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Question #40725

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function defined by $f(x) = \int_0^x \left( t^3 \sin\left(\frac{1}{t}\right) + t^2 \right) dt \quad \text{for } x \neq 0, \text{ and } f(0) = 0.$ For each $x \in \left(0, \frac{1}{3}\right)$ , let $c_x \in (0, x)$ be the unique point guaranteed by the Lagrange Mean Value Theorem such that the tangent to the curve $y = f(x)$ at $x = c_x$ is parallel to the chord joining the points $(0, 0)$ and $(x, f(x))$ . Find the value of $\lim_{x \to 0^+} \frac{c_x}{x}$

Options

A.
$\frac{1}{\sqrt{3}}$
B.
$\frac{1}{2}$
C.
0
D.
\text{Does not exist}

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

Locus of a Moving Point|LMVT Tangent Parallelism|The L'Hôpital's Rule Misuse

Why This Question Is Tricky

Common Mistakes

This question tests the asymptotic behavior of the LMVT locus near a coordinate singularity, specifically targeting the blind application of L'Hôpital's rule.

Students often attempt to evaluate the limit of the ratio $c_x/x$ by assuming it equals $\lim c_x'$ and differentiating the LMVT condition $x f'(c_x) = f(x)$ .

However, $f''(x)$ contains an oscillating term $\cos(1/x)$ that does not converge as $x \to 0^+$ .

Consequently, L'Hôpital's rule fails because the limit of the derivatives does not exist, leading students to incorrectly choose 'Does not exist'.

The correct approach requires recognizing that the original limit can be evaluated using asymptotic bounding (Big-O notation) and the Squeeze Theorem, bypassing the non-existent second derivative at the singularity.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Locus of a Moving Point
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free