Before You Begin

Question #61504

Difficulty: 8/10 | 🟢 Free

The Question

Let $p(x)$ be the probability density function of a continuous random variable $X$ defined on the interval $[0, a]$ . It is given that $p(x)$ satisfies the differential equation $x p'(x) - p(x) = \frac{6}{5a^3} x^2$ The variance of $X$ is:

Options

A.
$\frac{a^2}{20}$
B.
$\frac{3a^2}{80}$
C.
$\frac{3a^2}{20}$
D.
$\frac{a^2}{18}$

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

Calculation of Variance for Grouped Data|Exact Differential Recognition|Integration Constant Trap

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the intersection of exact differential equations and continuous probability distributions.

The primary trap is the 'Integration Constant Trap'.

Students often solve the differential equation and take the particular solution $p(x) = \frac{6}{5a^3}x^2$ by implicitly assuming $C=0$ .

If they calculate variance directly from this unnormalized function, they get $\frac{3a^2}{20}$ .

If they realize it doesn't integrate to 1 and normalize it to $p(x) = \frac{3}{a^3}x^2$ , they get $\frac{3a^2}{80}$ .

Both are incorrect because the true distribution requires finding the integration constant $C$ using the fundamental boundary condition $\int p(x) dx = 1$ , which fundamentally alters the shape of the distribution and its moments.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Statistics - Class 11
Concept	Calculation of Variance for Grouped Data
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free