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Question #41324

Difficulty: 8/10 | 🟢 Free

The Question

A curve is defined parametrically by $x(t) = t - \sin t$ and $y(t) = \frac{1}{3}(1 - \cos t)^{3/2}$ for $t \in (-\pi, \pi)$ . Let $y = f(x)$ be the continuous real-valued function representing this curve. If $L$ and $R$ denote the left-hand and right-hand derivatives of $f(x)$ at $x = 0$ respectively, find the value of $R - L$ .

Options

A.
$\sqrt{2}$
B.
0
C.
$\frac{1}{\sqrt{2}}$
D.
$2\sqrt{2}$

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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

Parametric Derivative Endpoint Trap|Chain Rule Inner Function Differentiability|Right-Hand vs Left-Hand Derivatives

Why This Question Is Tricky

Common Mistakes

This question targets the 'Analytical Blind Spot' in parametric differentiation involving fractional powers or roots.

Students frequently compute the derivative of $(1-\cos t)^{3/2}$ mechanically, implicitly assuming $\sqrt{x^2} = x$ and dropping the absolute value.

This algebraic oversight leads to the incorrect conclusion that $\frac{dy}{dx} = \frac{1}{\sqrt{2}}\cos(t/2)$ , which implies the derivative is continuous at $t=0$ (yielding $R-L=0$ ).

By evaluating at a point where $x'(t)=0$ and the inner function changes sign, it tests the rigorous application of LHD/RHD limits and domain constraints in chain rule compositions.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Parametric Derivative Endpoint Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free