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