Start Learning Free

Before You Begin

Question #41907

Difficulty: 8/10 | 🟢 Free

The Question

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a continuous function such that f′(x)=∣x2−5x+6∣x2−5x+6⋅2x−5x2−5x+8f'(x) = \frac{|x^2 - 5x + 6|}{x^2 - 5x + 6} \cdot \frac{2x - 5}{x^2 - 5x + 8} for all x∈R∖{2,3}x \in \mathbb{R} \setminus \{2, 3\}. If f(52)=0f\left(\frac{5}{2}\right) = 0, then the value of ef(1)+ef(4)e^{f(1)} + e^{f(4)} is

Options

  • A.

    72\frac{7}{2}

  • B.

    327\frac{32}{7}

  • C.

    14

  • D.

    87\frac{8}{7}

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

Modulus Function Breakdown|Logarithmic Integration Modulus|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the 'Logarithmic Integration Modulus' trap.

Students must recognize that the modulus creates a piecewise derivative, requiring separate constants of integration for each interval defined by the roots of the quadratic.

The core trap is failing to enforce continuity at the non-differentiable points (x=2,3x=2, 3) to match these constants.

Ignoring the modulus entirely or assuming the constant of integration remains unchanged across intervals leads directly to the incorrect exponential sums provided in the distractors.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterContinuity and Differentiability - Class 12
ConceptModulus Function Breakdown
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree