under Integral
Difficulty8
Cognitive Load7
Target Solve Time170 sec
AccessFree