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Question #68625

Difficulty: 8/10 | 🟢 Free

The Question

Let y=y(x)y = y(x) be a thrice differentiable function on (0,π)(0, \pi) satisfying the differential equation: ∣ysin⁡xcos⁡xy′′′−cos⁡xsin⁡xy′cos⁡x−sin⁡x∣sin⁡x+∣ysin⁡xcos⁡xy′cos⁡x−sin⁡xy′′−sin⁡x−cos⁡x∣cos⁡x=0\begin{vmatrix} y & \sin x & \cos x \\ y''' & -\cos x & \sin x \\ y' & \cos x & -\sin x \end{vmatrix} \sin x + \begin{vmatrix} y & \sin x & \cos x \\ y' & \cos x & -\sin x \\ y'' & -\sin x & -\cos x \end{vmatrix} \cos x = 0 Given that y(π2)=1y\left(\frac{\pi}{2}\right) = 1, y′(π2)=1y'\left(\frac{\pi}{2}\right) = 1, and y′′(π2)=0y''\left(\frac{\pi}{2}\right) = 0, find the value of y(π6)y\left(\frac{\pi}{6}\right).

Options

  • A.

    12−32+π312\frac{1}{2} - \frac{\sqrt{3}}{2} + \frac{\pi \sqrt{3}}{12}

  • B.

    12−32+π36−12ln⁡2\frac{1}{2} - \frac{\sqrt{3}}{2} + \frac{\pi \sqrt{3}}{6} - \frac{1}{2} \ln 2

  • C.

    12+32+π312\frac{1}{2} + \frac{\sqrt{3}}{2} + \frac{\pi \sqrt{3}}{12}

  • D.

    12−32−π36+12ln⁡2\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{\pi \sqrt{3}}{6} + \frac{1}{2} \ln 2

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

Integration Constant Trap|Differentiation of Determinants|Row/Column Exchange Parity|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the differentiation of determinants and the row-exchange parity trap.

Students must recognize that differentiating the determinant D(x)D(x) row-by-row yields a determinant where the second and third rows are swapped compared to the first term in the given equation.

Failing to account for this parity flip (or incorrectly integrating cot⁡x\cot x with the wrong sign) leads to the erroneous differential equation D′(x)sin⁡x+D(x)cos⁡x=0  ⟹  D(x)=Kcsc⁡xD'(x) \sin x + D(x) \cos x = 0 \implies D(x) = K \csc x.

This forces them to solve y′′+y=csc⁡xy'' + y = \csc x, leading to a completely different particular solution involving logarithmic terms, perfectly matching the distractor option B.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptIntegration Constant Trap
Difficulty8
Cognitive Load7
Target Solve Time170 sec
AccessFree