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Question #41603

Difficulty: 8/10 | 🟢 Free

The Question

Let a curve be defined parametrically by $x(t) = t^2 + t$ and $y = y(t)$ for $t \ge 0$ . It is given that $y(x)$ satisfies the differential equation $(4x+1)\frac{dy}{dx} - 2y = \frac{1}{2}(4x+1)^{3/2}$ with $y(0) = \frac{1}{2}$ . According to Lagrange's Mean Value Theorem, there exists a point on the curve for $x \in [0, 2]$ where the tangent is parallel to the chord joining the endpoints of the curve on this interval. The value of the parameter $t$ corresponding to this point is:

Options

A.
$\frac{1+\sqrt{7}}{6}$
B.
$\frac{-3+\sqrt{39}}{6}$
C.
$\frac{4+2\sqrt{13}}{9}$
D.
$\frac{-3+\sqrt{15}}{6}$

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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

Linear Differential Equation Structure|Parametric Form Manipulation|LMVT Tangent Parallelism

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of Linear Differential Equations and Lagrange's Mean Value Theorem (LMVT) within a parametric context.

The primary trap lies in the domain of application: students frequently confuse the independent variable $x$ with the parameter $t$ .

If they incorrectly apply the interval $[0, 2]$ to $t$ instead of $x$ , they will calculate the wrong secant slope and arrive at an incorrect parameter value (Option C).

Furthermore, it requires careful application of the chain rule $\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$ to equate the parametric derivative to the secant slope.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Differential Equations - Class 12
Concept	Linear Differential Equation Structure
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free