Question
If $$p, q, r$$ are any real numbers, then
A.
$${\text{max}} (p, q) < {\text{max}} (p, q, r)$$
B.
$${\text{min}}\left( {p,q} \right) = \frac{1}{2}\left( {p + q - \left| {p - q} \right|} \right)$$
C.
$${\text{max}} (p, q) < {\text{min}} (p, q, r)$$
D.
none of these
Answer :
$${\text{min}}\left( {p,q} \right) = \frac{1}{2}\left( {p + q - \left| {p - q} \right|} \right)$$
Solution :
$$\eqalign{
& {\text{if }}p = 5,q = 3,r = 2 \cr
& {\text{max }}\left( {p,q} \right) = 5\,;\,\,\,\max \left( {p,q,r} \right) = 5 \cr
& \Rightarrow \,\,{\text{max}}\left( {p,q} \right) = \max \,\left( {p,q,r} \right) \cr} $$
∴ (A) is not true. Similarly we can show that (C) is not true.
$$\eqalign{
& {\text{Also min}}\left( {p,q} \right) = \frac{1}{2}\left( {p + q - \left| {p - q} \right|} \right) \cr
& {\text{Let }}p < q\,\,{\text{then L}}{\text{.H}}{\text{.S}} = p \cr
& {\text{and R}}{\text{.H}}{\text{.S = }}\frac{1}{2}\left( {p + q - q + p} \right) = p \cr} $$
Similarly, we can prove that $$(B)$$ is true for $$q < p$$ too.