Explanation

(a)
Let $I = k\int_{0}^{1}{xf(3x)dx}$
Let $3x = t \Rightarrow dx = \frac{dt}{3}$
$\therefore I = k\int_{0}^{3}{\frac{t}{3}.f(t).\frac{dt}{3} = \frac{k}{9}\int_{0}^{3}t.f(t)dt}$
Now, $\frac{k}{9}\int_{0}^{3}{t\ .f(t)dt = \int_{0}^{3}t.f(t)dt\ }\ \ \ \$[given]
$\Rightarrow \frac{k}{9} = 1 \Longrightarrow k = 9$