Find $ P(X \le 1) $ The random variable X takes values on interval $ [0, 2] $ , where
$ P( X =1 ) =0.25            $
$ F( x|x<1) = \ x^2 $
$ F( x|x>1) = \ x-1 $
$ E(X) = 1 $
Find $  P(X \le 1) $
My attempt
Since we know that in a continuous distribution  the the probability of getting a specific  random variable x is 0 , so how is it here $ P( X =1 ) =0.25            ? $
Please clarify

I think you need to write down the cdf $F(x)$. There will be an unknown probability $a$ for $x<1$ and an unknown probability $b$ for $x>1$. You get one equation from $F(2)=1$ and another equation from $E[X]=1$, giving two equations in two unknowns to solve
