Question

$f\left(x\right)=\left[\frac{x(x-p)}{(q-p)}\right]+\left[\frac{x(x-q)}{(p-q)}\right],p\ne q.$ What is the value of $f\left(p\right)+f\left(q\right)$?

• A. $f(p-q)$
• B. $f(p+q)$
• C. $f\left(p\right(p+q\left)\right)$
• D. $f\left(q\right(p-q\left)\right)$

Answer 1

The correct option is B

$f(p+q)$

Explanation for the correct option:

Find the value of $f\left(p\right)+f\left(q\right)$:

$f\left(x\right)=\left[\frac{x(x-p)}{(q-p)}\right]+\left[\frac{x(x-q)}{(p-q)}\right]$

$=\left[\frac{-x(x-p)}{(p-q)}\right]+\left[\frac{x(x-q)}{(p-q)}\right]$

$=\frac{(-x2+px+x2–qx)}{(p-q)}$

$=\frac{(px–qx)}{(p-q)}$

$=\frac{x(p-q)}{(p-q)}$

$=x$

$\therefore$$f\left(x\right)=x$

$\Rightarrow$$f\left(p\right)=p$

and $f\left(q\right)=q$

Then, $f\left(p\right)+f\left(q\right)=p+q$

$=\mathit{f}\mathbf{(}\mathit{p}\mathbf{+}\mathit{q}\mathbf{)}$

Hence, Option ‘B’ is Correct.