JEE Main & Advanced Mathematics Functions Question Bank Self Evaluation Test - Relations and Functions-I

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

A) \[f(p-q)\]

B) \[f(p+q)\]

C) \[f(p(p+q))\]

D) \[f(q(p-q))\]

Correct Answer: B

Solution :

[b] In the definition of function.

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

Putting p and q in place of x, we get

\[f(p)=\frac{p(p-p)}{q-p}+\frac{p(p-q)}{(p-q)}=p\]

\[\Rightarrow f(p)=p\]

and \[f(q)=\frac{q(q-p)}{q-p}+\frac{q(p-q)}{(p-q)}=q\]

\[\Rightarrow f(q)=q\]

Putting \[x=(p+q)\]

\[f(p+q)=\frac{(p+q)(p+q-p)}{(q-p)}+\frac{(p+q)(p+q-q)}{(p-q)}\]

\[=\frac{(p+q)q}{(q-p)}+\frac{(p+q)(p)}{(p-q)}=\frac{pq+{{q}^{2}}-{{p}^{2}}-pq}{(q-p)}\]

\[=\frac{{{q}^{2}}-{{p}^{2}}}{q-p}=\frac{(q-p)(q+p)}{(q-p)}=p+q=f(q)+f(p)\]

So, \[f(p)+f(q)=f(p+q)\]