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Question #40451

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous even function such that $f(x+4) = f(x)$ for all $x \in \mathbb{R}$ , and $\int_{0}^{2} f(x) dx = 5$ . If $F(x) = \int_{x}^{x+4} t f(t) dt$ , then the value of $\int_{-10}^{10} F(x) dx$ is:

Options

A.
0
B.
200
C.
400
D.
800

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

Symmetry and Periodicity|Even and Odd Function Symmetry|Newton-Leibniz Formula with variable limits

Why This Question Is Tricky

Common Mistakes

This question tests the application of Leibniz's rule on an integral with variable limits and a periodic integrand.

The primary trap is the 'hidden constant of integration.' Students often differentiate $F(x)$ to get $4f(x)$ and incorrectly assume $F(x) = 4 \int_0^x f(t) dt$ , completely missing the initial value $F(0) = C$ .

This leads them to conclude $F(x)$ is an odd function, yielding an erroneous answer of 0.

Furthermore, evaluating $F(0)$ requires combining King's property with the even and periodic nature of $f(x)$ , creating a multi-layered conceptual barrier typical of JEE Advanced.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Symmetry and Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free