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Question #41272

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=∣x+1x2xx+1+cos⁡(1x)x2+sin⁡(1x)x+cos⁡(1x)x+2x2−2xx+1∣f(x) = \begin{vmatrix} x+1 & x^2 & x \\ x+1+\cos\left(\frac{1}{x}\right) & x^2 + \sin\left(\frac{1}{x}\right) & x + \cos\left(\frac{1}{x}\right) \\ x+2 & x^2 - 2x & x + 1 \end{vmatrix} for x≠0x \neq 0. Evaluate the limit: L=lim⁡x→01x∫0xf(t)dtL = \lim_{x \to 0} \frac{1}{x} \int_0^x f(t) dt

Options

  • A.

    0

  • B.

    1

  • C.

    -1

  • D.

    \text{Does not exist}

Check Your Answer

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

Elementary Row Operation Invariance|Exact Differential Recognition|The L'Hôpital's Rule Misuse|L'Hopital's Pitfall

Why This Question Is Tricky

Common Mistakes

This question tests the invariance of determinants under elementary operations to reveal an exact differential, coupled with a classic L'Hôpital's rule trap.

Students evaluating the limit of the integral divided by xx will encounter a 0/00/0 indeterminate form.

A blind application of L'Hôpital's rule yields lim⁡x→0f(x)\lim_{x \to 0} f(x), which oscillates and does not exist, leading students to incorrectly choose 'Does not exist'.

However, L'Hôpital's rule is only conclusive if the limit of the derivatives exists; when it fails, the original limit must be evaluated by explicitly integrating the exact differential first, which correctly yields 0 via the Squeeze Theorem.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDeterminants - Class 12
ConceptElementary Row Operation Invariance
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree