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Question #41152

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=(cos⁡x+([sin⁡x∣x∣]+2)sin⁡x)1∣x∣+[x+1]tan⁡xf(x) = \left( \cos x + \left( \left[ \frac{\sin x}{|x|} \right] + 2 \right) \sin x \right)^{\frac{1}{|x| + [x+1] \tan x}}, where [⋅][ \cdot ] denotes the greatest integer function. If L1=lim⁡x→0−f(x)L_1 = \lim_{x \to 0^-} f(x) and L2=lim⁡x→0+f(x)L_2 = \lim_{x \to 0^+} f(x), then the value of L1L2L_1 L_2 is:

Options

  • A.

    1

  • B.

    e

  • C.

    e^2

  • D.

    e^{-1}

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

Greatest Integer Function Traps|Indeterminate Forms|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle the Greatest Integer Function (GIF) and Modulus function in the context of one-sided limits.

The primary trap lies in evaluating [sin⁡x∣x∣]\left[ \frac{\sin x}{|x|} \right] as x→0−x \to 0^-.

Students often incorrectly assume sin⁡x∣x∣→1\frac{\sin x}{|x|} \to 1 implies the GIF is 11 or 00, missing that for x<0x < 0, the ratio approaches −1-1 from above, making the GIF −1-1.

Furthermore, the modulus sign dependency flips the denominator in the exponent, and the [x+1][x+1] term acts as an annihilator for the tan⁡x\tan x term in the Left-Hand Limit.

Failing to bifurcate the LHL and RHL properly leads to incorrect indeterminate form evaluations.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 12
ConceptGreatest Integer Function Traps
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree