Question

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

• A. f (p - q)
• B. f (p + q)
• C. f (p (p + q))
• D. f (q (p - q))

Answer 1

Answer: (B)

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