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Question #41620

Difficulty: 8/10 | 🟢 Free

The Question

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a continuously differentiable function such that f(2)=2f(2) = 2, f(3)=2f(3) = 2, and for all x∈Rx \in \mathbb{R}, (f(x)−∣x−2∣−∣x−4∣)(f′(x)+6x2−36x+48)=0(f(x) - |x-2| - |x-4|) \left( f'(x) + 6x^2 - 36x + 48 \right) = 0 Find the value of f(1)+f(5)f(1) + f(5).

Options

  • A.

    4

  • B.

    8

  • C.

    12

  • D.

    -4

Check Your Answer

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

Integration Constant Trap|Triangle Inequality Mind Games|Locus of a Moving Point|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Zero-Factor Cancellation Trap' combined with piecewise differential equations.

Students often blindly cancel the (f(x)−g(x))(f(x) - g(x)) term, assuming f′(x)=h(x)f'(x) = h(x) globally, which leads to an incorrect integration constant (distractor 12).

Alternatively, they might assume f(x)=g(x)f(x) = g(x) everywhere (distractor 8).

The core trap forces students to realize that continuous differentiability allows the solution to be piecewise, switching branches only where the function values and derivatives match.

The boundary conditions f(2)=2f(2)=2 and f(3)=2f(3)=2 act as a domain restrictor, forcing the singular solution f(x)=2f(x) = 2 on [2,4][2, 4], while the general solution dictates the behavior outside, perfectly blending Class 11 modulus concepts with Class 12 differential equations.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptIntegration Constant Trap
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree