P(|X|< 1) = ? A1/2B1/8C1/4D3/4Answer: C. 1/4 Read Explanation: ∫02f(x)dx=1\int_0^2 f(x)dx=1∫02f(x)dx=1∫02kx=1\int_0^2 kx =1∫02kx=1k[x22]02=1k[\frac{x^2}{2}]_0^2=1k[2x2]02=1k[42]=1k[\frac{4}{2}]=1k[24]=12k=12k=12k=1k=12k=\frac{1}{2}k=21P(|X|<1) = P(0<x<1)=∫01f(x)dx=\int_0^1 f(x)dx=∫01f(x)dx=k∫01xdx=k\int_0^1xdx =k∫01xdx=k×12=k \times \frac{1}{2}=k×21=12×12=14=\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}=21×21=41 Read more in App