Question

# 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) $\dfrac{\left( x+2 \right)f\left( x+1 \right)}{x}$

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 $f\left( x+2 \right)$.

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 $f(x)$.