Let $X$ have distribution function $F(x)$ expressed by
\begin{cases} 0 & \mbox{if } x < 0 \\ x/2 & \mbox{if } 0\leq x \leq 2 \\ 1 & \mbox{if } x \geq 2 \end{cases}
and let $Y = X^{2}$. Find
(a) $\textbf{P}(\frac{1}{2}\leq X\leq \frac{3}{2})$
(b) $\textbf{P}(1\leq X < 2)$
(c) $\textbf{P}(Y\leq X)$
(d) $\textbf{P}(X\leq 2Y)$
(e) $\textbf{P}(X+Y\leq\frac{3}{4})$
(f) The distribution of $Z = \sqrt{X}$.
MY ATTEMPT
(a) $\displaystyle\textbf{P}(0.5\leq X\leq 1.5) = F(1.5) - \lim_{x\uparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$
(b) $\displaystyle\textbf{P}(1\leq X < 2) = \lim_{x\uparrow 2}F(x) - \lim_{x\uparrow 1}F(x) = 1 - 0.5 = 0.5$
(c) $\displaystyle\textbf{P}(Y\leq X) = \textbf{P}(X^{2} \leq X) = \textbf{P}(X(X-1) \leq 0) = \textbf{P}(0 \leq X\leq 1)$
$\displaystyle\therefore \textbf{P}(Y\leq X) = F(1) - \lim_{x\uparrow 0}F(x) = 0.5 - 0 = 0.5$
(d) $\displaystyle\textbf{P}(X \leq 2Y) = \textbf{P}(X\leq 2X^{2}) = \textbf{P}(X(2X-1)\geq0) = 1 - \textbf{P}(X<0.5)$
$\displaystyle\therefore \textbf{P}(X\leq 2Y) = 1 - \lim_{x\uparrow 0.5}F(x) = 1 - 0 .25 = 0.75$
(e) $\textbf{P}(X+Y\leq\frac{3}{4}) = \textbf{P}(4X^{2} + 4X - 3\leq 0) = \textbf{P}(-1.5 \leq X\leq 0.5)$
$\displaystyle\therefore \textbf{P}(X+Y\leq 0.75) = F(0.5) - \lim_{x\uparrow-1.5}F(x) = 0.25 - 0 = 0.25$
(f) I am not sure how to solve it. Here is my attempt
\begin{align*} \textbf{P}(Z\leq z) = \textbf{P}(\sqrt{X}\leq z) = \textbf{P}(X\leq z^{2}) = F_{X}(z^{2}) \end{align*}
More precisely, the distribution function of $Z$ is given by
\begin{cases} 0 & \mbox{if } z < 0 \\ z^{2}/2 & \mbox{if } 0\leq z \leq \sqrt{2} \\ 1 & \mbox{if } z \geq\sqrt{2} \end{cases}
Could someone please double-check my results? I thank in advance for any contribution.
EDIT
I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.
I think there is a mistake in e). $4X^{2}+4X-3 \leq 0$ gives $-\frac 3 2 \leq X \leq \frac 1 2$. In f) you should replace the condition $0 \leq z \leq 2$ by $0 \leq z^{2} \leq 2$ or $0 \leq z \leq \sqrt 2$.