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Question #41239

Difficulty: 8/10 | 🟢 Free

The Question

Let M(x)=(∫0xt22t2t+2x−t[t]dtx2sin⁡xecos⁡x∫0x(x−t)22t2t+2x−t[x−t]dt)M(x) = \begin{pmatrix} \int_{0}^{x} \frac{t^2 2^t}{2^t + 2^{x-t}} [t] dt & x^2 \sin x \\ e^{\cos x} & \int_{0}^{x} \frac{(x-t)^2 2^t}{2^t + 2^{x-t}} [x-t] dt \end{pmatrix}, where [⋅][ \cdot ] denotes the greatest integer function.

If F(x)=Tr(M(x))F(x) = \text{Tr}(M(x)), find the value of: lim⁡h→0+F(3+h)−F(3−h)h\lim_{h \to 0^+} \frac{F(3+h) - F(3-h)}{h}

Options

  • A.

    45

  • B.

    54

  • C.

    36

  • D.

    27

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

Matrix Trace Properties|Definite Integral Properties (King's Property)|Step Function Discontinuities

Why This Question Is Tricky

Common Mistakes

This question tests the application of properties of definite integrals (a substitution variant of King's Rule) alongside the Leibniz Integral Rule in the presence of jump discontinuities.

The primary trap lies in blindly applying L'Hôpital's rule or assuming differentiability of the trace function F(x)F(x) at integer points.

Students often incorrectly assume F′(3)=32[3]=27F'(3) = 3^2[3] = 27 and evaluate the limit as 2F′(3)=542F'(3) = 54.

However, because the Greatest Integer Function creates a jump discontinuity in the integrand at t=3t=3, F(x)F(x) is non-differentiable at x=3x=3.

The limit must be split into right-hand and left-hand derivatives, yielding F′(3+)=27F'(3^+) = 27 and F′(3−)=18F'(3^-) = 18, making the correct sum 45.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterMatrices - Class 12
ConceptMatrix Trace Properties
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree