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

Answer 1

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

& 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{aligned}$$ Thus we have $$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{aligned} & 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{aligned}$$ Since in this case the value of $$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{aligned} & 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{aligned}$$ which is again not equal to the value obtained in the equation (1), hence option (b) is wrong. Now let if possible $$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{aligned} & 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{aligned}$$ which is again cannot be expressed as the value obtained in the equation (1), hence option (c) is wrong. Now let if possible $$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{aligned} & 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{aligned}$$ which is again not equal to the value obtained in the equation (1), hence option (d) is wrong. Now let if possible $$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{aligned} & 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{aligned}$$ which is equal to the value obtained in the equation (1). Hence we have that $$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)$$.