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Question #41909

Difficulty: 8/10 | 🟢 Free

The Question

Let {⋅}\{ \cdot \} denote the fractional part function. Evaluate the limit: lim⁡x→π∫πx(({tπ}+{−tπ})sec⁡2(t4)+tan⁡(t4))dtsin⁡(x−π)\lim_{x \to \pi} \frac{\int_{\pi}^{x} \left( \left( \left\{ \frac{t}{\pi} \right\} + \left\{ -\frac{t}{\pi} \right\} \right) \sec^2\left(\frac{t}{4}\right) + \tan\left(\frac{t}{4}\right) \right) dt}{\sin(x - \pi)}

Options

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    0

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

The L'Hôpital's Rule Misuse|Newton-Leibniz Formula with variable limits|Integral Part and Fractional Part

Why This Question Is Tricky

Common Mistakes

This question tests the misuse of the Newton-Leibniz formula (Fundamental Theorem of Calculus) when an integrand has a removable discontinuity at the limit point.

Students commonly assume that the derivative of ∫axh(t)dt\int_a^x h(t) dt is always h(x)h(x).

However, this is only true if h(t)h(t) is continuous at xx.

Here, the integrand contains {t/π}+{−t/π}\{t/\pi\} + \{-t/\pi\}, which equals 1 almost everywhere but drops to 0 at t=πt=\pi.

Blindly applying L'Hôpital's Rule and substituting x=πx=\pi into the discontinuous integrand yields the trap answer 1.

The correct approach recognizes that altering an integrand at a single point does not change the integral function, allowing the use of the continuous equivalent function to find the true derivative and limit.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterContinuity and Differentiability - Class 12
ConceptThe L'Hôpital's Rule Misuse
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree