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Question #41383

Difficulty: 8/10 | 🟢 Free

The Question

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a function defined by f(x)={1+(x−1)2+(x−1)42−e(x−1)2(x−1)2if x<1x5−6x4+14x3−16x2+9x−2if x≥1f(x) = \begin{cases} \frac{1 + (x-1)^2 + \frac{(x-1)^4}{2} - e^{(x-1)^2}}{(x-1)^2} & \text{if } x < 1 \\ x^5 - 6x^4 + 14x^3 - 16x^2 + 9x - 2 & \text{if } x \ge 1 \end{cases} Let SS be the set of all points of inflection of the curve y=f(x)y = f(x). Find the sum of the elements of SS.

Options

  • A.

    85\frac{8}{5}

  • B.

    135\frac{13}{5}

  • C.

    35\frac{3}{5}

  • D.

    1

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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

Point of Inflection Conditions|Polynomial Continuity/Differentiability Link|Closed vs Open Intervals|L'Hopital's Pitfall

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous definition of a point of inflection, specifically the necessity of a sign change in the second derivative.

The trap is set by a piecewise function that is twice differentiable at the junction x=1x=1 with f′′(1)=0f''(1)=0.

Students often mistakenly assume f′′(c)=0f''(c)=0 is sufficient for an inflection point and will incorrectly include x=1x=1 in their set SS, leading to the trap answer 13/5.

The problem requires careful analysis of the sign of f′′(x)f''(x) on both sides of the junction, utilizing Taylor series for the transcendental part to prove it remains strictly negative, thus revealing x=1x=1 as a 'ghost' inflection point.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Derivatives - Class 12
ConceptPoint of Inflection Conditions
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree