Question

If $f(x) = \dfrac{{x(x - p)}}{{q - p}} + \dfrac{{x(x - q)}}{{p - q}}$,then what is the value of $f(p) + f(q)$?
A. $f(p - q)$
B. $f(p + q)$
C. $f(p(p + q))$
D. $f(q(p - q))$

Answer 1

Hint: Simplify the given function into a simple function, then substitute the values of the function and try to get the answer in the requested format using algebraic manipulations.

Complete step by step solution:
Given-$f(x) = \dfrac{{x(x - p)}}{{q - p}} + \dfrac{{x(x - q)}}{{p - q}}$
Taking x common from the expression
$f(x) = x\left( {\dfrac{{(x - p)}}{{q - p}} + \dfrac{{(x - q)}}{{p - q}}} \right)$
Taking $\dfrac{1}{{p - q}}$ common from the expression
$f(x) = \dfrac{x}{{p - q}}\left( { - (x - p) + (x - q)} \right)$
Simplifying the expression
$f(x) = \dfrac{x}{{p - q}}\left( { - x + p + x - q} \right)$
$ \Rightarrow f(x) = \dfrac{x}{{p - q}}\left( {p - q} \right)$
$ \Rightarrow f(x) = x$
Hence,$f(p) = p$ and $f(q) = q$
Thus $f(p) + f(q) = p + q$
But $p + q = f(p + q)$
$\therefore f(p) + f(q) = f(p + q)$

Option ‘B’ is correct

Additional information:
- Function is denoted the relationship between two variables. A function is generally written as $f\left(x\right)$.
- We calculate the value of the function at $x=a$, substitute $x=a$ in the expression of the function.
- The value of the function $f\left(x\right)$ at $x=a$ is $f\left(a\right)$.

Note: We can solve the given question by another method. That is by, substituting $x=p$ in the given function $f(p) = \dfrac{{p(p - p)}}{{q - p}} + \dfrac{{p(p - q)}}{{p - q}}=p$. Again substitute $x=q$ in the given function $f(q) = \dfrac{{q(q - p)}}{{q - p}} + \dfrac{{q(q - q)}}{{p - q}}=q$. Now, $f(p) + f(q)=p+q$. Again substitute $x=p+q$ in the given function $f(p+q) = \dfrac{{(p+q)(p+q - p)}}{{q - p}} + \dfrac{{(p+q)(p+q - q)}}{{p - q}}=p+q$. Therefore, $f(p+q) =f(p)+f(q)$.