Before You Begin

Question #40664

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(\alpha) = \int_0^{\sqrt{3}} \lfloor x^2 \rfloor (1+\alpha x^2)^3 (1+\alpha - \alpha x^2)^3 x \, dx$ , where $\lfloor \cdot \rfloor$ denotes the greatest integer function. The value of $\left. \frac{d^3 f}{d\alpha^3} \right|_{\alpha=0}$ is:

Options

A.
-144
B.
-72
C.
-288
D.
9

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Vandermonde's Identity|Leibniz Rule for Differentiation under Integral|Substitution (Change of Variable)|Step Function Discontinuities

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of the Leibniz Rule, algebraic manipulation (related to Vandermonde's identity/multinomial expansion), and definite integration with step functions.

The primary trap lies in the step function $\lfloor x^2 \rfloor$ .

After applying the substitution $t=x^2$ , students often incorrectly apply the Fundamental Theorem of Calculus across the entire interval $[0, 3]$ , ignoring the internal discontinuities at $t=1$ and $t=2$ , which leads to the distractor -72.

A secondary trap is the differentiation of the product; expanding blindly is tedious, but recognizing the algebraic structure $(1+\alpha x^2)(1+\alpha(1-x^2)) = 1+\alpha+\alpha^2 x^2(1-x^2)$ provides an elegant shortcut.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Vandermonde's Identity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free