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Question #41411

Difficulty: 8/10 | 🟢 Free

The Question

Let $y = f(x)$ be the continuous real-valued function defined on $[-2, 2]$ satisfying the implicit equation $y^3 - 3y + x = 0$ and $f(0) = 0$ . Evaluate the integral: $I = \int_{-2}^{2} \frac{f(x)}{\sqrt{2-x}} \, dx$

Options

A.
$\frac{12}{5}(2\sqrt{3} - 3)$
B.
$\frac{12}{5}(3 - 2\sqrt{3})$
C.
$\frac{12}{5}(2\sqrt{3} + 3)$
D.
$\frac{4}{5}(2\sqrt{3} - 3)$

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

Substitution (Change of Variable), Implicit to Explicit Conversion, Parametric Derivative Endpoint Trap, First Derivative Test Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Parametric Derivative Endpoint Trap' combined with the 'First Derivative Test Trap' during implicit-to-explicit integral substitution.

When converting the integral using $x = 3y - y^3$ , the term $\sqrt{2-x}$ factors into $\sqrt{(y-1)^2(y+2)}$ .

The critical trap lies at the endpoint $y=1$ where the parametric derivative $dx/dy$ vanishes.

Students frequently ignore the absolute value in $\sqrt{(y-1)^2} = |y-1|$ and blindly simplify it to $y-1$ .

Because the domain of substitution is $y \in [-1, 1]$ , the correct simplification is $1-y$ .

Failing to apply this first derivative sign analysis reverses the sign of the canceled factor, leading directly to the negative of the correct answer (Option B).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Substitution (Change of Variable)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free