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Question #41384

Difficulty: 8/10 | 🟢 Free

The Question

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a function defined by f(x)=∫0x(t2−4)3(∫−tt(u4ln⁡(3+sin⁡u3−sin⁡u)e∣u∣+55u4−330u2)du)dtf(x) = \int_{0}^{x} (t^2 - 4)^3 \left( \int_{-t}^{t} \left( u^4 \ln\left(\frac{3+\sin u}{3-\sin u}\right) e^{|u|} + 55u^4 - 330u^2 \right) du \right) dt Find the sum of the squares of the abscissae of all points of inflection of the curve y=f(x)y = f(x).

Options

  • A.

    10

  • B.

    20

  • C.

    28

  • D.

    38

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

Point of Inflection Conditions|Leibniz Rule for Differentiation under Integral|Even and Odd Function Symmetry|Integration by Parts (LIATE Rule)

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Leibniz rule combined with parity (even/odd) properties of definite integrals, while setting a multi-layered trap.

The primary trap is the intimidating inner integrand, which tempts students into complex Integration by Parts (LIATE rule); however, recognizing the odd symmetry immediately annihilates the non-elementary term.

The secondary, more subtle trap lies in the definition of a point of inflection.

Students often incorrectly assume that any root of f′′(x)=0f''(x) = 0 is a point of inflection.

By engineering the derivative to have roots with even multiplicity (x=0,±2x=0, \pm 2), students who blindly sum the squares of all roots will arrive at the incorrect trap answer of 28, rather than the correct sum of 20 from the roots where concavity actually changes.

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