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Question #40925

Difficulty: 8/10 | 🟢 Free

The Question

Let a curve CC be defined parametrically by x(t)=(t−1)3x(t) = (t-1)^3 y(t)=lim⁡n→∞t2n(t−1)3(t2+4)+∣t−1∣3(t3+1)t2n+1y(t) = \lim_{n \to \infty} \frac{t^{2n} (t-1)^3 (t^2 + 4) + |t-1|^3 (t^3 + 1)}{t^{2n} + 1} for t∈(0,∞)t \in (0, \infty). Let L=lim⁡x→0+dydxL = \lim_{x \to 0^+} \frac{dy}{dx} and R=lim⁡x→0−dydxR = \lim_{x \to 0^-} \frac{dy}{dx}. Find the value of 10L+R10L + R.

Options

  • A.

    52

  • B.

    48

  • C.

    -22

  • D.

    55

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Indeterminate Forms|Polynomial Continuity/Differentiability Link|Modulus Sign Dependency|Parametric Derivative Endpoint Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Parametric Derivative Endpoint Trap' combined with 'Modulus Sign Dependency' in limit-defined functions.

Students often blindly apply L'Hopital's rule or differentiate the parametric equations without analyzing the piecewise nature of y(t)y(t) caused by the lim⁡n→∞t2n\lim_{n \to \infty} t^{2n} term.

Furthermore, the modulus ∣t−1∣3|t-1|^3 requires careful sign evaluation for t<1t < 1.

Failing to recognize that x→0−x \to 0^- corresponds to t→1−t \to 1^- and dropping the negative sign from the modulus leads to the incorrect right-hand derivative, trapping students who rely on rote differentiation.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLimits and Derivatives - Class 11
ConceptIndeterminate Forms
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree