Question

If a random variable $X$ has p.d.f. $f(x) = \begin{cases} \frac{ax^2}{2} + bx & , \text{if } 1 \leq x \leq 3 \\ 0 & , \text{otherwise} \end{cases}$ and $f(2) = 2$, then the values of $a$ and $b$ are, respectively}

• A. 11, -10
• B. -9, 10
• C. $\frac{1}{6}, \frac{5}{6}$
• D. 9, -8

Hint:

Sum of probabilities (Area under PDF) must always equal 1.

Answer 1

The Correct Option is D

Step 1: Point Condition
$f(2) = 2 \implies \frac{a(2^2)}{2} + b(2) = 2 \implies 2a + 2b = 2 \implies a + b = 1$.
Step 2: Total Probability Condition
$\int_1^3 (\frac{ax^2}{2} + bx) dx = 1$.
$[\frac{ax^3}{6} + \frac{bx^2}{2}]_1^3 = 1 \implies (\frac{27a}{6} + \frac{9b}{2}) - (\frac{a}{6} + \frac{b}{2}) = 1$.
$\frac{26a}{6} + \frac{8b}{2} = 1 \implies \frac{13a}{3} + 4b = 1$.
Step 3: Solve System
From $b = 1-a$: $\frac{13a}{3} + 4(1-a) = 1 \implies \frac{13a + 12 - 12a}{3} = 1 \implies a + 12 = 3 \implies a = -9$.
Wait, let's re-calculate: $13a/3 - 4a = 1 - 4 \implies a/3 = -3 \implies a = -9, b = 10$.
Final Answer: (B)