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Question #41531

Difficulty: 8/10 | 🟢 Free

The Question

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a twice differentiable function such that for all x≠0x \neq 0, ∫01f(xt)dt=f(x)−xf′(x)+(x2+1−1x)ex+1x\int_0^1 f(xt) dt = f(x) - x f'(x) + \left(x^2 + 1 - \frac{1}{x}\right) e^x + \frac{1}{x} If f(1)=e+2f(1) = e + 2, then the value of f(2)f(2) is:

Options

  • A.

    2e2+22e^2 + 2

  • B.

    2e22e^2

  • C.

    2e2+ln⁡2+22e^2 + \ln 2 + 2

  • D.

    2e2+2e2e^2 + 2e

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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

Integrating f(x) + xf'(x) forms|Leibniz Rule for Differentiation under Integral|Arbitrary Constants vs Parameters

Why This Question Is Tricky

Common Mistakes

The question tests the application of the Leibniz rule to an integral equation where the parameter is hidden in the upper limit after substitution.

The core anchor is recognizing xf′′(x)+f′(x)x f''(x) + f'(x) as the exact derivative of xf′(x)x f'(x).

The primary trap lies in the 'Arbitrary Constants vs Parameters' concept: students often ignore the domain constraint (f:R→Rf: \mathbb{R} \to \mathbb{R}) and carry an arbitrary constant CC, leading to incorrect logarithmic terms, or they incorrectly assume all integration constants are zero (yielding 2e22e^2).

The condition f(1)=e+2f(1) = e+2 specifically isolates the second constant KK, rewarding only those who correctly distinguish between the boundary conditions for CC (implicit via differentiability at x=0x=0) and KK (explicit).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptIntegrating f(x) + xf'(x) forms
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree