If a continuous random variable \(\mathrm{X}\) has probability density function \(\mathrm{f}(x)\) given by
\(
f(x)=\left\{\begin{array}{cc}
a x , \text { if } 0 \leq x < 1 \
a , \text { if } 1 \leq x < 2 \
3 a-a x
\end{array},\right.
\) \(\text { if } 2 \leq x \leq 3 \
0 \text {, otherwise }\)
then a has the value

Since \(\mathrm{f}(x)\) is the p.d.f. of \(\mathrm{X}\),
\(
\begin{aligned}
& \int_{-\infty}^{\infty} \mathrm{f}(x) \mathrm{d} x=1 \\
& \Rightarrow \int_0^1 \mathrm{a} x \mathrm{~d} x+\int_1^2 \mathrm{ad} x+\int_2^3(3 \mathrm{a}-\mathrm{ax}) \mathrm{d} x=1 \\
& \Rightarrow \mathrm{a}\left[\frac{x^2}{2}\right]_0^1+\mathrm{a}[x]_1^2+\left[3 \mathrm{a} x-\frac{\mathrm{a} x^2}{2}\right]_2^3=1
\end{aligned}
\)
\(\begin{aligned} & \Rightarrow \mathrm{a}\left(\frac{1}{2}\right)+\mathrm{a}(1)+\left(\frac{9 \mathrm{a}}{2}-4 \mathrm{a}\right)=1 \\ & \Rightarrow 2 \mathrm{a}=1 \\ & \Rightarrow \mathrm{a}=\frac{1}{2}\end{aligned}\)