Q99 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics

Let X be a continuous random variable with the probability density function(p.d.f.) given by
$f(x) = \begin{cases} kx, & 0 \leq x < 1 \\ k, & 1 \leq x < 2 \\ -kx + 3k, & 2 \leq x < 3 \\ 0, & \text{otherwise} \end{cases}$
$P(2 < X \leq 3) = \cdots$

  • A
    $\dfrac{1}{2}$
  • B
    $\dfrac{1}{3}$
  • Correct answer
    C
    $\dfrac{1}{4}$
  • D
    $\dfrac{1}{5}$

Solution

Since $f(x)$ is a probability density function, the total probability must be $1$.

$\int_{-\infty}^{\infty} f(x) dx = 1$

$\int_{0}^{1} kx dx + \int_{1}^{2} k dx + \int_{2}^{3} (-kx + 3k) dx = 1$

$k \left[ \dfrac{x^2}{2} \right]_0^1 + k [x]_1^2 + k \left[ -\dfrac{x^2}{2} + 3x \right]_2^3 = 1$

$\dfrac{k}{2} + k(2 - 1) + k \left( \left(-\dfrac{9}{2} + 9\right) - \left(-\dfrac{4}{2} + 6\right) \right) = 1$

$\dfrac{k}{2} + k + k \left( \dfrac{9}{2} - 4 \right) = 1$

$\dfrac{k}{2} + k + \dfrac{k}{2} = 1$

$2k = 1 \Rightarrow k = \dfrac{1}{2}$

Now, we need to find $P(2 < X \leq 3)$:

$P(2 < X \leq 3) = \int_{2}^{3} f(x) dx = \int_{2}^{3} (-kx + 3k) dx$

$P(2 < X \leq 3) = k \left[ -\dfrac{x^2}{2} + 3x \right]_2^3 = \dfrac{k}{2}$

Substituting $k = \dfrac{1}{2}$, we get:

$P(2 < X \leq 3) = \dfrac{1}{4}$

Answer: $\dfrac{1}{4}$

Hence, the correct answer is option: 'C'.

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