Answer: The correct option is (b) [tex]\dfrac{361}{4}.[/tex]
Step-by-step explanation: We are given to select the correct value of 'n' such that [tex] x^2-19x+n[/tex] becomes a perfect square trinomial.
The standard form of a perfect square trinomial is
[tex](x+a)^2=x^2+2a+a^2.[/tex]
Now, we can write
[tex]x^2-19x+n\\\\=x^2-2\times x\times \dfrac{19}{2}+\dfrac{361}{4}+n-\dfrac{361}{4}\\\\\\=(x-\dfrac{19}{2})^2+n-\dfrac{361}{4}.[/tex]
So, for the given expression to be perfect trinomial,
[tex]n-\dfrac{361}{4}=0\\\\\Rightarrow n=\dfrac{361}{4}.[/tex]
Thus, (b) is the correct option.
