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Question #40627

Difficulty: 8/10 | 🟢 Free

The Question

Let $n = \int_{0}^{\pi/2} \frac{18 \sin 2x}{(1 + \sin 2x)^2} dx$ . If the term independent of $x$ in the binomial expansion of $\left( x^2 \arcsin(\alpha) + \frac{\arccos(\alpha)}{x} \right)^n$ is equal to $\frac{15 \pi^6}{2}$ , then the number of real values of $\alpha$ is

Options

A.
0
B.
1
C.
2
D.
4

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

Term Independent of x|Substitution (Change of Variable)|Monotonicity and Boundedness

Why This Question Is Tricky

Common Mistakes

The question tests the integration of trigonometric functions using the tangent substitution to determine the binomial index $n$ .

The primary trap lies in the domain restriction of inverse trigonometric functions.

Students must recognize that $y = \arcsin(\alpha)$ is bounded within $[-\pi/2, \pi/2]$ .

By setting the independent term to a value strictly greater than its absolute maximum on this restricted domain, the equation yields no real solutions.

A common mistake is to treat $y$ as an unrestricted real variable, which leads to falsely identifying roots for the degree-6 polynomial $y^2(\pi/2 - y)^4 = \pi^6/2$ , thereby falling for the distractor options.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Term Independent of x
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free