Question

When $x\ge 2$, the function $f\left( x \right)=2\left| x-2 \right|-\left| x+1 \right|+x$ is reduced to:
(a) $f\left( x \right)=-2x+3$
(b) $f\left( x \right)=2x-5$
(c) $f\left( x \right)=5$
(d) $f\left( x \right)=-1$
(e) $f\left( x \right)=2x+4$

Answer 1

Hint: We know that according to the definition of modulus, $\left| h\left( x \right) \right|=\left\{ \begin{matrix}
   h\left( x \right) & \text{when }h\left( x \right)\ge 0 \\
   -h\left( x \right) & \text{when }h\left( x \right)<0 \\
\end{matrix} \right.$. So, we must first find the values of given two modulus expressions separately, and then substitute their values in the equation for $f\left( x \right)$.

Complete step by step answer:
We all know very well that the modulus or absolute value of any function $h\left( x \right)$ is defined as the positive part of function $h\left( x \right)$, and it is represented as $\left| h\left( x \right) \right|$. We can write this definition in mathematical terms as
$\left| h\left( x \right) \right|=\left\{ \begin{matrix}
   h\left( x \right) & \text{when }h\left( x \right)\ge 0 \\
   -h\left( x \right) & \text{when }h\left( x \right)<0 \\
\end{matrix} \right.$
In the question, we are given that
$x\ge 2...\left( i \right)$
Let us subtract 2 on both sides of inequation (i). Now, we get
$x-2\ge 2-2$
Thus, we have
$x-2\ge 0$
Now, since we know that $x-2\ge 0$, we can easily write that
$\left| x-2 \right|=x-2...\left( ii \right)$
Also, by adding 1 on both sides of equation (i), we get
$x+1\ge 2+1$
Thus, we get that
$x+1\ge 3$
Now, since $x+1\ge 3$, we can easily say that $\left( x+1 \right)$ will always be greater than 0, that is, $\left( x+1 \right)$ will always have a positive value.
Thus, we get
$\left| x+1 \right|=x+1...\left( iii \right)$
We are given that the function $f\left( x \right)=2\left| x-2 \right|-\left| x+1 \right|+x$.
By putting the values of $\left| x-2 \right|$ and $\left| x+1 \right|$ from equation (ii) and equation (iii) respectively into the function $f\left( x \right)$, we can write
$f\left( x \right)=2\left( x-2 \right)-\left( x+1 \right)+x$
We can now easily simplify the above equation as
$f\left( x \right)=2x-4-x-1+x$
Thus, the simplified equation becomes
$f\left( x \right)=2x-5$

So, the correct answer is “Option b”.

Note: We know that $\left| x \right|$ is replaced by $x$, when the value of $x$ is greater than or equal to 0, and it is replaced by $-x$ when the value of $x$ is less than 0. We must note here that to solve any equation involving the modulus operator, we must equate the expression inside the modulus to 0.