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  4. f(x) = (x(x-p)/q-p) + (x(x-q)/p-q) , p ≠ q . What is the value of f (p) + f (q) ?

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

Relations and Functions

Solution:

In the definition of function
$f\left(x\right) = \frac{x\left(x-p\right)}{q-p} + \frac{x\left(p-q\right)}{\left(p-q\right)} =p$
Putting p and q in place x, we get
$ f\left(p\right) = \frac{p\left(p-p\right)}{q-p} + \frac{p\left(p-q\right)}{\left(p-q\right)} =p$
$ \Rightarrow f\left(p\right) =p $
and $f\left(q\right) = \frac{q\left(q-p\right)}{q-p} + \frac{q\left(p-q\right)}{\left(p-q\right)} = q $
$\Rightarrow f\left(q\right)=q $
Putting $x =\left(p+q\right)$
$ f\left(p+q\right) = \frac{\left(p+q\right)\left(p+q-p\right)}{\left(q-p\right)} + \frac{\left(p+q\right)\left(p+q-q\right)}{\left(p-q\right)} $
$= \frac{\left(p+q\right)q}{\left(q-p\right)} + \frac{\left(p+q\right)\left(p\right)}{\left(p-q\right)}$
$ = \frac{pq+q^{2}-p^{2}-pq}{\left(q-p\right)} $
$= \frac{q^{2}-p^{2}}{q-p} = \frac{\left(p-q\right)\left(q+p\right)}{\left(q-p\right)} $
$= q+q$
$ = f\left(q\right) + f\left(p\right) $
So, $f\left(p\right)+f\left(q\right) = f\left(p+q\right)$