Q.If the function $f(x)=\dfrac{1}{12}$ for $a<x<b$, represents a probability density function of a continuous random variable X, then which of the following cannot be the value of a and b ?

The pdf normalization condition forces $b-a=12$; only one option violates this.

1. For $f(x)=\dfrac1{12}$ on $a<x<b$ to be a valid probability density function, the total probability must be $1$: $\displaystyle\int_a^bf(x)\,dx=1$.
2. $\displaystyle\int_a^b\dfrac1{12}\,dx=\dfrac{b-a}{12}=1\Rightarrow b-a=12$. …

For $f(x)=2x$ on $[0,a]$ to be a valid pdf, the total probability must equal $1$, which forces $a=1$.

1. A probability density function must satisfy $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx=1$.
2. Since $f(x)=2x$ only on $[0,a]$ and $0$ elsewhere, this becomes $\displaystyle\int_{0}^{a}2x\,dx=1$. …

A cumulative distribution function $F(x)$ is always non-decreasing, never decreasing — so option (c) is the false statement.

1. The distribution function of a discrete random variable $X$ is $F(x)=P(X\le x)$.
2. By definition, as $x$ increases the event $\{X\le x\}$ can only include more outcomes, so $F(x)$ is a non-decreasing (monotonically increasing) function of $x$ — never decreasing.
3. Since probabilities lie between $0$ and $1$, we always have $0\le F(x)\le 1$ for all $x\in\mathbb{R}$ — this makes (b) a true statement.
4. As $x\to\infty$, all outcomes are included, so $\displaystyle\lim_{x\to\infty}F(x)=F(\infty)=1$ — this makes (a) a true statement. …

Recall the defining axioms of a probability density function: non-negativity everywhere, and total area 1 — there is no requirement that $f(x)\le1$, since density values can exceed 1 for narrow distributions.

1. By definition, a function $f(x)$ is a probability density function (p.d.f.) of a continuous random variable $X$ if and only if:
   1. $f(x)\ge 0$ for all $x\in\mathbb R$, and
   2. $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$. …