Functions Question 399
Question: $ f(x)=\frac{x(x-p)}{q-p}+\frac{x(x-q)}{p-q}, $ $ p\ne q $ . What is the value of $ f( q )+f( q ) $ ?
Options:
A) $ f(p-q) $
B) $ f(p+q) $
C) $ f(p(p+q)) $
D) $ f(q(p-q)) $
Show Answer
Answer:
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) $
BETA
