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If \[f\left( x \right)=\dfrac{x\left( x+1 \right)}{2}\] is given, then determine the expression for the value of \[f\left( x+2 \right)\].
(a) \[f\left( x \right)+f\left( 2 \right)\]
(b) \[\left( x+2 \right)f\left( x \right)\]
(c) \[x\left( x+2 \right)f\left( x \right)\]
(d) \[\dfrac{xf\left( x \right)}{x+2}\]
(e) \[\dfrac{\left( x+2 \right)f\left( x+1 \right)}{x}\]

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Hint: In this question, We are given with the function \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\]. We will then evaluate the expression of \[f\left( x+2 \right)\] by \[x\] by \[x+2\] in \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\]. And then we will check which of the given options satisfy the obtained expression of \[f\left( x+2 \right)\].

Complete step-by-step answer:
The function \[f\left( x \right)\] is given by \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\].
In order to find the expression of \[f\left( x+2 \right)\] in terms of some value of the function \[f\left( x \right)\], we will first evaluate the value of \[f\left( x+2 \right)\].
Now replacing \[x\] by \[x+2\] in the expression \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\], we get
\[f\left( x+2 \right)=\dfrac{\left( x+2 \right)\left( \left( x+2 \right)-1 \right)}{2}\]
On simplifying the above expression for \[f\left( x+2 \right)\], we get
\[\begin{align}
  & f\left( x+2 \right)=\dfrac{\left( x+2 \right)\left( x+2-1 \right)}{2} \\
 & =\dfrac{\left( x+2 \right)\left( x+1 \right)}{2}
\end{align}\]
Thus we have
\[f\left( x+2 \right)=\dfrac{\left( x+2 \right)\left( x+1 \right)}{2}\]
Now if we multiply and divide the left hand side of the above equation with \[x\], then we will get
\[f\left( x+2 \right)=\dfrac{\left( x+2 \right)\left( x+1 \right)\left( x \right)}{2x}...........(1)\]
Now we can check which of the following options will satisfy equation (1)
Let if possible \[f\left( x+2 \right)=f\left( x \right)+f\left( 2 \right)\].
Now the function \[f(x)\] is given by \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\]. Using this in the above expression for \[f\left( x+2 \right)\], we get
\[\begin{align}
  & f\left( x+2 \right)=f\left( x \right)+f\left( 2 \right) \\
 & =\dfrac{x\left( x-1 \right)}{2}+\dfrac{2\left( 2-1 \right)}{2} \\
 & =\dfrac{x\left( x-1 \right)}{2}+1 \\
 & =\dfrac{x\left( x-1 \right)+2}{2}
\end{align}\]
Since in this case the value of \[f\left( x+2 \right)\] cannot be expressed as the value obtained in the equation (1), hence option (a) is wrong.
Now let if possible \[f\left( x+2 \right)=\left( x+2 \right)f\left( x \right)\]. Then again using \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\], we get
\[\begin{align}
  & f\left( x+2 \right)=\left( x+2 \right)f\left( x \right) \\
 & =\left( x+2 \right)\left( \dfrac{x\left( x-1 \right)}{2} \right) \\
 & =\dfrac{x\left( x-1 \right)\left( x+2 \right)}{2}
\end{align}\]
which is again not equal to the value obtained in the equation (1), hence option (b) is wrong.
Now let if possible \[f\left( x+2 \right)=x\left( x+2 \right)f\left( x \right)\]. Then again using \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\], we get
\[\begin{align}
  & f\left( x+2 \right)=x\left( x+2 \right)f\left( x \right) \\
 & =x\left( x+2 \right)\left( \dfrac{x\left( x-1 \right)}{2} \right) \\
 & =\dfrac{{{x}^{2}}\left( x-1 \right)\left( x+2 \right)}{2}
\end{align}\]
which is again cannot be expressed as the value obtained in the equation (1), hence option (c) is wrong.
Now let if possible \[f\left( x+2 \right)=\dfrac{xf\left( x \right)}{\left( x+2 \right)}\]. Then again using \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\], we get
\[\begin{align}
  & f\left( x+2 \right)=\dfrac{xf\left( x \right)}{\left( x+2 \right)} \\
 & =\left( \dfrac{\left( x \right)}{x+2} \right)\left( \dfrac{x\left( x-1 \right)}{2} \right) \\
 & =\dfrac{{{x}^{2}}\left( x-1 \right)}{2\left( x+2 \right)}
\end{align}\]
which is again not equal to the value obtained in the equation (1), hence option (d) is wrong.
Now let if possible \[f\left( x+2 \right)=\dfrac{\left( x+2 \right)f\left( x+1 \right)}{x}\]. Then again using \[f\left( x \right)=\dfrac{x\left( x-1 \right)}{2}\], we get
\[\begin{align}
  & f\left( x+2 \right)=\dfrac{\left( x+2 \right)f\left( x+1 \right)}{x} \\
 & =\left( \dfrac{x+2}{x} \right)\left( \dfrac{\left( x+1 \right)\left( \left( x+1 \right)-1 \right)}{2} \right) \\
 & =\dfrac{\left( x+2 \right)\left( x+1 \right)x}{2x}
\end{align}\]
which is equal to the value obtained in the equation (1).
Hence we have that \[f\left( x+2 \right)=\dfrac{\left( x+2 \right)f\left( x+1 \right)}{x}\].

So, the correct answer is “Option (e)”.

Note: In this problem, we can also evaluate the expression for \[f\left( x+2 \right)\] obtained in equation (1) and simplify it and express it in terms of some value of the function \[f(x)\].