A) 9
B) 3
C) \[\frac{1}{9}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
[a]: Given. \[k\int\limits_{0}^{1}{x}.f(3x)dx=\int\limits_{0}^{3}{t.f(t)dt}\] ?.(i) Let\[I=k\int\limits_{0}^{1}{x}.f(3x)dx\] Put\[3x=t\Rightarrow dx=dt/3\]. So,\[x=0\Rightarrow t=0;x=1\Rightarrow t=3\] \[\therefore \]\[I=k\int\limits_{0}^{3}{\frac{1}{3}\frac{t}{3}f(t)}dt\] \[=\frac{k}{9}\int\limits_{0}^{3}{t}.f(t)dt=\int\limits_{0}^{3}{t}.f(t)dt\] [Using (i)] \[\Rightarrow \]\[\frac{k}{9}=1\Rightarrow k=9\]
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