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Question #40516

Difficulty: 8/10 | 🟢 Free

The Question

Let f:(0,∞)→Rf: (0, \infty) \to \mathbb{R} be a continuous function satisfying the differential equation xf′(x)+f(x)=15x2∣x−2∣x−2x f'(x) + f(x) = 15x^2 \frac{|x-2|}{x-2} for all x∈(0,∞)∖{2}x \in (0, \infty) \setminus \{2\}. If f(1)=27f(1) = 27, then the value of f(8)f(8) is

Options

  • A.

    312

  • B.

    314

  • C.

    316

  • D.

    324

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

Sign Flipping Rule|Integrating f(x) + xf'(x) forms|Integration Constant Trap|General vs Particular Solutions

Why This Question Is Tricky

Common Mistakes

The question tests the 'Integration Constant Trap' within a piecewise continuous differential equation.

Students easily identify the exact differential ddx(xf(x))\frac{d}{dx}(xf(x)), but the modular condition forces a sign flip at x=2x=2.

The trap occurs when students assume a single integration constant CC for the entire domain or incorrectly transfer C2C_2 to the x>2x > 2 domain.

To find the correct particular solution, they must enforce continuity at the boundary x=2x=2, solving for a new constant C1C_1 that bridges the discontinuity.

Distractor 312 catches a common arithmetic error during the boundary evaluation, while 324 catches the false assumption that C1=C2C_1 = C_2.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLinear Inequalities - Class 11
ConceptSign Flipping Rule
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree