If a, b, c, d are positive real numbers such that $a+b+c+d=2$, then $M=\left( a+b \right)\left( c+d \right)$ satisfies the relation
A. $0< M\le 1$
B. $1\le M\le 2$
C. $2\le M\le 3$
D. $3\le M\le 4$

Hint: We try to form the value of M as a square form. As we know the minimum value of a square, we can form the maximum value of M using its negative form. Also using the condition of a, b, c, d being positive real numbers we find the minimum value of M. So, we can start by rewriting the given condition $a+b+c+d=2$ such that we get the value of (a+b) and then substitute it in $M=\left( a+b \right)\left( c+d \right)$ and solve further.

Complete step-by-step solution:
It’s given that a, b, c, d are positive real numbers such that $a+b+c+d=2$.
We need to find out the range of the number $M=\left( a+b \right)\left( c+d \right)$.
From $a+b+c+d=2$, we can get value of $a+b$ as
$\begin{align}
  & a+b+c+d=2 \\
 & \Rightarrow a+b=2-\left( c+d \right) \\
\end{align}$
We place the value in the equation $M=\left( a+b \right)\left( c+d \right)$ and get
$\begin{align}
  & M=\left( a+b \right)\left( c+d \right) \\
 & \Rightarrow M=\left\{ 2-\left( c+d \right) \right\}\left( c+d \right) \\
\end{align}$
Now we try to form a square term of M. The square will be of $\left( c+d \right)$ and 1.
$\begin{align}
  & M=\left\{ 2-\left( c+d \right) \right\}\left( c+d \right) \\
 & \Rightarrow M=2\left( c+d \right)-{{\left( c+d \right)}^{2}} \\
 & \Rightarrow M-1=-{{\left( 1 \right)}^{2}}+2\left( c+d \right).1-{{\left( c+d \right)}^{2}} \\
\end{align}$
The square is in the form of ${{\left( x \right)}^{2}}-2xy+{{\left( y \right)}^{2}}={{\left( x-y \right)}^{2}}$.
$\begin{align}
  & M-1=-\left\{ {{\left( 1 \right)}^{2}}-2\left( c+d \right).1+{{\left( c+d \right)}^{2}} \right\} \\
 & \Rightarrow M-1=-{{\left( c+d-1 \right)}^{2}} \\
\end{align}$
Now for any value of c and d we will get the square value as greater than or equal to 0.
So, ${{\left( c+d-1 \right)}^{2}}\ge 0$ which means $-{{\left( c+d-1 \right)}^{2}}\le 0$.
This gives the upper limit of M as $M-1=-{{\left( c+d-1 \right)}^{2}}\le 0$ which gives $M\le 1$.
Now for the lower limit we have addition and multiplication of all 4 positive numbers. So, the value of $M=\left( a+b \right)\left( c+d \right)$ will be greater than 0.
Therefore, $0< M\le 1$. The correct option is A.

Note: Multiplication of a negative number changes the inequality sign of the relation. Both sides also change their sign. Students might try to open the brackets in $M=\left( a+b \right)\left( c+d \right)$ and simplify it. But, it will complicate things as they will end up with terms like ac, ad, … which cannot be simplified further. So, always look for conditions given in question for simplification.