MHT CET202519 Apr 2025Morning ShiftMathematicsProbabilityActual
If a random variable X has the p.d.f. f (x)= array cl k x^2+1 & , if 0 < x < 0 & , otherwise array . then c.d.f. of X is
Options
- A2 ⁻¹ x
- B2 ⁻¹ x
- C2 ⁻¹ x
- D⁻¹ x
Correct answer
C. 2 ⁻¹ x
Step-by-step solution
The probability density function integrates to 1 over its domain, requiring ₀^ k x^2+1 dx = 1 . Evaluating the integral gives k [ x ]₀^ = k ( 2 - 0 ) = 1 and thus k = 2 . The cumulative distribution function for x > 0 is F(x) = ₀^ x 2/ t^2+1 dt = 2 x . For x 0 , F(x) = 0 . The resulting c.d.f. is F(x) = cases 0 & x 0 2 x & x > 0 cases which corresponds to option C .